Unit 10 — End of Topic Tests (reviewer copy: all keys and model answers visible)
End of Topic Test: What a solution is and how dissolving works & Electrolytes and like dissolves like (u10-t1 · 5 forms × 21 items · 105 total)
David's notes for James:
FRQ COMPOSITION - frq_per_form = 1 (the baseline; your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - zero of the 13 scope lessons are quant (atomKinds: 5 discrimination, 5 explanation, 2 fact, 1 representation), zero of your 39 pool seeds are numeric, and no scope lesson carries a displayEquation; Unit 10's calculation strands (molarity, dilution/solution stoichiometry) belong to LATER U10 gates, where the 2-3-FRQ calc split applies instead. The single FRQ is the scope's production capstone: SOL-006 'Sketch dissolving at the particle level', prod-seeded from YOUR two gate production items p15/p17 ('gate prods are fine as FRQ seeds'; U01-G1 MPM-015 drawing-FRQ canon shape: draw instruction + rubric summing to 100 + described-diagram model answer). D5 scaling evaluated and NOT triggered: min(20, 6 x 13 lessons) = 20; your target_size=20 honored. Total 20 objective + 1 FRQ = 21 items/form.
SEED FRESHENINGS UNDER THE STRICT SEED RULING (no freshness exemption; validator never weakened) - 14 dispositions to confirm, every one preserving your item structure and misconception keys with all fields swept (QB 3.9): (a) p0@A#1 key -> brewed black coffee, creek water -> creek sample (own-lesson base tokens). (b) p4@A#2 hydrogen peroxide -> glucose drip 3 g/97 g ('hydrogen' is reserved rule-content vocabulary of the SOL-004 charged-ends mechanism family). (c) p6@A#7 KNO3 -> LiNO3 (KNO3 is owned by seq 14 in the home-seq ledger); your answer-contract shape kept. (d) p9@A#8 magnesium chloride -> zinc chloride (Zn2+/Cl-). (e) p13@A#10 calcium bromide -> strontium iodide (CaBr2 is owned by your FRQ seed at 21B). (f) p14@B#10 sodium bromide -> potassium chloride rework (pairs/molecules/metal-lump/charge-swap distractor family kept; key verbatim-generic). (g) p25@B#13 glycerin -> ethylene glycol, 'a molecular compound' stated, a/an grammar fixed. (h) p30@A#15 beeswax -> lanolin (turpentine key kept - recorded other-lesson echo). (i) p33@A#19 wax -> petroleum jelly, plus a 6-character key trim ('each group stays' -> 'each stays') to hold the deterministic longest-key bound after the freshen. (j) p36@A#16 distractors sodium chloride -> borax and potassium nitrate -> washing soda; your stem and key verbatim. (k) p37@A#20 vegetable oil -> kerosene; the thickness distractor reworded to the thin instance. (l) p38@B#16 REBUILT on the 16A hexane system (its wax/salt/mineral-oil instances are all owned elsewhere on the ledger); the same-seq hexane/paraffin/borax sharing is recorded by-design (WARN lane, not a fail). (m) p15@A#21 sodium bromide -> potassium bromide, 3+3 ions (FRQ; your drawing conventions kept verbatim). (n) STAGE-2 REWORK: the planned fresh 8D instance 'lithium sulfate' was REJECTED because SO4(2-) normalizes to a digit colliding with your 13A MgSO4 seed - 8D is barium iodide (Ba2+/I-) instead.
SEED KEY WORDING AMENDED AT AUTHORITATIVE QC - B#4 (confirm): the item's key said 'compound' where the course rule line defines an electrolyte as 'a SUBSTANCE whose water solution conducts electricity'; gpt-5.6-sol (authoritative cross-vendor QC) failed the mismatch, so the key now carries the rule-line wording, and the absurd pourability distractor was replaced with a tester-hardware misconception (the metal strips carry electricity into the liquid).
DISTRACTOR REDESIGN AT BLIND QC - D#5 (confirm): the 'evenly cloudy throughout' distractor was ruled genuinely defensible under the taught same-throughout rule (the course teaches no clear-vs-cloudy discriminator at SOL-001), so it was replaced with the taught settling counter-behavior (the muddy-water counter-example from the lesson's own design).
FRQ RUBRIC HARDENING (all five forms): the positive-ion-shell rubric criteria now REQUIRE the taught 3-4-water shell count (sol caught the omission on four forms; the fifth was hardened as a deterministic class sweep - 'never trust a judge to re-catch'). Rubrics still sum to 100 with your <=3-criterion drawing-production shape.
REALISM AMENDMENTS at QC (confirm): B#2 cream-of-tartar amounts reduced to 1 g in 99 g of hot water (the draft's 8 g/92 g exceeds real solubility); C#16 menthol -> naphthalene -> SULFUR (naphthalene is named in SOL-013's own spec text and the untouched own-lesson material scan correctly hard-failed it; sulfur is genuinely nonpolar, spec-clean, and bank-unique); E#6 the ethyl-acetate/water 91%/9% system is not actually miscible at that ratio -> 90% acetone / 10% water (acetone = recorded taught-elsewhere WARN echo, SOL-007/009 surfaces).
ADJUDICATED INSTANCE FAMILIES (confirm; the gate's structural whitelists, enforced by a gate-local wrapper over the UNTOUCHED shared validator; slot lists enumerated from the scan): (1) AQUEOUS-SOLVENT-WATER - this gate's topic IS dissolving in water ('water' is the taught solvent in the rule lines of 12 of 13 scope lessons; SOL-004 is literally 'Why water dissolves so many substances'); cross-slot material exclusivity is structurally unsatisfiable for the word, so water-anchored windows are excused at 77 enumerated slots while the instance-bearing SOLUTES stay globally unique under the home-seq ledger. (2) WATER-POLARITY-OXYGEN-HYDROGEN - 'partially negative oxygen end / partially positive hydrogen ends' is SOL-004's taught mechanism sentence; at seqs 8 and 10 the O/H end-language IS the assessed content on every form (parts of the water molecule, not sample substances; 8 enumerated slots). (3) SEED-CABR2-FRQ-B21 + (4) SEED-METHANOL-C12 - SPEC-INTERNAL reuse where YOUR text sits on both sides: SOL-010's own spec text names 'calcium bromide (CaBr2)' (drawn by form B's FRQ) and 'methanol, a molecular compound' (C#12 = your SOL-007 gate item placed verbatim). Everything outside the enumerated slots stays fully armed.
GATE_CONFIG TASK-IDIOM WHITELIST (8 recorded entries, confirm) - each preserves YOUR wording verbatim where the auto-derivation cannot see it, zero instance signal in every gram: SOL-006's production-instruction sentences 'On paper, sketch a particle diagram' and 'Then select Continue to compare your sketch with the model answer' (your p15/p17 open and close with exactly this wording, and the seq-21 FRQ runs it on every form BY DESIGN); 'light(s) the bulb because' (your SOL-008/009 conductivity option idiom); 'The figure shows a student' (your SOL-006/007 error-find stem idiom); 'The student is (nearly) right' (your claim-evaluation option idiom, SOL-010/012/013); 'Write the formula with the symbol' (your SOL-003 write-direction task instruction, run at the seq-7 text row on every form by design). Shared validator untouched.
FIGURE-STYLE HYGIENE (disclosure): your hydration-sketch seed figureSpec 'style' strings NARRATED the deliberate error ('one deliberate orientation error around the positive ions' @p16; 'solute incorrectly drawn as charged pieces' @p20) - a key-blind render of those strings would leak the answer, so every error-carrying style string was neutralized to convention wording ('clean particle sketch in the lesson figure's conventions'); the drawn CONTENT strings are unchanged and the error lives only in the drawn content, exactly as a sighted student sees it.
BLIND-PACKET DEFECT DISCLOSURE (QC provenance, no action needed): the round-1 key-blind packets rendered the hydration-sketch/particle-panels figures style-only (7 items momentarily unanswerable blind) and dumped the seq-21 POST-ANSWER model-answer diagrams verbatim; both defects were fixed per-gate (content-rendering + model-answer-figure stripping), the tainted round-1 seq-21 cold answers were DISCARDED, and the FRQ was re-blinded cold on fresh reviewer tags. Final blind ledger: rounds 1-5 = 246/250 option-item key picks with round-1's only 2 misses both packet-defect items (re-blind rounds 58/58), 0 open hard fails; all four fresh-tag reviewers reproduced the full FRQ model answer cold on every form (hand-graded with quoted evidence in frq_verdicts.json). Sol authoritative ledger: loops 1-5 = 18 -> 2 -> 1 -> 0 -> 0 FAILs, final 105/105 pass, 5 recorded honest-survivor minors (feedback/style class, no key affected).
SOL-013 SPEC-INTERNAL NOTE (seed-file observation, no action): your specNotes for SOL-013 claim disjoint instance sets, yet hexane appears in both the article rule line and gate item p36 ('gates paraffin-in-hexane'). The reuse is deterministically silent (hexane is not a base MATERIALS token) and no reviewer raised memory-answerability on it; C#16's replacement instance deliberately avoided naphthalene for exactly this class.
NOTA/AOTA: zero literal none-of-the-above/all-of-the-above hits across your 39-seed pool (V-T5 regex clone at stage 1); zero replace_distractor flags needed. V-T5 PASS on all forms.
SPARE-SEED NOTE: 7 spares, none dropped on merit - p22/p23 (SOL-008) and p32 (SOL-011), p34/p35 (SOL-012) are parity/coverage-barred duplicates of case families already placed; your two SOL-006 in-lesson fluency productions are student-seen and therefore spares-only by standing rule (the FRQ seeds came from the GATE production items p15/p17 instead).
TEXT-ROW CASE CONTRACT (seq 7, all forms - schema-gap carrier note, same class as the U9-G2 text-row note): the typed-formula answers are graded CASE-SENSITIVELY on the element symbols (source answer.caseSensitive = true - capitalization is a taught feature of formulas at SOL-003; 'LiNO3' vs 'lino3' must not both pass), but the section-6 answer shape has no caseSensitive lane, so the flag is carried here and restated inside each item's exported gradingNote ('graded case-sensitively on the element symbols'). Unicode subscripts are NOT required to type - every item accepts the plain-digit form and a space before (aq) via acceptedAlternates; nothing typed requires sub/superscripts.
BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings; classroom-safe instance set throughout (household solutions, classroom salts, conductivity-tester bench scenarios). Your 39-seed pool scanned clean at stage 1; re-swept at packaging over forms_final AND this export body - the only matches are the word 'dose' in C#18's claim-evaluation distractor ('a large enough dose of any solute...' - amount-of-solute sense, benign and non-radiological, same recorded class as U11-G3's 'dye dose') and these disclosure sentences themselves.
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u10-t1 — notes/queries from David's side (U10-G1)
1. FRQ COMPOSITION - frq_per_form = 1 (the baseline; your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - zero of the 13 scope lessons are quant (atomKinds: 5 discrimination, 5 explanation, 2 fact, 1 representation), zero of your 39 pool seeds are numeric, and no scope lesson carries a displayEquation; Unit 10's calculation strands (molarity, dilution/solution stoichiometry) belong to LATER U10 gates, where the 2-3-FRQ calc split applies instead. The single FRQ is the scope's production capstone: SOL-006 'Sketch dissolving at the particle level', prod-seeded from YOUR two gate production items p15/p17 ('gate prods are fine as FRQ seeds'; U01-G1 MPM-015 drawing-FRQ canon shape: draw instruction + rubric summing to 100 + described-diagram model answer). D5 scaling evaluated and NOT triggered: min(20, 6 x 13 lessons) = 20; your target_size=20 honored. Total 20 objective + 1 FRQ = 21 items/form.
2. SEED FRESHENINGS UNDER THE STRICT SEED RULING (no freshness exemption; validator never weakened) - 14 dispositions to confirm, every one preserving your item structure and misconception keys with all fields swept (QB 3.9): (a) p0@A#1 key -> brewed black coffee, creek water -> creek sample (own-lesson base tokens). (b) p4@A#2 hydrogen peroxide -> glucose drip 3 g/97 g ('hydrogen' is reserved rule-content vocabulary of the SOL-004 charged-ends mechanism family). (c) p6@A#7 KNO3 -> LiNO3 (KNO3 is owned by seq 14 in the home-seq ledger); your answer-contract shape kept. (d) p9@A#8 magnesium chloride -> zinc chloride (Zn2+/Cl-). (e) p13@A#10 calcium bromide -> strontium iodide (CaBr2 is owned by your FRQ seed at 21B). (f) p14@B#10 sodium bromide -> potassium chloride rework (pairs/molecules/metal-lump/charge-swap distractor family kept; key verbatim-generic). (g) p25@B#13 glycerin -> ethylene glycol, 'a molecular compound' stated, a/an grammar fixed. (h) p30@A#15 beeswax -> lanolin (turpentine key kept - recorded other-lesson echo). (i) p33@A#19 wax -> petroleum jelly, plus a 6-character key trim ('each group stays' -> 'each stays') to hold the deterministic longest-key bound after the freshen. (j) p36@A#16 distractors sodium chloride -> borax and potassium nitrate -> washing soda; your stem and key verbatim. (k) p37@A#20 vegetable oil -> kerosene; the thickness distractor reworded to the thin instance. (l) p38@B#16 REBUILT on the 16A hexane system (its wax/salt/mineral-oil instances are all owned elsewhere on the ledger); the same-seq hexane/paraffin/borax sharing is recorded by-design (WARN lane, not a fail). (m) p15@A#21 sodium bromide -> potassium bromide, 3+3 ions (FRQ; your drawing conventions kept verbatim). (n) STAGE-2 REWORK: the planned fresh 8D instance 'lithium sulfate' was REJECTED because SO4(2-) normalizes to a digit colliding with your 13A MgSO4 seed - 8D is barium iodide (Ba2+/I-) instead.
3. SEED KEY WORDING AMENDED AT AUTHORITATIVE QC - B#4 (confirm): the item's key said 'compound' where the course rule line defines an electrolyte as 'a SUBSTANCE whose water solution conducts electricity'; gpt-5.6-sol (authoritative cross-vendor QC) failed the mismatch, so the key now carries the rule-line wording, and the absurd pourability distractor was replaced with a tester-hardware misconception (the metal strips carry electricity into the liquid).
4. DISTRACTOR REDESIGN AT BLIND QC - D#5 (confirm): the 'evenly cloudy throughout' distractor was ruled genuinely defensible under the taught same-throughout rule (the course teaches no clear-vs-cloudy discriminator at SOL-001), so it was replaced with the taught settling counter-behavior (the muddy-water counter-example from the lesson's own design).
5. FRQ RUBRIC HARDENING (all five forms): the positive-ion-shell rubric criteria now REQUIRE the taught 3-4-water shell count (sol caught the omission on four forms; the fifth was hardened as a deterministic class sweep - 'never trust a judge to re-catch'). Rubrics still sum to 100 with your <=3-criterion drawing-production shape.
6. REALISM AMENDMENTS at QC (confirm): B#2 cream-of-tartar amounts reduced to 1 g in 99 g of hot water (the draft's 8 g/92 g exceeds real solubility); C#16 menthol -> naphthalene -> SULFUR (naphthalene is named in SOL-013's own spec text and the untouched own-lesson material scan correctly hard-failed it; sulfur is genuinely nonpolar, spec-clean, and bank-unique); E#6 the ethyl-acetate/water 91%/9% system is not actually miscible at that ratio -> 90% acetone / 10% water (acetone = recorded taught-elsewhere WARN echo, SOL-007/009 surfaces).
7. ADJUDICATED INSTANCE FAMILIES (confirm; the gate's structural whitelists, enforced by a gate-local wrapper over the UNTOUCHED shared validator; slot lists enumerated from the scan): (1) AQUEOUS-SOLVENT-WATER - this gate's topic IS dissolving in water ('water' is the taught solvent in the rule lines of 12 of 13 scope lessons; SOL-004 is literally 'Why water dissolves so many substances'); cross-slot material exclusivity is structurally unsatisfiable for the word, so water-anchored windows are excused at 77 enumerated slots while the instance-bearing SOLUTES stay globally unique under the home-seq ledger. (2) WATER-POLARITY-OXYGEN-HYDROGEN - 'partially negative oxygen end / partially positive hydrogen ends' is SOL-004's taught mechanism sentence; at seqs 8 and 10 the O/H end-language IS the assessed content on every form (parts of the water molecule, not sample substances; 8 enumerated slots). (3) SEED-CABR2-FRQ-B21 + (4) SEED-METHANOL-C12 - SPEC-INTERNAL reuse where YOUR text sits on both sides: SOL-010's own spec text names 'calcium bromide (CaBr2)' (drawn by form B's FRQ) and 'methanol, a molecular compound' (C#12 = your SOL-007 gate item placed verbatim). Everything outside the enumerated slots stays fully armed.
8. GATE_CONFIG TASK-IDIOM WHITELIST (8 recorded entries, confirm) - each preserves YOUR wording verbatim where the auto-derivation cannot see it, zero instance signal in every gram: SOL-006's production-instruction sentences 'On paper, sketch a particle diagram' and 'Then select Continue to compare your sketch with the model answer' (your p15/p17 open and close with exactly this wording, and the seq-21 FRQ runs it on every form BY DESIGN); 'light(s) the bulb because' (your SOL-008/009 conductivity option idiom); 'The figure shows a student' (your SOL-006/007 error-find stem idiom); 'The student is (nearly) right' (your claim-evaluation option idiom, SOL-010/012/013); 'Write the formula with the symbol' (your SOL-003 write-direction task instruction, run at the seq-7 text row on every form by design). Shared validator untouched.
9. FIGURE-STYLE HYGIENE (disclosure): your hydration-sketch seed figureSpec 'style' strings NARRATED the deliberate error ('one deliberate orientation error around the positive ions' @p16; 'solute incorrectly drawn as charged pieces' @p20) - a key-blind render of those strings would leak the answer, so every error-carrying style string was neutralized to convention wording ('clean particle sketch in the lesson figure's conventions'); the drawn CONTENT strings are unchanged and the error lives only in the drawn content, exactly as a sighted student sees it.
10. BLIND-PACKET DEFECT DISCLOSURE (QC provenance, no action needed): the round-1 key-blind packets rendered the hydration-sketch/particle-panels figures style-only (7 items momentarily unanswerable blind) and dumped the seq-21 POST-ANSWER model-answer diagrams verbatim; both defects were fixed per-gate (content-rendering + model-answer-figure stripping), the tainted round-1 seq-21 cold answers were DISCARDED, and the FRQ was re-blinded cold on fresh reviewer tags. Final blind ledger: rounds 1-5 = 246/250 option-item key picks with round-1's only 2 misses both packet-defect items (re-blind rounds 58/58), 0 open hard fails; all four fresh-tag reviewers reproduced the full FRQ model answer cold on every form (hand-graded with quoted evidence in frq_verdicts.json). Sol authoritative ledger: loops 1-5 = 18 -> 2 -> 1 -> 0 -> 0 FAILs, final 105/105 pass, 5 recorded honest-survivor minors (feedback/style class, no key affected).
11. SOL-013 SPEC-INTERNAL NOTE (seed-file observation, no action): your specNotes for SOL-013 claim disjoint instance sets, yet hexane appears in both the article rule line and gate item p36 ('gates paraffin-in-hexane'). The reuse is deterministically silent (hexane is not a base MATERIALS token) and no reviewer raised memory-answerability on it; C#16's replacement instance deliberately avoided naphthalene for exactly this class.
12. NOTA/AOTA: zero literal none-of-the-above/all-of-the-above hits across your 39-seed pool (V-T5 regex clone at stage 1); zero replace_distractor flags needed. V-T5 PASS on all forms.
13. SPARE-SEED NOTE: 7 spares, none dropped on merit - p22/p23 (SOL-008) and p32 (SOL-011), p34/p35 (SOL-012) are parity/coverage-barred duplicates of case families already placed; your two SOL-006 in-lesson fluency productions are student-seen and therefore spares-only by standing rule (the FRQ seeds came from the GATE production items p15/p17 instead).
14. TEXT-ROW CASE CONTRACT (seq 7, all forms - schema-gap carrier note, same class as the U9-G2 text-row note): the typed-formula answers are graded CASE-SENSITIVELY on the element symbols (source answer.caseSensitive = true - capitalization is a taught feature of formulas at SOL-003; 'LiNO3' vs 'lino3' must not both pass), but the section-6 answer shape has no caseSensitive lane, so the flag is carried here and restated inside each item's exported gradingNote ('graded case-sensitively on the element symbols'). Unicode subscripts are NOT required to type - every item accepts the plain-digit form and a space before (aq) via acceptedAlternates; nothing typed requires sub/superscripts.
15. BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings; classroom-safe instance set throughout (household solutions, classroom salts, conductivity-tester bench scenarios). Your 39-seed pool scanned clean at stage 1; re-swept at packaging over forms_final AND this export body - the only matches are the word 'dose' in C#18's claim-evaluation distractor ('a large enough dose of any solute...' - amount-of-solute sense, benign and non-radiological, same recorded class as U11-G3's 'dye dose') and these disclosure sentences themselves.
Brewed black coffee — evenly colored, with nothing that ever settles. KEY
Fresh-squeezed lemonade with flecks of pulp swirling through it.
A creek sample with fine grit that slowly sinks to the bottom.
Chicken broth with small fat droplets floating on the surface.
SOL-002 · easy · recall · seq 2 · seed:U10-G1/pool[4] (= SOL-002 gateItems[1]) — FRESHENED (SEED RULING): 'HYDROGEN peroxide' carries base token 'hydrogen', reserved for the SOL-004/005/006 mechanism family (allocation note 5): instance -> 3 g of glucose dissolved in 97 g of water; roles/distractor structure unchanged (97 stays the unique big number); all fields swept · mcq4
A hospital glucose drip is made from 3 g of glucose dissolved in 97 g of water. Which statement names the two parts correctly?
Glucose is the solute, and water is the solvent. KEY
Glucose is the solvent, and water is the solute.
Both substances are solutes, because both are in the mixture.
Glucose is the solute, and the finished mixture is the solvent.
A student shakes a cloudy mixture hard until it looks even all the way through, and says, 'Now it is a solution.' Ten minutes later, particles begin settling to the bottom. Which response is correct?
It is not a solution — the settling shows the particles were hanging in the liquid, not spread evenly among the liquid's particles. KEY
It was a solution while it looked even all the way through, and it stopped being one when the particles began to settle.
It is a solution, because any mixture that contains more than one substance counts as a solution.
It is a solution, because the particles are small enough to swirl through every part of the liquid.
Lithium nitrate, LiNO₃, is dissolved in water. Write the formula with the symbol that shows it is dissolved in water.
Answer: LiNO₃(aq) KEY
also accepted: LiNO3(aq), LiNO₃ (aq), LiNO3 (aq)
wrongValueDiagnostics
[
{
"value": "LiNO₃(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted lithium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted lithium nitrate; dissolved in water is (aq)"
},
{
"value": "LiNO₃(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid lithium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid lithium nitrate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
feedback
{
"keyExplanation": "The solute's formula comes first, and (aq) after it marks the substance as dissolved in water: LiNO₃(aq). The (aq) always belongs to the dissolved substance — the water, as the solvent, never takes the label.",
"anticipatedWrong": [
{
"value": "LiNO₃(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted lithium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted lithium nitrate; dissolved in water is (aq)"
},
{
"value": "LiNO₃(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid lithium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid lithium nitrate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
}
The figure shows a student's sketch of dissolved potassium fluoride. The sketch has one error. What is the error?
FIGURE (hydration-sketch · stimulus) — content: "two separated gray circles labeled K⁺ each ringed by three water molecules (red O, white H) with their H sides facing the ion, and two separated pale-teal c… · style: "clean particle sketch in the lesson figure's conventions"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "hydration-sketch",
"content": "two separated gray circles labeled K⁺ each ringed by three water molecules (red O, white H) with their H sides facing the ion, and two separated pale-teal circles labeled F⁻ each ringed by three water molecules with their H sides facing the ion",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
The water molecules around each K⁺ point the wrong ends inward. KEY
The ions should be drawn touching, in pairs.
There are too few water molecules in each shell.
The shells around the F⁻ ions should face their O sides inward.
Beaker 1 holds dissolved magnesium sulfate, which separates into Mg²⁺ and SO₄²⁻ ions. Beaker 2 holds dissolved urea, which stays as neutral whole molecules. Which beaker lights the conductivity bulb, and why?
Beaker 1 — its free-moving charged ions carry the current. KEY
Beaker 2 — whole molecules carry current better than charged ions.
Both beakers — every solution conducts once something is dissolved in it.
Neither beaker — dissolved particles of any kind block the current.
A beaker holds hexane, a nonpolar liquid. Which solute is predicted to dissolve well in it: borax (ionic), citric acid (polar), paraffin (nonpolar), or washing soda (ionic)?
Paraffin KEY
Borax
Citric acid
Washing soda
SOL-009 · hard · transfer · seq 17 · seed:U10-G1/pool[26] (= SOL-009 gateItems[2]) — placed verbatim (spec-sourced Unicode) · mcq4
Two clear, look-alike solutions are tested: one lights the bulb and one does not. Which statement about their particles MUST be true?
The conducting solution holds free-moving charged particles; the dark one holds only neutral particles. KEY
The conducting solution simply holds more dissolved substance than the dark solution does.
The dark solution must have nothing dissolved in it at all — it is just pure water.
The conducting solution is warmer than the dark one, and warmth is what lets current pass.
SOL-010 · hard · transfer · seq 18 · seed:U10-G1/pool[29] (= SOL-010 gateItems[2]) — placed verbatim (spec-sourced Unicode) · mcq4
A student claims, 'Every compound that dissolves in water gives a solution that conducts electricity.' Which correction is right?
Only ionic compounds reliably conduct when dissolved — most molecular compounds dissolve as neutral whole molecules. KEY
The claim is right, because the act of dissolving always produces free-moving ions of some kind.
The claim is backwards — dissolved compounds never conduct electricity; only solid compounds carry current.
The claim fails only for colored solutions, whose color blocks the current from crossing.
SOL-012 · hard · transfer · seq 19 · seed:U10-G1/pool[33] (= SOL-012 gateItems[0]) — FRESHENED (SEED RULING): own-lesson base token 'wax' (wax-crayon FLU0) -> petroleum jelly (nonpolar, household, fresh tokens); no-repulsion key + magnet/temperature/size distractor family unchanged; all fields swept · mcq4
A student says, 'Petroleum jelly and water refuse to mix because their particles repel each other.' Which correction is right?
Nothing repels — the attractions petroleum jelly could form with water are different in kind from the attractions each substance's particles already form, so each stays with its own kind. KEY
The student is right — particles of opposite polarity push one another apart, in just the way two like magnetic poles push each other apart when brought close together.
The student is nearly right — the particles do repel each other, but the repulsion only switches on when the water is cold.
The real reason is that petroleum jelly particles are larger than water particles, so they cannot fit into the spaces between them.
SOL-013 · hard · transfer · seq 20 · seed:U10-G1/pool[37] (= SOL-013 gateItems[1]) — FRESHENED (SEED RULING): own-lesson base token 'oil' (vegetable oil vs cooking-oil FLU2) -> kerosene (nonpolar liquid, base token home-seq 20A; classroom-safe lamp-fuel-class instance); dissolves-poorly key + universal-solvent/liquids-always-mix/thickness distractor family kept, the thickness cue reworded to the fresh instance (kerosene is thin, not thick); all fields swept · mcq4
A student says kerosene, a nonpolar liquid, will dissolve in water because water dissolves almost everything. What is the correct prediction?
The kerosene dissolves poorly, because nonpolar kerosene and polar water are unlike in polarity. KEY
The kerosene dissolves well, because water dissolves almost every substance.
The kerosene dissolves well, because two liquids always mix.
The kerosene dissolves poorly, but only because it is so much thinner than water.
SOL-006 · medium · application · seq 21 · seed:U10-G1/pool[15] (= SOL-006 gateItems[0]; demo_curated[0], dual ref) — FRESHENED (SEED RULING): 'SODIUM bromide' trips own-lesson base token 'sodium' (SOL-006 FLU0 NaF) -> potassium bromide, three K+ and three Br- (3+3 shape, checklist and model-answer conventions kept; KBr other-lesson presence is non-base + gram-distinct, recorded); rubric/model answer in ASCII ion labels per the V-T12 contract; U01-G1 MPM-015 drawing-FRQ canon shape · frq
On paper, sketch a particle diagram of dissolved potassium bromide: three K⁺ ions and three Br⁻ ions, each hydrated. Label every ion, draw each water molecule as one large O circle with two small H circles, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
FIGURE (model_answer_diagram · model_answer_diagram) — see raw specraw spec
{
"figureShape": "model_answer_diagram",
"modelAnswerFigureSpecs": {
"a": {
"figureType": "hydration-sketch",
"content": "three separated circles labeled K⁺ each ringed by three water molecules with O sides inward, and three separated circles labeled Br⁻ each ringed by three water molecules with H sides inward",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
}
On paper, sketch a particle diagram of dissolved potassium bromide: three K⁺ ions and three Br⁻ ions, each hydrated. Label every ion, draw each water molecule as one large O circle with two small H circles, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
Part a. On paper, sketch a particle diagram of dissolved potassium bromide: three K⁺ ions and three Br⁻ ions, each hydrated. Label every ion, draw each water molecule as one large O circle with two small H circles, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
Model: Six well-separated circles — three labeled K+ and three labeled Br-. Each K+ sits inside a ring of three or four water molecules with O sides facing it; each Br- sits inside a ring of three or four water molecules with H sides facing it. No ion touches another ion, because the dissolved lattice is fully taken apart.
Full model answer. Six well-separated circles — three labeled K+ and three labeled Br-. Each K+ sits inside a ring of three or four water molecules with O sides facing it; each Br- sits inside a ring of three or four water molecules with H sides facing it. No ion touches another ion, because the dissolved lattice is fully taken apart.
The figure shows particle diagrams of two mixtures, Panel A and Panel B. Which panel shows a solution?
FIGURE (particle-panels · stimulus) — panels: [{"panel": "A", "description": "a box of many small light circles with a few small dark circles spread evenly through the whole box, a clear gap between ever… · style: "clean two-panel particle diagram, boxes labeled A and B in one figure"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-panels",
"panels": [
{
"panel": "A",
"description": "a box of many small light circles with a few small dark circles spread evenly through the whole box, a clear gap between every dark circle and its light neighbors"
},
{
"panel": "B",
"description": "a box of many small light circles with the same number of small dark circles gathered into one cluster in the bottom corner"
}
],
"style": "clean two-panel particle diagram, boxes labeled A and B in one figure"
}
}
A sweetened iced-tea mix contains both sugar and instant-tea powder, and a scoop of it is dissolved in a glass of water. Which statement is correct?
Water is the solvent, and the sugar and the instant-tea powder are both solutes. KEY
The sugar is the solute and the instant-tea powder is the solvent.
Water and sugar are the solvents, and the instant-tea powder is the solute.
The mixture is not a solution, because a solvent can hold only one solute.
SOL-003 · medium · application · seq 7 · authored (stage 2) — fresh write-direction formula (blueprint seq-7 candidate list); barium nitrate home-seq 7B ('barium' also at 21D BaCl₂ — non-base token, distinct compound, recorded) · text
Barium nitrate, Ba(NO₃)₂, is dissolved in water. Write the formula with the symbol that shows it is dissolved in water.
Answer: Ba(NO₃)₂(aq) KEY
also accepted: Ba(NO3)2(aq), Ba(NO₃)₂ (aq), Ba(NO3)2 (aq)
wrongValueDiagnostics
[
{
"value": "Ba(NO₃)₂(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted barium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted barium nitrate; dissolved in water is (aq)"
},
{
"value": "Ba(NO₃)₂(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid barium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid barium nitrate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
feedback
{
"keyExplanation": "The solute's formula comes first, and (aq) after it marks the substance as dissolved in water: Ba(NO₃)₂(aq). The (aq) always belongs to the dissolved substance — the water, as the solvent, never takes the label.",
"anticipatedWrong": [
{
"value": "Ba(NO₃)₂(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted barium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted barium nitrate; dissolved in water is (aq)"
},
{
"value": "Ba(NO₃)₂(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid barium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid barium nitrate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
}
A student says water pulls an ionic compound apart because water 'soaks into the solid like a sponge soaks up juice.' Which correction states the real reason?
Because water's partially negative oxygen end attracts positive ions and its partially positive hydrogen ends attract negative ions. KEY
The student is nearly right — water seeps into tiny cracks in the solid and physically wedges the ions apart from one another.
Water pulls the compound apart because every liquid attracts every solid a little, and water is simply an ordinary liquid.
Water pulls the compound apart because its molecules collide with the solid over and over until the crystal crumbles.
A crystal of an ionic compound is dropped into a tall beaker of water. What happens FIRST?
Water molecules attach to the ions on the crystal's surface. KEY
Complete hydration shells form instantly around every ion, even those deep inside the crystal.
All of the ions break away from the crystal at the same moment.
The crystal melts into a liquid, which then blends into the water.
SOL-005 · medium · application · seq 10 · seed:U10-G1/pool[14] (= SOL-005 gateItems[2]) — FRESHENED (SEED RULING): 'SODIUM bromide' = base token 'sodium' (home 8C) AND NaBr collided with the original FRQ-A seed -> rebuilt on potassium chloride K⁺/Cl⁻ with the same pairs/molecules/metal-lump/charge-swap distractor family; key kept verbatim (generic); explanation swept · mcq4
A student says that when potassium chloride dissolves, the K⁺ and Cl⁻ travel through the water stuck together as K⁺Cl⁻ pairs. What actually happens?
Each ion is pulled away from the lattice separately and ends up surrounded by its own shell of water molecules. KEY
The K⁺ and Cl⁻ leave the lattice joined together as whole, uncharged KCl molecules.
The Cl⁻ ions dissolve into the water, but the K⁺ ions stay behind as a lump of solid potassium metal.
The ions swap their charges with the surrounding water molecules as they leave the crystal surface.
SOL-006 · medium · application · seq 11 · authored (stage 2) — fresh error-find: wrong ends around the NEGATIVE ion (blueprint seq-11 error rotation); rubidium chloride home-seq 11B; correct-feature foil kept (seed pattern) · mcq4
In the figure, dissolved rubidium chloride has been sketched by a student. The sketch contains one error. Which statement identifies it?
FIGURE (hydration-sketch · stimulus) — content: "two separated circles labeled Rb⁺ each ringed by three water molecules with their O sides facing the ion, and two separated circles labeled Cl⁻ each ringed … · style: "clean particle sketch in the lesson figure's conventions"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "hydration-sketch",
"content": "two separated circles labeled Rb⁺ each ringed by three water molecules with their O sides facing the ion, and two separated circles labeled Cl⁻ each ringed by three water molecules with their O sides facing the ion",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
The water molecules around each Cl⁻ point the wrong ends inward. KEY
The water molecules around each Rb⁺ point the wrong ends inward.
A student says, 'All compounds split into ions when they dissolve — that is what dissolving means.' Which correction is right?
Only ionic compounds separate into ions; a molecular compound's molecules stay whole and spread among the water molecules. KEY
No compounds at all split into ions when they dissolve — every dissolved solute spreads as whole molecules.
The student is right, but the ions rejoin into whole molecules once the solution has been left to sit still.
Whether a compound splits into ions depends on how much of it is dissolved, not on what kind of compound it is.
SOL-009 · medium · application · seq 13 · seed:U10-G1/pool[25] (= SOL-009 gateItems[1]) — FRESHENED (SEED RULING): glycerin is owned by seq 12A (cross-form different-seq base token) -> ethylene glycol, 'a molecular compound' stated in the stem; key/distractor structure unchanged; a/an articles fixed (H.32 class); all fields swept · mcq4
A student says an ethylene glycol solution fails to light the bulb because ethylene glycol, a molecular compound, 'absorbs the electricity.' What is the real reason?
Dissolved ethylene glycol is neutral whole molecules — and a solution conducts only when charged particles are free to move through it. KEY
The student is right — the dissolved molecules soak up the current just the way a dry sponge soaks up spilled water.
The ethylene glycol coats the two metal strips and seals the electric current inside them.
The ethylene glycol's ions are locked inside its molecules and cannot escape to carry the current.
Without testing it, a student must predict whether a solution of strontium chloride, SrCl₂ — an ionic compound — will light a conductivity bulb. Which prediction and reason are correct?
The bulb lights — dissolved SrCl₂ is free-moving Sr²⁺ and Cl⁻ ions, and a solution conducts only when charged particles are free to move through it. KEY
The bulb lights — dissolved SrCl₂ stays as whole, uncharged SrCl₂ molecules, and whole neutral molecules are what carry current best.
The bulb stays dark — the Sr²⁺ and Cl⁻ ions stay locked together in bound pairs as they drift through the water.
The bulb stays dark — and in any case no prediction is possible without actually running the tester.
A nonpolar bicycle-chain lubricant will not mix into a bucket of water, and a student explains: 'their particles push each other away.' Which correction is right?
No pushing happens — the attractions are mismatched in kind, so each substance keeps to its own kind. KEY
There truly is a push — nonpolar particles carry a charge that shoves polar particles back.
The pushing exists, but only while the liquid is moving — stillness switches it off.
The student is wrong — the lubricant simply weighs too much to blend into water.
Beef tallow, a nonpolar solid, does not dissolve when stirred into water. A student says, 'That is only because it is a solid — melt the tallow first and it will dissolve fine.' What is the correct evaluation?
Melting will not help — melted tallow is still nonpolar, so the mismatch remains. KEY
Melting really is the fix — any melted substance dissolves in water.
Melting works, as long as the melted tallow is poured into the water slowly.
The rule runs the other way — solids always dissolve more easily than liquids.
SOL-006 · medium · application · seq 21 · seed:U10-G1/pool[17] (= SOL-006 gateItems[2]; demo_curated[2], dual ref) — placed VERBATIM (prompt + figureSpec from the spec, Unicode); rubric/model answer rebuilt in ASCII ion labels with the calcium charge in words per the V-T12 contract ('calcium' home-seq 21B); U01-G1 MPM-015 canon shape · frq
On paper, sketch a particle diagram of dissolved calcium bromide, CaBr₂, showing one formula unit's ions hydrated. (CaBr₂ gives Ca²⁺ and Br⁻ ions.) Label every ion with its symbol and full charge, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
FIGURE (model_answer_diagram · model_answer_diagram) — see raw specraw spec
{
"figureShape": "model_answer_diagram",
"modelAnswerFigureSpecs": {
"a": {
"figureType": "hydration-sketch",
"content": "one separated gray circle labeled Ca²⁺ ringed by four water molecules (red O, white H) with O sides inward, and two separated brown circles labeled Br⁻ each ringed by three water molecules with H sides inward",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
}
On paper, sketch a particle diagram of dissolved calcium bromide, CaBr₂, showing one formula unit's ions hydrated. (CaBr₂ gives Ca²⁺ and Br⁻ ions.) Label every ion with its symbol and full charge, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
Part a. On paper, sketch a particle diagram of dissolved calcium bromide, CaBr₂, showing one formula unit's ions hydrated. (CaBr₂ gives Ca²⁺ and Br⁻ ions.) Label every ion with its symbol and full charge, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
Model: Three well-separated circles — one labeled with the calcium symbol and a 2+ charge, and two labeled Br-. The calcium ion sits inside a ring of four water molecules with O sides facing it; each Br- sits inside a ring of three water molecules with H sides facing it. No ion touches another ion — the formula gives one calcium ion and two Br- ions, so the sketch shows exactly those three.
Full model answer. Three well-separated circles — one labeled with the calcium symbol and a 2+ charge, and two labeled Br-. The calcium ion sits inside a ring of four water molecules with O sides facing it; each Br- sits inside a ring of three water molecules with H sides facing it. No ion touches another ion — the formula gives one calcium ion and two Br- ions, so the sketch shows exactly those three.
The figure shows particle diagrams of two mixtures, Panel R and Panel S. Which panel could show a solution?
FIGURE (particle-panels · stimulus) — panels: [{"panel": "R", "description": "a box of many small light circles with a handful of small dark circles gathered in a band along the top edge of the box"}, {"… · style: "clean two-panel particle diagram, boxes labeled R and S in one figure"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-panels",
"panels": [
{
"panel": "R",
"description": "a box of many small light circles with a handful of small dark circles gathered in a band along the top edge of the box"
},
{
"panel": "S",
"description": "a box of many small light circles with the same number of small dark circles spread evenly through the whole box, a clear gap between every dark circle and its light neighbors"
}
],
"style": "clean two-panel particle diagram, boxes labeled R and S in one figure"
}
}
A teacher stirs two different powders, labeled X and Y, into one beaker of water — far more water than powder — until both dissolve completely. Which statement assigns the roles correctly?
The water is the solvent, and X and Y are both solutes. KEY
X is the solute and Y is the solvent, because Y dissolved second.
The powders are the solvents, because they changed the water.
The mixture cannot be a solution, because one solvent cannot hold two solutes.
SOL-003 · medium · application · seq 7 · authored (stage 2) — fresh write-direction formula (blueprint seq-7 candidate list); potassium sulfate home-seq 7C ('potassium' recurs in other seqs' compounds — non-base token, distinct compounds, recorded) · text
Potassium sulfate, K₂SO₄, is dissolved in water. Write the formula with the symbol that shows it is dissolved in water.
Answer: K₂SO₄(aq) KEY
also accepted: K2SO4(aq), K₂SO₄ (aq), K2SO4 (aq)
wrongValueDiagnostics
[
{
"value": "K₂SO₄(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted potassium sulfate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted potassium sulfate; dissolved in water is (aq)"
},
{
"value": "K₂SO₄(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid potassium sulfate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid potassium sulfate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
feedback
{
"keyExplanation": "The solute's formula comes first, and (aq) after it marks the substance as dissolved in water: K₂SO₄(aq). The (aq) always belongs to the dissolved substance — the water, as the solvent, never takes the label.",
"anticipatedWrong": [
{
"value": "K₂SO₄(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted potassium sulfate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted potassium sulfate; dissolved in water is (aq)"
},
{
"value": "K₂SO₄(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid potassium sulfate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid potassium sulfate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
}
Sodium fluoride, made of Na⁺ and F⁻ ions, is added to some public water supplies and dissolves readily. Which feature of the water molecule makes it so good at dissolving ionic compounds like sodium fluoride?
Its partially negative oxygen end and partially positive hydrogen ends, which attract the ions. KEY
Its small molecular size, which lets it slide into the gaps between the ions in the solid.
Its overall negative charge, which pulls every positive ion out of the solid toward it.
Its ability to split apart into separate H⁺ and OH⁻ ions that grab hold of the individual ions.
Rubidium fluoride, made of Rb⁺ and F⁻ ions, dissolves in water. Which statement describes the water shells correctly?
The shells around Rb⁺ aim oxygen ends inward, and the shells around F⁻ aim hydrogen ends inward. KEY
The shells around Rb⁺ aim hydrogen ends inward, and the shells around F⁻ aim oxygen ends inward.
Every shell aims oxygen ends inward, whatever the ion's charge.
Every shell aims hydrogen ends inward, because two hydrogens beat one oxygen.
SOL-006 · medium · application · seq 11 · authored (stage 2) — fresh error-find: wrong ion count for a 1:2 formula (blueprint seq-11 error rotation); nickel(II) chloride home-seq 11C (nickel = base token, home-seq 11C); two correct-feature foils + wrong-fix foil (seed pattern) · mcq4
A drawing of dissolved nickel(II) chloride, NiCl₂, appears in the figure. The drawing has one error. What is it?
FIGURE (hydration-sketch · stimulus) — content: "one circle labeled Ni²⁺ ringed by four water molecules with their O sides facing the ion, and one circle labeled Cl⁻ ringed by three water molecules with th… · style: "clean particle sketch in the lesson figure's conventions"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "hydration-sketch",
"content": "one circle labeled Ni²⁺ ringed by four water molecules with their O sides facing the ion, and one circle labeled Cl⁻ ringed by three water molecules with their H sides facing the ion, the two ions well separated",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
The ion count is wrong — NiCl₂ needs two Cl⁻ ions for every Ni²⁺. KEY
The shell around the Ni²⁺ points the wrong ends inward.
The shell around the Cl⁻ points the wrong ends inward.
The figure shows a student's particle diagram of methanol, a molecular compound, dissolved in water — with the methanol drawn as separated circles marked + and −. What is the error?
FIGURE (particle-panels · stimulus) — content: "a box of water molecules with the methanol drawn as single separated circles, half marked with plus signs and half marked with minus signs, spread through t… · style: "clean particle diagram"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "particle-panels",
"content": "a box of water molecules with the methanol drawn as single separated circles, half marked with plus signs and half marked with minus signs, spread through the box",
"style": "clean particle diagram"
}
}
The methanol should be drawn as whole, uncharged molecules — a molecular compound does not separate into ions. KEY
The methanol pieces should be drawn touching one another in a single cluster at the center of the box.
The plus and minus signs should be swapped between the two kinds of circles.
Nothing is wrong — dissolved methanol really does spread as separated charged pieces.
Strontium nitrate, an ionic compound, and mannitol, a molecular compound, are each dissolved in its own beaker of water. Which classification pair is correct?
Strontium nitrate is an electrolyte; mannitol is a nonelectrolyte. KEY
Mannitol is an electrolyte; strontium nitrate is a nonelectrolyte.
Both are electrolytes, because both dissolved completely.
Both are nonelectrolytes, because neither compound is a metal.
Two look-alike clear liquids sit side by side; only the first one lights a conductivity tester. Which difference between the liquids is CERTAIN?
Only the first liquid contains charged particles that are free to move. KEY
The two liquids are at different temperatures.
The first liquid holds more dissolved material than the second.
Up close, the two liquids would look different from each other.
SOL-010 · hard · transfer · seq 18 · authored (stage 2) — fresh claim-eval: amount-decides polarity (blueprint seq-18 rotation); generic — instance-free; MOST-hedge held (D6) · mcq4
A student claims, 'Any dissolved compound will conduct if you simply dissolve enough of it.' Which correction is right?
Compound type decides, not amount — neutral molecules carry no current at any concentration. KEY
A large enough dose of any solute really will light the bulb eventually.
Enough solute conducts, provided the solution is also stirred while it is tested.
The rule is backwards — adding more solute always makes a solution conduct less.
SOL-012 · hard · transfer · seq 19 · authored (stage 2) — fresh repel-claim correction, coated-surface transfer (blueprint seq-19 rotation); waterproofing spray home-seq 19C · mcq4
Rain beads up on a jacket treated with a nonpolar waterproofing spray. A student says the coating 'repels each raindrop the way a magnet repels a magnet.' Which correction is right?
No repulsion — the drops' particles attract one another more than the coating does, so they bead up. KEY
Waterproof coatings really do work by pushing charged raindrops away from the fabric.
The coating dissolves each drop so quickly that the drops only appear to bounce off.
The drops bead up only because rain falls too fast to soak into any surface.
Charcoal lighter fluid is a thin, fast-pouring nonpolar liquid. A student predicts it will dissolve in water 'because thin liquids blend into water easily.' What is the correct evaluation?
The prediction fails — thinness is not the test; the polarity mismatch predicts poor dissolving. KEY
Thinness really is the rule — the thinner the liquid, the better it dissolves.
A fast-pouring liquid slips between water's particles, so the prediction holds.
The real blocker is weight — the fluid is too lightweight to sink and mix.
SOL-006 · medium · application · seq 21 · authored (stage 2) — fresh FRQ instance per blueprint seq-21 plan (lithium iodide 2+2, stated-count variant of the A-form skill); rubric/model answer ASCII per the V-T12 contract · frq
On paper, sketch a particle diagram of dissolved lithium iodide: two Li⁺ ions and two I⁻ ions, each inside its own hydration shell. Label every ion, draw each water molecule as one large O circle with two small H circles, and point every water shell the right way. Then select Continue to compare your sketch with the model answer.
FIGURE (model_answer_diagram · model_answer_diagram) — see raw specraw spec
{
"figureShape": "model_answer_diagram",
"modelAnswerFigureSpecs": {
"a": {
"figureType": "hydration-sketch",
"content": "two separated circles labeled Li⁺ each ringed by three water molecules with O sides inward, and two separated circles labeled I⁻ each ringed by three water molecules with H sides inward",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
}
On paper, sketch a particle diagram of dissolved lithium iodide: two Li⁺ ions and two I⁻ ions, each inside its own hydration shell. Label every ion, draw each water molecule as one large O circle with two small H circles, and point every water shell the right way. Then select Continue to compare your sketch with the model answer.
Part a. On paper, sketch a particle diagram of dissolved lithium iodide: two Li⁺ ions and two I⁻ ions, each inside its own hydration shell. Label every ion, draw each water molecule as one large O circle with two small H circles, and point every water shell the right way. Then select Continue to compare your sketch with the model answer.
Model: Four well-separated circles — two labeled Li+ and two labeled I-. Each Li+ sits inside a ring of three or four water molecules with O sides facing it; each I- sits inside a ring of three or four water molecules with H sides facing it. No ion touches another ion.
Full model answer. Four well-separated circles — two labeled Li+ and two labeled I-. Each Li+ sits inside a ring of three or four water molecules with O sides facing it; each I- sits inside a ring of three or four water molecules with H sides facing it. No ion touches another ion.
To tint frosting, 2 g of green food coloring is stirred into 98 g of water, making one evenly colored liquid. Which statement names the two parts correctly?
The food coloring is the solute, and the water is the solvent. KEY
The water is the solute, because it takes on the green color.
The food coloring is the solvent, because its color spreads through everything.
The green liquid itself is the solvent, and nothing in it is a solute.
Right after hard stirring, a cloudy gray mixture looks even all the way through. Which LATER observation would confirm that the mixture really is a solution?
Hours later, with no further stirring, every part is still identical and nothing has settled. KEY
Hours later a thin gray layer has collected on the bottom, while the liquid above it looks even.
The mixture pours smoothly, without any lumps catching in the spout.
The mixture was stirred twice as long as the recipe required.
A food-grade fog fluid is 85% propylene glycol and 15% water, mixed completely into one uniform liquid. Which substance is the solvent?
The propylene glycol, because it is the substance present in the larger amount. KEY
The water, because water is the solvent in every solution.
Neither substance — two liquids cannot form a solution.
The fog fluid itself, because it is the finished product.
SOL-003 · medium · application · seq 7 · authored (stage 2) — fresh write-direction formula; cesium nitrate home-seq 7D ('cesium' also at 11E CsBr — non-base token, distinct compound, recorded; blueprint's zinc-nitrate candidate rejected: zinc = base token owned by 8A) · text
Cesium nitrate, CsNO₃, is dissolved in water. Write the formula with the symbol that shows it is dissolved in water.
Answer: CsNO₃(aq) KEY
also accepted: CsNO3(aq), CsNO₃ (aq), CsNO3 (aq)
wrongValueDiagnostics
[
{
"value": "CsNO₃(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted cesium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted cesium nitrate; dissolved in water is (aq)"
},
{
"value": "CsNO₃(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid cesium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid cesium nitrate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
feedback
{
"keyExplanation": "The solute's formula comes first, and (aq) after it marks the substance as dissolved in water: CsNO₃(aq). The (aq) always belongs to the dissolved substance — the water, as the solvent, never takes the label.",
"anticipatedWrong": [
{
"value": "CsNO₃(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted cesium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted cesium nitrate; dissolved in water is (aq)"
},
{
"value": "CsNO₃(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid cesium nitrate; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid cesium nitrate; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
}
SOL-004 · medium · application · seq 8 · authored (stage 2) — fresh which-end-attracts-which-ion pairing (blueprint seq-8 rotation); barium iodide home-seq 8D (sulfate REJECTED at stage 2: SO₄²⁻ normalizes to a false-instance '42' colliding with the 13A MgSO₄ seed — V-T7 numeric class; 'barium' also at 7B/21D, non-base, distinct compounds, recorded); oxygen/hydrogen end-language = pre-flagged rule-content family (allocation note 5) · mcq4
Barium iodide, made of Ba²⁺ and I⁻ ions, dissolves in water. Which statement pairs each end of the water molecule with the ion it attracts?
The oxygen ends turn toward the Ba²⁺ ions, and the hydrogen ends turn toward the I⁻ ions. KEY
The oxygen ends turn toward the I⁻ ions, and the hydrogen ends turn toward the Ba²⁺ ions.
Both ends turn toward the Ba²⁺ ions, because positive ions attract everything nearby.
Neither end attracts either ion — the ions separate on their own once wet.
An ionic compound has fully dissolved. A student claims its positive and negative ions now drift through the solution joined as neutral two-ion pairs. What is actually true?
The ions travel separately — every ion moves inside its own hydration shell. KEY
Opposite charges must stay stuck together even in solution, so the paired drifting is real.
The positive ions travel in pairs with each other, and the negative ions do the same.
The ions travel separately at first, then re-join into pairs once the stirring stops.
SOL-006 · medium · application · seq 11 · authored (stage 2) — fresh error-find: lattice not taken apart (blueprint seq-11 error rotation); lithium fluoride home-seq 11D · mcq4
A classmate sketches lithium fluoride after it has fully dissolved, as shown in the figure. The sketch has one error. What is the error?
FIGURE (hydration-sketch · stimulus) — content: "a tight block of alternating circles labeled Li⁺ and F⁻ drawn in rows touching one another, ringed on the outside by water molecules whose O sides face the … · style: "clean particle sketch in the lesson figure's conventions"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "hydration-sketch",
"content": "a tight block of alternating circles labeled Li⁺ and F⁻ drawn in rows touching one another, ringed on the outside by water molecules whose O sides face the block near Li⁺ circles and whose H sides face the block near F⁻ circles",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
The crystal is still drawn as one solid block — dissolved ions separate, each in its own shell. KEY
The water molecules on the Li⁺ side of the block point the wrong ends inward.
The ions' charge labels should be removed, because dissolved ions lose their charges.
There are too many water molecules around the block.
SOL-007 · medium · application · seq 12 · authored (stage 2) — fresh whole-molecule read (blueprint seq-12 rotation); sorbitol home-seq 12D; 'a molecular compound' stated per specNotes · mcq4
Sorbitol, a molecular compound, dissolves fully in water. Which statement correctly identifies the dissolved sorbitol particles?
Whole, neutral sorbitol molecules — dissolving spread them out without breaking them apart. KEY
Sorbitol ions — every dissolved substance separates into charged particles.
Individual atoms — dissolving takes molecules apart into their atoms.
One large clump of sorbitol molecules drifting through the liquid.
A compound dissolves completely, yet its solution leaves the bulb dark. Two students explain it. Student 1: 'The solution blocks electricity.' Student 2: 'Nothing in the solution can carry charge.' Which judgment is right?
Student 2 — the dissolved particles are neutral, so nothing is available to carry current between the strips. KEY
Student 1 — some solutions actively stop current that would otherwise flow.
Both students — blocking the current and lacking carriers are the same thing.
Neither student — the bulb stays dark only because too little compound was dissolved.
A solution lights the conductivity bulb brightly. Which of these does the lit bulb NOT tell you?
Which compound is dissolved in the solution. KEY
That charged particles are present in the solution.
That the charged particles are free to move through the liquid.
That the solution conducts electricity.
SOL-010 · hard · transfer · seq 18 · authored (stage 2) — fresh claim-eval: same-dissolving-same-behavior polarity (blueprint seq-18 rotation); generic — instance-free; MOST-hedge held (D6) · mcq4
A student claims, 'If two different compounds both dissolve completely in water, their solutions must behave the same way in a conductivity test.' Which correction is right?
Dissolving alike is not conducting alike — ionic solutions conduct; most molecular ones do not. KEY
Complete dissolving really is what makes any solution conduct.
The claim has it backwards — dissolved molecular compounds conduct and dissolved ionic ones do not.
The claim is wrong, because solutions cannot conduct electricity at all.
Melted lard and water separate into layers no matter how hard they are shaken. A student concludes the two liquids 'repel each other like matching magnet poles.' Which correction is right?
No push exists — the attractions are mismatched in kind, so each liquid gathers with its own kind. KEY
The repulsion is real, and stronger shaking would eventually overcome it.
The layers form because the two liquids' particles are exactly the same size.
The layers form because one liquid is hot and the other is cold.
SOL-013 · hard · transfer · seq 20 · authored (stage 2) — fresh claim-eval: amount-of-water basis (blueprint seq-20 rotation); pine rosin home-seq 20D · mcq4
A lump of pine rosin, a nonpolar solid, sits undissolved in a beaker. A student says, 'Keep adding water — with enough water, anything dissolves.' What is the correct evaluation?
More water does not change the prediction — it is more of the same polar solvent, so the mismatch remains. KEY
The plan works, because adding water raises the mixture's polarity until the rosin matches it.
Switching to hot water would dissolve the rosin — temperature, not polarity, is what decides dissolving.
More water just dilutes the water's dissolving power, so the lump becomes even less likely to dissolve.
SOL-006 · medium · application · seq 21 · authored (stage 2) — fresh FRQ instance per blueprint seq-21 plan (barium chloride, ion count derived from the 1:2 formula — the p17 skill); rubric/model answer ASCII with the 2+ charge in words per the V-T12 contract · frq
On paper, sketch a particle diagram of dissolved barium chloride, BaCl₂, showing one formula unit's ions hydrated. (BaCl₂ gives Ba²⁺ and Cl⁻ ions.) Label every ion with its symbol and full charge, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
FIGURE (model_answer_diagram · model_answer_diagram) — see raw specraw spec
{
"figureShape": "model_answer_diagram",
"modelAnswerFigureSpecs": {
"a": {
"figureType": "hydration-sketch",
"content": "one separated circle labeled Ba²⁺ ringed by four water molecules with O sides inward, and two separated circles labeled Cl⁻ each ringed by three water molecules with H sides inward",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
}
On paper, sketch a particle diagram of dissolved barium chloride, BaCl₂, showing one formula unit's ions hydrated. (BaCl₂ gives Ba²⁺ and Cl⁻ ions.) Label every ion with its symbol and full charge, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
Part a. On paper, sketch a particle diagram of dissolved barium chloride, BaCl₂, showing one formula unit's ions hydrated. (BaCl₂ gives Ba²⁺ and Cl⁻ ions.) Label every ion with its symbol and full charge, and aim every water shell correctly. Then select Continue to compare your sketch with the model answer.
Model: Three well-separated circles — one labeled with the barium symbol and a 2+ charge, and two labeled Cl-. The barium ion sits inside a ring of four water molecules with O sides facing it; each Cl- sits inside a ring of three water molecules with H sides facing it. No ion touches another ion — the formula gives one barium ion for every two Cl- ions.
Full model answer. Three well-separated circles — one labeled with the barium symbol and a 2+ charge, and two labeled Cl-. The barium ion sits inside a ring of four water molecules with O sides facing it; each Cl- sits inside a ring of three water molecules with H sides facing it. No ion touches another ion — the formula gives one barium ion for every two Cl- ions.
A liquid on a shelf looks uniform, but tiny flecks can be seen drifting everywhere inside it when it is held up to the light. A student calls it a solution. Which response is correct?
Not a solution — the flecks are undissolved pieces, so it is not truly the same throughout. KEY
It is a solution, because the flecks are spread through every part of the liquid.
It is a solution, because it looks uniform from a distance.
It is not a solution, because a true solution must be completely colorless.
A lab glass-cleaning rinse is 90% acetone and 10% water, blended into one uniform liquid. Which statement assigns the roles correctly?
The acetone is the solvent, and the water is the solute. KEY
The water is the solvent, because water always does the dissolving.
The acetone is the solute, because it evaporates first.
Both substances are solvents, because each could dissolve things on its own.
SOL-003 · medium · application · seq 7 · authored (stage 2) — fresh write-direction formula; iron(III) chloride home-seq 7E (iron = base MATERIALS token, home-seq 7E) · text
Iron(III) chloride, FeCl₃, is dissolved in water. Write the formula with the symbol that shows it is dissolved in water.
Answer: FeCl₃(aq) KEY
also accepted: FeCl3(aq), FeCl₃ (aq), FeCl3 (aq)
wrongValueDiagnostics
[
{
"value": "FeCl₃(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted iron(iii) chloride; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted iron(iii) chloride; dissolved in water is (aq)"
},
{
"value": "FeCl₃(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid iron(iii) chloride; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid iron(iii) chloride; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
feedback
{
"keyExplanation": "The solute's formula comes first, and (aq) after it marks the substance as dissolved in water: FeCl₃(aq). The (aq) always belongs to the dissolved substance — the water, as the solvent, never takes the label.",
"anticipatedWrong": [
{
"value": "FeCl₃(l)",
"misconception": "you used the pure-liquid symbol — (l) means melted iron(iii) chloride; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the pure-liquid symbol — (l) means melted iron(iii) chloride; dissolved in water is (aq)"
},
{
"value": "FeCl₃(s)",
"misconception": "you used the solid symbol — (s) means undissolved solid iron(iii) chloride; dissolved in water is (aq)",
"feedback": "This answer is wrong — you used the solid symbol — (s) means undissolved solid iron(iii) chloride; dissolved in water is (aq)"
},
{
"value": "H₂O(aq)",
"misconception": "you attached the label to the water — the (aq) follows the solute's formula, not the solvent's",
"feedback": "This answer is wrong — you attached the label to the water — the (aq) follows the solute's formula, not the solvent's"
}
]
}
A student claims an ionic crystal dissolves by breaking apart all at once, 'like a sandcastle collapsing.' Which correction is right?
Ions leave the crystal one at a time, each pulled away by the water molecules attached to it. KEY
The collapse picture is right, except that it happens only in hot liquids.
The crystal first splits into a few large chunks, and each chunk then dissolves in one burst.
The crystal dissolves from the inside out, hollowing before its surface changes.
SOL-005 · medium · application · seq 10 · authored (stage 2) — fresh shell-orientation probe, polyatomic 2- ion (blueprint seq-10 rotation); potassium carbonate home-seq 10E (same-seq 'potassium' sharing with 10B = A5 class, recorded) · mcq4
Potassium carbonate, made of K⁺ and CO₃²⁻ ions, dissolves in water. Which description of the hydration shells is correct?
Shells around each K⁺ point oxygen ends inward, and shells around each CO₃²⁻ point hydrogen ends inward. KEY
Shells around each K⁺ point hydrogen ends inward, and shells around each CO₃²⁻ point oxygen ends inward.
Only the K⁺ ions get shells — negative ions repel water and stay bare.
Every shell points oxygen ends inward, because the oxygen end always leads.
SOL-006 · medium · application · seq 11 · authored (stage 2) — fresh error-find: ions drawn touching in pairs (blueprint seq-11 error rotation); cesium bromide home-seq 11E ('cesium' also at 7D CsNO₃ — non-base token, distinct compound, recorded) · mcq4
The figure reproduces a lab-notebook sketch of dissolved cesium bromide. There is one error in the sketch. Which statement names it?
FIGURE (hydration-sketch · stimulus) — content: "three pairs of circles, each pair one circle labeled Cs⁺ touching one circle labeled Br⁻, each pair ringed together by four water molecules with O sides tow… · style: "clean particle sketch in the lesson figure's conventions"raw spec
{
"figureShape": "stimulus",
"figureSpec": {
"figureType": "hydration-sketch",
"content": "three pairs of circles, each pair one circle labeled Cs⁺ touching one circle labeled Br⁻, each pair ringed together by four water molecules with O sides toward the Cs⁺ side of the pair and H sides toward the Br⁻ side",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
The ions are drawn touching in pairs — dissolved ions separate, each in its own shell. KEY
Each ring holds too many water molecules.
The O sides of the water molecules should face the Br⁻ ions.
The sketch should show the ions gathered back into one crystal block.
Solution P contains free-moving dissolved ions. Solution Q contains only dissolved neutral molecules. A conductivity tester is dipped into each. Which result and reason are correct?
P lights the bulb — its moving ions carry charge across; Q stays dark — neutral molecules carry none. KEY
Q is the one that lights the bulb, since whole molecules are bigger and push more current along.
Both bulbs light, because each solution contains dissolved particles.
Neither lights the bulb, because liquids cannot conduct electricity.
A bottle's label says its compound is molecular, and the compound dissolves completely in water. Without a tester, which prediction follows the taught rule?
Most likely a nonconducting solution — most molecular compounds dissolve as neutral whole molecules. KEY
The solution certainly conducts, because every dissolved compound conducts.
No prediction is possible, because compound type says nothing about conduction.
The solution conducts once enough of the compound has dissolved.
A generous scoop of a compound has dissolved completely, yet the tester's bulb stays dark. Which statement MUST be true?
Every particle the compound put into the solution is electrically neutral. KEY
The compound failed to dissolve after all.
The solvent soaked up the charges the compound released.
Too little of the compound was used for the bulb to light.
SOL-010 · hard · transfer · seq 18 · authored (stage 2) — fresh claim-eval: dissolving-charges-the-molecules polarity (blueprint seq-18 rotation); generic — instance-free; MOST-hedge held (D6) · mcq4
A student claims, 'Molecular compounds conduct after dissolving, because the act of dissolving gives their molecules electric charges.' Which correction is right?
Dissolving adds no charge — most molecular compounds spread out as the same neutral molecules they always were. KEY
It really works that way — rubbing against water molecules charges the solute, as rubbing a balloon does.
In warm solutions the extra energy knocks charges loose, so the claim holds there.
The claim is wrong only because molecular compounds never dissolve in the first place.
SOL-012 · hard · transfer · seq 19 · authored (stage 2) — fresh repel-claim correction (blueprint seq-19 rotation); rubber cement home-seq 19E · mcq4
Rubber cement, a nonpolar adhesive, stays separate from water in the same jar. A student says, 'their particles must repel, or they would mix.' Which correction is right?
No particles repel — the two substances' attractions are mismatched in kind, so each stays with its own. KEY
Refusing to mix is itself the proof — repelling is exactly what non-mixing means.
The two stay separate because the adhesive's particles are stickier than everything else.
Hard shaking would turn the pair into one lasting, uniform solution.
SOL-013 · hard · transfer · seq 20 · authored (stage 2) — fresh claim-eval: water-as-universal-solvent basis, parallel to the 20A seed family by design (blueprint seq-20 rotation); squalane home-seq 20E · mcq4
Squalane, a nonpolar skin-care liquid, is poured into water. A student predicts it will dissolve 'because water is the universal solvent — it dissolves almost everything.' What is the correct evaluation?
The prediction fails — the nickname covers polar and ionic solutes, and nonpolar squalane is neither. KEY
Universal really means universal — the nickname promises that every substance dissolves.
Skin-care liquids are designed to blend with everything, so the prediction holds here.
The only mistake is the temperature — squalane must be warmed before it can dissolve.
SOL-006 · medium · application · seq 21 · authored (stage 2) — fresh FRQ instance per blueprint seq-21 plan (calcium iodide, 1:2 formula-count skill; same-seq 'calcium' sharing with the 21B seed = A5 class, recorded); rubric/model answer ASCII with the 2+ charge in words per the V-T12 contract · frq
On paper, sketch a particle diagram showing one formula unit of calcium iodide, CaI₂, dissolved with every ion hydrated. (CaI₂ gives Ca²⁺ and I⁻ ions.) Label every ion with its symbol and full charge, and point each water shell in the correct direction. Then select Continue to compare your sketch with the model answer.
FIGURE (model_answer_diagram · model_answer_diagram) — see raw specraw spec
{
"figureShape": "model_answer_diagram",
"modelAnswerFigureSpecs": {
"a": {
"figureType": "hydration-sketch",
"content": "one separated circle labeled Ca²⁺ ringed by four water molecules with O sides inward, and two separated circles labeled I⁻ each ringed by three water molecules with H sides inward",
"style": "clean particle sketch in the lesson figure's conventions"
}
}
}
On paper, sketch a particle diagram showing one formula unit of calcium iodide, CaI₂, dissolved with every ion hydrated. (CaI₂ gives Ca²⁺ and I⁻ ions.) Label every ion with its symbol and full charge, and point each water shell in the correct direction. Then select Continue to compare your sketch with the model answer.
Part a. On paper, sketch a particle diagram showing one formula unit of calcium iodide, CaI₂, dissolved with every ion hydrated. (CaI₂ gives Ca²⁺ and I⁻ ions.) Label every ion with its symbol and full charge, and point each water shell in the correct direction. Then select Continue to compare your sketch with the model answer.
Model: Three well-separated circles — one labeled with the calcium symbol and a 2+ charge, and two labeled I-. The calcium ion sits inside a ring of four water molecules with O sides facing it; each I- sits inside a ring of three water molecules with H sides facing it. No ion touches another ion — the formula gives one calcium ion for every two I- ions.
Full model answer. Three well-separated circles — one labeled with the calcium symbol and a 2+ charge, and two labeled I-. The calcium ion sits inside a ring of four water molecules with O sides facing it; each I- sits inside a ring of three water molecules with H sides facing it. No ion touches another ion — the formula gives one calcium ion for every two I- ions.
End of Topic Test: Solubility, temperature, and saturation & Solubility curves & Speed of dissolving vs amount dissolved (u10-t2 · 5 forms × 21 items · 105 total)
David's notes for James:
FRQ COMPOSITION - frq_per_form = 1 (your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - this scope (SOL-014..028) teaches exactly ONE quantitative routine, SOL-024's worded-equation curve-difference (cooling and warming are one routine per D10: the same difference of two canonical readings; the atom's equation is worded, no symbols, no rearrangement). U10's calc-heavy scopes (molarity SOL-034..040, dilution/solution stoichiometry SOL-041..045) sit in U10-G3/G4, where the 2-3-FRQ split applies instead. One FRQ per distinct routine (the wave-5 rule) = ONE FRQ, homed at SOL-024, terminal at seq 21 per D6, parts (a) both curve readings (b) the R15-shaped worded-equation subtraction (c) a cap-direction check in words; rubric 30/40/30 = 100. The form already carries three numeric-entry free-calculation rows (seq 7/11/16), so a second FRQ would re-measure the same routine for zero construct gain. D5 scaling NOT engaged: the 15-lesson scope fills the full 20-objective blueprint (10 lessons x 1 row + the SOL-020/024/028 capstones and SOL-021/023 multi-case lessons x 2). prod_seeds is EMPTY in your gate data, so all five FRQs are authored fresh (the standing typed-plan flag below applies).
NUMERIC CONTRACT + TOLERANCE CARRIAGE (disclosure - the section-6 answer shape has no `tolerance` lane): 15 numeric ENTRY rows (seq 7/16 forward curve reads, seq 11 curve-difference). Our internal contracts carry an explicit tolerance field - +/-2 on single curve reads (the SOL-021 law) and +/-4 on seq-11 differences (a difference of two +/-2 readings legitimately compounds; parity with the FRQ rubric's carried-tolerance rule - recorded contract adjudication at sol loop 1, raised from the drafted +/-2). Since the export schema has no tolerance lane, EVERY numeric item's gradingNote states the governing tolerance in words - please carry it into grading. sigFigs is exported as display metadata only (V-T6 requires the field; your specNotes say sigFigs null for curve reads; the tolerance governs - the recorded adjudication is restated in each gradingNote). wrongValueDiagnostics [{value, misconception, feedback}] are carried on the numeric rows; internal check-expressions are authoring machinery and are not exported.
SOL-021 FIGURE LAW (the gate's structural spine, confirm): every solubility-curve figureSpec draws the unit's canonical fixed curve data VERBATIM (your SOL-021 extraNotes: real published values, used verbatim in every figure, 10-80 degree gridlines), item temperatures always sit on gridlines, and answers are the canonical curve values AT gridline temperatures with +/-2 reading tolerance and NO rounding instruction - your own fluency items grade 55, 88, and 138 as numeric answers on 10/20-g y-grids under exactly this contract. sol (authoritative QC) twice demanded y-gridline answers or stem-stated precision; the law is course authority, so content is unchanged and the gradingNotes now quote the law + your fluency-item precedent. The residual graph-precision and sigFigs minors on file (B#7/B#11/B#16/E#7/E#16 class) are this recorded tension, not defects.
SEED FRESHENINGS UNDER THE STRICT SEED RULING (validator never weakened) - 8 dispositions to confirm, every one preserving your structure and misconception keys with all fields swept (QB 3.9); the first three touch James-authored text: (a) B#12 (SOL-025 GT1) 'aquarium salt' -> 'water-conditioner crystals' ('salt' is SOL-025's own WE instance). (b) B#13 (SOL-026 GT1) 'bath salts' -> 'bath crystals' (kills the B-form seq-12/13 adjacent 'salt' pairing and the SOL-027 bath-salt fluency echo). (c) A#12 (SOL-025 GT0) '200 mL' -> '250 mL' (200 is SOL-014's sugar-200 corpus value) and 'potassium chloride' -> 'KCl' (the '10 g of potassium' 4-gram collides with scope text; your explanation already used KCl). (d) C#1 authored: 18 g -> 16 g. (e) E#19 authored: reported value 160 -> 155 (160 sits in corpus axis text). (f) D#19 authored: rebuilt on KClO3 30->80 (report 28) from the drafted 20->60/17 (17 in corpus); distractors made number-free. (g) C#11 authored: KClO3 10->50 -> 20->60 so no two forms share the mandated 'warmed from 10 degrees C' stem opening (form B's seeded seq-11 and FRQ own it). (h) Cross-form wording de-collisions with no instance change (B15/E12/E14/D6/D5/E6/D17/E20/E13).
NOTA PURGE-BY-REPLACEMENT x2 (ANSWERS #2, confirm): A#10 (your pool 19, SOL-020 GT1) 'None of the above' -> 'Either saturated or supersaturated - resting crystals fit both states.' (keyed to the growth-trigger misconception); C#8 (your pool 26, SOL-022 GT2) 'None of the above' -> 'It cannot be classified until the point is drawn on the graph.' (keyed to the demanded-drawn-point error). Note: spare pool 15 (SOL-019 GT0) also carries a NOTA distractor - any later promotion of that spare must repeat the replacement.
ADJUDICATED INSTANCE FAMILIES x7 (confirm; the gate's structural whitelists, enforced by a gate-local wrapper over the UNTOUCHED shared validator, slot lists enumerated from the scan): (1) AQUEOUS-SOLVENT-WATER (78 slots) - the topic IS solubility in water; the g-per-100-g-of-water reporting basis (SOL-014 rule line) plus your SOL-024 stem law put 'water' in nearly every stem BY DESIGN; solutes and numbers still scan through their own lanes. (2) TAUGHT-GAS-INSTANCES (oxygen/air, 8 slots) - SOL-016's rule lines name dissolved oxygen and dissolved air as THE taught gas pattern; on the seq-3/6 rows these are the lessons' own assessed content. (3) SEED-SALT-INSTANCES (A#13/A#15/B#17) - your verbatim gate seeds name table/rock salt on both sides (spec-internal reuse). (4) SEED-SPOON-C12 - C#12 is SOL-025's own gate item and its assessed claim IS about the spoon. (5) TAUGHT-GRID-CONSTANTS (100 + the 20..80 gridline lattice, 50 slots) - course law (the 100-g basis and the gridline temperatures), not instance choices; instance identity on curve rows is the (substance, temperature) PAIR, kept fresh by authoring (pair ledger in fix_log.md). (6) CURVE-VALUES-BY-CONSTRUCTION (B#8/C#19/E#19) - on-curve point values and error-analysis caps/sums ARE canonical curve readings by design; the fixed table cannot be freshened without breaking the figure law. (7) SEED-TEMP-COINCIDENCE-A13 ('55') - your SOL-026 seed's 5/55 contrast coincides with an unrelated series value; single slot. Everything outside the enumerated slots stays fully armed.
GATE_CONFIG TASK-IDIOM WHITELIST (12 recorded entries, confirm) - each preserves YOUR wording verbatim with zero instance signal: 'Using the <substance> curve on the graph' x5 + 'Using the same KCl and KNO3 graph' (SOL-024/023's MANDATED stem-law instructions; the substance inside each clause still scans as the item's instance), 'is warmed from 10 degrees C to' (the stem law's warming clause, colliding with SOL-024's own fluency wording; 10 sits in the validator's ignore band), 'Classify the two solutions' (SOL-018's imperative task clause), 'Which order ranks them from most to least soluble' (SOL-023's ranking clause), 'that can now dissolve' / 'crystallizes out, in grams' (SOL-024's warming/cooling question wording), 'dissolved in the same conditions'. Shared validator untouched.
BLIND HARD-FAIL REWORK - D#6 (SOL-017, confirm): the drafted stem compared warm SURFACE water with cold DEEP water, so the pressure distractor invoked the real depth/pressure dependence of gas solubility (two-defensible, F2). Reworked to a same-water seasonal comparison (late summer vs early spring); the pressure distractor became a density misconception and the sink-to-cold-bottom option a settling misconception. A deterministic sweep over every SOL-016/017 item proved D#6 was the only carrier of the confound.
JAMES-SEED AMENDMENT AT AUTHORITATIVE QC - C#12 (SOL-025 gate item, confirm): sol showed the bare stem left mechanical fragmentation defensible, so the stem now reads 'gently stirring ... with a spoon' and the grinding feedback was softened. Construct and key unchanged.
CHEMICAL-REALISM AMENDMENTS at QC (confirm): D#17 alum probe -> potassium sulfate (aluminum alum reacts with washing soda - aluminum hydroxide precipitation would contaminate the saturation-test scenario; sodium sulfate was rejected because 'sodium' is SOL-020 lexicon); E#1 'alum ~6 g at 20 degrees C' was hydrate/anhydrous-ambiguous -> the unit's own canonical KClO3@20 = 7 g; E#13 bouillon (heterogeneous) then borax (a decahydrate that melts in its own hydration water near 75 degrees C - sol right twice) -> cream of tartar; B#2 cream-of-tartar options made parallel-qualified and its speed-vs-amount distractor re-fenced inside SOL-028's scope; B#13's 'rising current' distractor invoked real convection -> inside-out soaking misconception; E#5 stem states equal solvent as 'the same solvent'; E#15 'the amount' pinned to the maximum that can dissolve; D#11's 60-g diagnostic acknowledged both numeric routes to 60.
LONGEST-OPTION REBALANCE + ACCEPTED ADVISORY CLASSES (recorded, no further fix): blind flagged answer-length cues repeatedly - 39 of 85 mcq keys were strictly longest; rebalanced to 25 with every margin <=11% (the residual 'key is the most complete explanation' style is structural to explanation MCQs). Accepted with rationales: canonical fixed-curve data recurring across a form's graph items = the SOL-021 figure LAW (every figure draws the same table verbatim; every student sees it on every item); 'item 19 reveals the subtraction routine' = the routine is taught course content and no seq-19 key discloses another item's values; graph-precision flags = the tolerance contract above; B#18 crossing-explicit = the tested skill; B#19/C#19/E#19-class 'the wrong report is checkable' = error-analysis by construction; B#17 'salt' shorthand = the stem itself defines table salt (your seed); definition-echo recall keys = keyed on the taught rule by design (V-T4 bounds margins deterministically).
BLIND-PACKET DEFECT DISCLOSURE (QC provenance, no action needed): the round-1 key-blind packets rendered all 40 solubility-curve figureSpecs style-only (no drawn data) - one reviewer honestly refused a form as unanswerable and the round was VOIDED pre-adjudication with fresh tags for the corrected round. Fixed per-gate (renders axes + every series point list + marked points with their drawn guides - exactly what a sighted student sees). Corrected full round r3/r4: key picks 170/170, 1 hard fail (D#6 above), 55 advisories; re-blind rounds r5/r6 64/64, r7/r8 18/18, r9/r10+g11 12/12 across three seats - 0 open hard fails. All 30 numeric cold answers EXACT against the contracts and all 10 FRQ cold answers match caps, routine, result, and direction reasoning (hand-graded with quoted evidence in frq_verdicts.json). Harness notes batched for the maintainer: generic solubility-curve rendering belongs in _shared fig_text; gateway gemini needed a 32k token budget for 21-item replies; the reviewer-tag matcher routes any r1-prefix tag to the sol seat.
SOL AUTHORITATIVE LEDGER: loops 1-4 = 11 -> 6 -> 0 -> 0 FAILs; final 105/105 pass, 0 open fails; 13 honest-survivor minors on file, all style/metadata with keys unaffected (the sigFigs-vs-gradingNote tension and no-precision-instruction notes mandated by the SOL-021 law; weaker-third-distractor plausibility notes B#6/C#1/C#6; B#14 hour-to-minutes phrasing; C#15 key wording; C#21 rubric-(b) strictness; E#18 crossing-range wording; E#20 sealed-jar scope).
SPARE-SEED NOTE: 9 spares, none dropped on merit - application-band seeds on lessons whose single blueprint row is recall (SOL-014 x2, SOL-015 x2, SOL-016 x2, SOL-019 pool 16), pool 13's warming-reclassify chains two lessons' content (kept spare to protect the one-lesson-per-item rule), and pool 15 carries the NOTA noted above.
BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose/radiological framings - the banned-token scan (uranium/plutonium/radium/actinide/enrich/reactor/weapon/radiation/fission/radioactive/meltdown/dose) over your 45-seed pool at stage 1, all 105 finalized items, and this export body = 0 hits everywhere (instances are household and classroom lab salts and sugars throughout).
EXPORTER VERIFICATION DISCLOSURE (new shared-suite scoping class, first occurrence - no action needed): the shared exporter's post-write NOTA/AOTA scan runs over FULL item payloads including the `source` provenance lane, while its own metadata-leak scan two lines later is already scoped to student-facing fields (source/seedRef = intentionally-carried provenance). This gate's two `source` strings for A#10/C#8 honestly QUOTE the replaced 'None of the above' literal in their freshening disclosures, so the scan tripped AFTER the export was written in full. The shared suite was NOT edited (never-weaken + wave isolation): a gate-local _verify_james_export.py re-runs the ENTIRE shared verification suite verbatim with the one scoping correction - the NOTA scan passes over every student-facing field, and the two provenance hits are enumerated and asserted to be exactly the recorded A#10/C#8 disclosure strings. Batched for the shared maintainer alongside the text-row scoping gap.
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u10-t2 — notes/queries from David's side (U10-G2)
1. FRQ COMPOSITION - frq_per_form = 1 (your gate data's frq_per_form=1 concurs): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - this scope (SOL-014..028) teaches exactly ONE quantitative routine, SOL-024's worded-equation curve-difference (cooling and warming are one routine per D10: the same difference of two canonical readings; the atom's equation is worded, no symbols, no rearrangement). U10's calc-heavy scopes (molarity SOL-034..040, dilution/solution stoichiometry SOL-041..045) sit in U10-G3/G4, where the 2-3-FRQ split applies instead. One FRQ per distinct routine (the wave-5 rule) = ONE FRQ, homed at SOL-024, terminal at seq 21 per D6, parts (a) both curve readings (b) the R15-shaped worded-equation subtraction (c) a cap-direction check in words; rubric 30/40/30 = 100. The form already carries three numeric-entry free-calculation rows (seq 7/11/16), so a second FRQ would re-measure the same routine for zero construct gain. D5 scaling NOT engaged: the 15-lesson scope fills the full 20-objective blueprint (10 lessons x 1 row + the SOL-020/024/028 capstones and SOL-021/023 multi-case lessons x 2). prod_seeds is EMPTY in your gate data, so all five FRQs are authored fresh (the standing typed-plan flag below applies).
2. NUMERIC CONTRACT + TOLERANCE CARRIAGE (disclosure - the section-6 answer shape has no `tolerance` lane): 15 numeric ENTRY rows (seq 7/16 forward curve reads, seq 11 curve-difference). Our internal contracts carry an explicit tolerance field - +/-2 on single curve reads (the SOL-021 law) and +/-4 on seq-11 differences (a difference of two +/-2 readings legitimately compounds; parity with the FRQ rubric's carried-tolerance rule - recorded contract adjudication at sol loop 1, raised from the drafted +/-2). Since the export schema has no tolerance lane, EVERY numeric item's gradingNote states the governing tolerance in words - please carry it into grading. sigFigs is exported as display metadata only (V-T6 requires the field; your specNotes say sigFigs null for curve reads; the tolerance governs - the recorded adjudication is restated in each gradingNote). wrongValueDiagnostics [{value, misconception, feedback}] are carried on the numeric rows; internal check-expressions are authoring machinery and are not exported.
3. SOL-021 FIGURE LAW (the gate's structural spine, confirm): every solubility-curve figureSpec draws the unit's canonical fixed curve data VERBATIM (your SOL-021 extraNotes: real published values, used verbatim in every figure, 10-80 degree gridlines), item temperatures always sit on gridlines, and answers are the canonical curve values AT gridline temperatures with +/-2 reading tolerance and NO rounding instruction - your own fluency items grade 55, 88, and 138 as numeric answers on 10/20-g y-grids under exactly this contract. sol (authoritative QC) twice demanded y-gridline answers or stem-stated precision; the law is course authority, so content is unchanged and the gradingNotes now quote the law + your fluency-item precedent. The residual graph-precision and sigFigs minors on file (B#7/B#11/B#16/E#7/E#16 class) are this recorded tension, not defects.
4. SEED FRESHENINGS UNDER THE STRICT SEED RULING (validator never weakened) - 8 dispositions to confirm, every one preserving your structure and misconception keys with all fields swept (QB 3.9); the first three touch James-authored text: (a) B#12 (SOL-025 GT1) 'aquarium salt' -> 'water-conditioner crystals' ('salt' is SOL-025's own WE instance). (b) B#13 (SOL-026 GT1) 'bath salts' -> 'bath crystals' (kills the B-form seq-12/13 adjacent 'salt' pairing and the SOL-027 bath-salt fluency echo). (c) A#12 (SOL-025 GT0) '200 mL' -> '250 mL' (200 is SOL-014's sugar-200 corpus value) and 'potassium chloride' -> 'KCl' (the '10 g of potassium' 4-gram collides with scope text; your explanation already used KCl). (d) C#1 authored: 18 g -> 16 g. (e) E#19 authored: reported value 160 -> 155 (160 sits in corpus axis text). (f) D#19 authored: rebuilt on KClO3 30->80 (report 28) from the drafted 20->60/17 (17 in corpus); distractors made number-free. (g) C#11 authored: KClO3 10->50 -> 20->60 so no two forms share the mandated 'warmed from 10 degrees C' stem opening (form B's seeded seq-11 and FRQ own it). (h) Cross-form wording de-collisions with no instance change (B15/E12/E14/D6/D5/E6/D17/E20/E13).
5. NOTA PURGE-BY-REPLACEMENT x2 (ANSWERS #2, confirm): A#10 (your pool 19, SOL-020 GT1) 'None of the above' -> 'Either saturated or supersaturated - resting crystals fit both states.' (keyed to the growth-trigger misconception); C#8 (your pool 26, SOL-022 GT2) 'None of the above' -> 'It cannot be classified until the point is drawn on the graph.' (keyed to the demanded-drawn-point error). Note: spare pool 15 (SOL-019 GT0) also carries a NOTA distractor - any later promotion of that spare must repeat the replacement.
6. ADJUDICATED INSTANCE FAMILIES x7 (confirm; the gate's structural whitelists, enforced by a gate-local wrapper over the UNTOUCHED shared validator, slot lists enumerated from the scan): (1) AQUEOUS-SOLVENT-WATER (78 slots) - the topic IS solubility in water; the g-per-100-g-of-water reporting basis (SOL-014 rule line) plus your SOL-024 stem law put 'water' in nearly every stem BY DESIGN; solutes and numbers still scan through their own lanes. (2) TAUGHT-GAS-INSTANCES (oxygen/air, 8 slots) - SOL-016's rule lines name dissolved oxygen and dissolved air as THE taught gas pattern; on the seq-3/6 rows these are the lessons' own assessed content. (3) SEED-SALT-INSTANCES (A#13/A#15/B#17) - your verbatim gate seeds name table/rock salt on both sides (spec-internal reuse). (4) SEED-SPOON-C12 - C#12 is SOL-025's own gate item and its assessed claim IS about the spoon. (5) TAUGHT-GRID-CONSTANTS (100 + the 20..80 gridline lattice, 50 slots) - course law (the 100-g basis and the gridline temperatures), not instance choices; instance identity on curve rows is the (substance, temperature) PAIR, kept fresh by authoring (pair ledger in fix_log.md). (6) CURVE-VALUES-BY-CONSTRUCTION (B#8/C#19/E#19) - on-curve point values and error-analysis caps/sums ARE canonical curve readings by design; the fixed table cannot be freshened without breaking the figure law. (7) SEED-TEMP-COINCIDENCE-A13 ('55') - your SOL-026 seed's 5/55 contrast coincides with an unrelated series value; single slot. Everything outside the enumerated slots stays fully armed.
7. GATE_CONFIG TASK-IDIOM WHITELIST (12 recorded entries, confirm) - each preserves YOUR wording verbatim with zero instance signal: 'Using the <substance> curve on the graph' x5 + 'Using the same KCl and KNO3 graph' (SOL-024/023's MANDATED stem-law instructions; the substance inside each clause still scans as the item's instance), 'is warmed from 10 degrees C to' (the stem law's warming clause, colliding with SOL-024's own fluency wording; 10 sits in the validator's ignore band), 'Classify the two solutions' (SOL-018's imperative task clause), 'Which order ranks them from most to least soluble' (SOL-023's ranking clause), 'that can now dissolve' / 'crystallizes out, in grams' (SOL-024's warming/cooling question wording), 'dissolved in the same conditions'. Shared validator untouched.
8. BLIND HARD-FAIL REWORK - D#6 (SOL-017, confirm): the drafted stem compared warm SURFACE water with cold DEEP water, so the pressure distractor invoked the real depth/pressure dependence of gas solubility (two-defensible, F2). Reworked to a same-water seasonal comparison (late summer vs early spring); the pressure distractor became a density misconception and the sink-to-cold-bottom option a settling misconception. A deterministic sweep over every SOL-016/017 item proved D#6 was the only carrier of the confound.
9. JAMES-SEED AMENDMENT AT AUTHORITATIVE QC - C#12 (SOL-025 gate item, confirm): sol showed the bare stem left mechanical fragmentation defensible, so the stem now reads 'gently stirring ... with a spoon' and the grinding feedback was softened. Construct and key unchanged.
10. CHEMICAL-REALISM AMENDMENTS at QC (confirm): D#17 alum probe -> potassium sulfate (aluminum alum reacts with washing soda - aluminum hydroxide precipitation would contaminate the saturation-test scenario; sodium sulfate was rejected because 'sodium' is SOL-020 lexicon); E#1 'alum ~6 g at 20 degrees C' was hydrate/anhydrous-ambiguous -> the unit's own canonical KClO3@20 = 7 g; E#13 bouillon (heterogeneous) then borax (a decahydrate that melts in its own hydration water near 75 degrees C - sol right twice) -> cream of tartar; B#2 cream-of-tartar options made parallel-qualified and its speed-vs-amount distractor re-fenced inside SOL-028's scope; B#13's 'rising current' distractor invoked real convection -> inside-out soaking misconception; E#5 stem states equal solvent as 'the same solvent'; E#15 'the amount' pinned to the maximum that can dissolve; D#11's 60-g diagnostic acknowledged both numeric routes to 60.
11. LONGEST-OPTION REBALANCE + ACCEPTED ADVISORY CLASSES (recorded, no further fix): blind flagged answer-length cues repeatedly - 39 of 85 mcq keys were strictly longest; rebalanced to 25 with every margin <=11% (the residual 'key is the most complete explanation' style is structural to explanation MCQs). Accepted with rationales: canonical fixed-curve data recurring across a form's graph items = the SOL-021 figure LAW (every figure draws the same table verbatim; every student sees it on every item); 'item 19 reveals the subtraction routine' = the routine is taught course content and no seq-19 key discloses another item's values; graph-precision flags = the tolerance contract above; B#18 crossing-explicit = the tested skill; B#19/C#19/E#19-class 'the wrong report is checkable' = error-analysis by construction; B#17 'salt' shorthand = the stem itself defines table salt (your seed); definition-echo recall keys = keyed on the taught rule by design (V-T4 bounds margins deterministically).
12. BLIND-PACKET DEFECT DISCLOSURE (QC provenance, no action needed): the round-1 key-blind packets rendered all 40 solubility-curve figureSpecs style-only (no drawn data) - one reviewer honestly refused a form as unanswerable and the round was VOIDED pre-adjudication with fresh tags for the corrected round. Fixed per-gate (renders axes + every series point list + marked points with their drawn guides - exactly what a sighted student sees). Corrected full round r3/r4: key picks 170/170, 1 hard fail (D#6 above), 55 advisories; re-blind rounds r5/r6 64/64, r7/r8 18/18, r9/r10+g11 12/12 across three seats - 0 open hard fails. All 30 numeric cold answers EXACT against the contracts and all 10 FRQ cold answers match caps, routine, result, and direction reasoning (hand-graded with quoted evidence in frq_verdicts.json). Harness notes batched for the maintainer: generic solubility-curve rendering belongs in _shared fig_text; gateway gemini needed a 32k token budget for 21-item replies; the reviewer-tag matcher routes any r1-prefix tag to the sol seat.
13. SOL AUTHORITATIVE LEDGER: loops 1-4 = 11 -> 6 -> 0 -> 0 FAILs; final 105/105 pass, 0 open fails; 13 honest-survivor minors on file, all style/metadata with keys unaffected (the sigFigs-vs-gradingNote tension and no-precision-instruction notes mandated by the SOL-021 law; weaker-third-distractor plausibility notes B#6/C#1/C#6; B#14 hour-to-minutes phrasing; C#15 key wording; C#21 rubric-(b) strictness; E#18 crossing-range wording; E#20 sealed-jar scope).
14. SPARE-SEED NOTE: 9 spares, none dropped on merit - application-band seeds on lessons whose single blueprint row is recall (SOL-014 x2, SOL-015 x2, SOL-016 x2, SOL-019 pool 16), pool 13's warming-reclassify chains two lessons' content (kept spare to protect the one-lesson-per-item rule), and pool 15 carries the NOTA noted above.
15. BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose/radiological framings - the banned-token scan (uranium/plutonium/radium/actinide/enrich/reactor/weapon/radiation/fission/radioactive/meltdown/dose) over your 45-seed pool at stage 1, all 105 finalized items, and this export body = 0 hits everywhere (instances are household and classroom lab salts and sugars throughout).
16. EXPORTER VERIFICATION DISCLOSURE (new shared-suite scoping class, first occurrence - no action needed): the shared exporter's post-write NOTA/AOTA scan runs over FULL item payloads including the `source` provenance lane, while its own metadata-leak scan two lines later is already scoped to student-facing fields (source/seedRef = intentionally-carried provenance). This gate's two `source` strings for A#10/C#8 honestly QUOTE the replaced 'None of the above' literal in their freshening disclosures, so the scan tripped AFTER the export was written in full. The shared suite was NOT edited (never-weaken + wave isolation): a gate-local _verify_james_export.py re-runs the ENTIRE shared verification suite verbatim with the one scoping correction - the NOTA scan passes over every student-facing field, and the two provenance hits are enumerated and asserted to be exactly the recorded A#10/C#8 disclosure strings. Batched for the shared maintainer alongside the text-row scoping gap.
A student claims that warming water raises how much of ANY substance it can dissolve. Which substance type proves the claim wrong, and what is the correct direction?
Gases — their solubility falls as the water warms. KEY
Typical solids — their solubility falls as the water warms.
Nothing proves the claim wrong — it holds for every substance.
Gases — their solubility stays exactly constant as the water warms.
The solubility of NaCl at 20 °C is 36 g per 100 g of water. Beaker A holds 36 g of NaCl dissolved in 100 g of water at 20 °C; beaker B holds 12 g dissolved in the same conditions. Classify the two solutions.
Beaker A is saturated and beaker B is unsaturated. KEY
Beaker A is unsaturated and beaker B is saturated.
[
{
"value": 110,
"misconception": "you read the KNO₃ curve instead of the NaCl curve",
"feedback": "This answer is wrong — you read the KNO₃ curve instead of the NaCl curve."
},
{
"value": 24,
"misconception": "you read the KClO₃ curve instead of the NaCl curve",
"feedback": "This answer is wrong — you read the KClO₃ curve instead of the NaCl curve."
}
]
feedback
{
"keyExplanation": "Pick the labeled NaCl curve first, then read it: find 60 °C on the bottom axis, go straight up to the NaCl curve, then straight across — the side axis reads 37. The solubility of NaCl at 60 °C is 37 g per 100 g of water. The steep curve belongs to KNO₃ and the low curve to KClO₃; reading either gives a different substance's solubility."
}
SOL-022 · medium · application · seq 8 · seed:U10-G2/pool[25] (= SOL-022 gateItems[1]) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · mcq4
The graph shows the KCl solubility curve with four plotted points, W, X, Y, and Z. Which point represents a saturated KCl solution at 60 °C?
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}] · markedPoints: [{"label": "W", "x": 60, "y": 46, "guides": "dashed lines to both axes"}, {"label": "X", "x": 60, "y": 30, "guides": "dashed lines to both axes"}, {"label": … · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 70, gridlines every 10" · style: "clean gridded solubility graph, curve and points labeled in-diagram"raw spec
{
"figureSpec": {
"figureType": "solubility-curve",
"series": [
{
"label": "KCl",
"points": [
[
10,
31
],
[
20,
34
],
[
30,
37
],
[
40,
40
],
[
50,
43
],
[
60,
46
],
[
70,
48
],
[
80,
51
]
]
}
],
"markedPoints": [
{
"label": "W",
"x": 60,
"y": 46,
"guides": "dashed lines to both axes"
},
{
"label": "X",
"x": 60,
"y": 30,
"guides": "dashed lines to both axes"
},
{
"label": "Y",
"x": 60,
"y": 60,
"guides": "dashed lines to both axes"
},
{
"label": "Z",
"x": 46,
"y": 60,
"guides": "dashed lines to both axes"
}
],
"xAxis": "temperature (°C), 0 to 90, gridlines every 10",
"yAxis": "solubility (g per 100 g of water), 0 to 70, gridlines every 10",
"style": "clean gridded solubility graph, curve and points labeled in-diagram"
}
}
Point W KEY
Point X
Point Y
Point Z
SOL-023 · medium · application · seq 9 · seed:U10-G2/pool[28] (= SOL-023 gateItems[1]) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · mcq4
A factory dissolves its raw material at 70 °C and wants, per 100 g of water, the substance that dissolves most at that temperature. The graph shows the curves of KBr, NH₄Cl, and KClO₃. Which substance should the factory choose?
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}, {"label": "NH₄Cl", "points": [[10, 33], [20, 3… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 100, gridlines every 10" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
The choice needs the mass of water to be known first.
SOL-020 · medium · application · seq 10 · seed:U10-G2/pool[19] (= SOL-020 gateItems[1], also demo_curated) — NOTA distractor REPLACED (ANSWERS #2): 'None of the above' -> 'either saturated or supersaturated' keyed to the growth-trigger misconception; recorded for DAVID_REVIEW; rest verbatim · mcq4
A clear solution has a few crystals of its solute resting on the bottom, unchanged for an hour at constant temperature. What is the liquid?
Saturated KEY
Unsaturated
Supersaturated
Either saturated or supersaturated — resting crystals fit both states.
SOL-024 · medium · application · seq 11 · seed:U10-G2/pool[30] (= SOL-024 gateItems[0], also demo_curated) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · numeric
A saturated KNO₃ solution in 100 g of water cools from 70 °C to 40 °C. Using the KNO₃ curve on the graph, find the mass of KNO₃ that crystallizes out, in grams.
FIGURE (solubility-curve · ) — series: [{"label": "KNO₃", "points": [[10, 21], [20, 32], [30, 46], [40, 64], [50, 85], [60, 110], [70, 138], [80, 169]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
[
{
"value": 202,
"misconception": "you added the two curve readings instead of subtracting them",
"feedback": "This answer is wrong — you added the two curve readings instead of subtracting them."
},
{
"value": 30,
"misconception": "you subtracted the temperatures instead of the solubilities",
"feedback": "This answer is wrong — you subtracted the temperatures instead of the solubilities."
},
{
"value": 64,
"misconception": "you gave the cap at the cooler temperature instead of the difference between the caps",
"feedback": "This answer is wrong — you gave the cap at the cooler temperature instead of the difference between the caps."
}
]
feedback
{
"keyExplanation": "A saturated solution holds exactly the cap at its temperature.\nRead the two caps from the curve:\nsolubility at 70 °C = 138 g per 100 g of water\nsolubility at 40 °C = 64 g per 100 g of water\nWrite down the equation:\nmass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature\nSubstitute in the values, and calculate:\nmass that crystallizes = 138 − 64\nmass that crystallizes = 74 g of KNO₃"
}
SOL-025 · medium · application · seq 12 · seed:U10-G2/pool[33] (= SOL-025 gateItems[0]) — FRESHENED (SEED RULING): '200 mL' -> '250 mL' (200 collides with SOL-014's 200 g sugar corpus value) and 'potassium chloride' -> 'KCl' (the '10 g of potassium' 4-gram collides with scope spec text); same substance, explanation already says KCl; full-field sweep for leftovers (QB 3.9) · mcq4
Two beakers each hold 250 mL of water at 20 °C and 10 g of KCl crystals. A magnetic stirrer spins in beaker 1; beaker 2 sits still. Which statement is correct?
The crystals in beaker 1 finish dissolving sooner. KEY
The crystals in beaker 2 finish dissolving sooner, because undisturbed water soaks into crystals better.
The crystals finish at the same time, because dissolving speed depends only on temperature.
Identical 5 g samples of table salt sit unstirred in two glasses of water, one at 5 °C and one at 55 °C. Which sample finishes dissolving first, and why?
The 55 °C sample, because its water particles strike the salt surface more often and harder KEY
The 5 °C sample, because cold water is denser and presses on the salt harder
The 55 °C sample, because heat melts the salt into liquid first
Both finish together, because without stirring the water's temperature makes no difference to dissolving.
Road crews dissolve calcium chloride into tank water to make de-icing brine. A crew member says flaked calcium chloride makes brine faster than pellets 'because flakes are lighter than pellets.' What is the real reason the flakes dissolve faster?
The thin flakes expose more surface to the water than compact pellets of the same mass. KEY
The crew member is right — the lighter a piece is, the faster it dissolves, whatever its shape.
Flakes float, and floating solutes dissolve faster than sunken ones.
Flakes are a more soluble form of calcium chloride than pellets.
[
{
"value": 43,
"misconception": "you read the KCl curve instead of the KBr curve",
"feedback": "This answer is wrong — you read the KCl curve instead of the KBr curve."
},
{
"value": 76,
"misconception": "you read the KBr curve at 40 °C instead of 50 °C",
"feedback": "This answer is wrong — you read the KBr curve at 40 °C instead of 50 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled KBr curve — the upper one — then read it: find 50 °C on the bottom axis, go straight up to the KBr curve, then straight across. The side axis reads 81, so the solubility of KBr at 50 °C is 81 g per 100 g of water."
}
SOL-020 · hard · transfer · seq 17 · seed:U10-G2/pool[18] (= SOL-020 gateItems[0], also demo_curated) — verbatim (field mapping only) · mcq4
Solutions A and B contain the same solute at the same temperature. An identical small crystal of the solute is dropped into each. In A the crystal dissolves; in B it sets off a rush of new crystals. Identify the two solutions.
A was unsaturated and B was supersaturated. KEY
A was supersaturated and B was unsaturated.
A was saturated and B was supersaturated.
A was unsaturated and B was saturated.
SOL-023 · hard · transfer · seq 18 · seed:U10-G2/pool[27] (= SOL-023 gateItems[0]) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · mcq4
The graph shows the solubility curves of KCl and KNO₃, which cross between 20 °C and 30 °C. Which statement does the graph support?
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}, {"label": "KNO₃", "points": [[10, 21], [20, 32… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram, crossing visible between 20 and 30 °C"raw spec
KCl is more soluble at 10 °C, and KNO₃ is more soluble at 40 °C. KEY
KNO₃ is more soluble at every temperature shown.
KCl is more soluble at every temperature shown.
The two substances have equal solubility at every temperature.
SOL-024 · hard · transfer · seq 19 · seed:U10-G2/pool[32] (= SOL-024 gateItems[2], also demo_curated) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · mcq4
A saturated NH₄Cl solution in 100 g of water cools from 80 °C to 30 °C. The NH₄Cl curve on the graph gives the caps at both temperatures. A student reports that 107 g of NH₄Cl crystallizes out. What went wrong?
FIGURE (solubility-curve · ) — series: [{"label": "NH₄Cl", "points": [[10, 33], [20, 37], [30, 41], [40, 46], [50, 50], [60, 55], [70, 60], [80, 66]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 80, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
The student added the two curve readings instead of subtracting them. KEY
The student subtracted the temperatures instead of the solubilities.
The student read the caps at the wrong temperatures.
Nothing went wrong — 107 g is correct.
SOL-028 · hard · transfer · seq 20 · seed:U10-G2/pool[42] (= SOL-028 gateItems[0], also demo_curated) — verbatim (field mapping only) · mcq4
A saturated sugar solution sits above a layer of undissolved sugar. The student stirs hard for five minutes, but the layer will not shrink. Which change would let more of the sugar dissolve?
Raising the water's temperature KEY
Stirring even faster
Crushing the undissolved sugar into powder
Shaking the container instead of stirring it
SOL-024 · medium · application · seq 21 · authored (blueprint row 21 form A — cooling FRQ, KClO₃ 80→40 (38−14=24); rubric 30/40/30; R15 shape: principle line → labeled readings → worded equation → substitute → calculate; no symbol algebra anywhere — the taught equation is worded) · frq
A saturated KClO₃ solution in 100 g of water cools from 80 °C to 40 °C. Use the KClO₃ curve on the graph.
FIGURE (solubility-curve · ) — series: [{"label": "KClO₃", "points": [[10, 5], [20, 7], [30, 10], [40, 14], [50, 19], [60, 24], [70, 30], [80, 38]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 50, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
A saturated KClO₃ solution in 100 g of water cools from 80 °C to 40 °C. Use the KClO₃ curve on the graph.
(a) Read the solubility of KClO₃ at each of the two temperatures, in g per 100 g of water.
(b) Find the mass of KClO₃ that crystallizes out of the solution, showing the worded equation, the substituted values, and the result with its unit.
(c) In one or two sentences, explain why the solute leaves the solution when it cools, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Read the solubility of KClO₃ at each of the two temperatures, in g per 100 g of water.
Model: A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve:
solubility at 80 °C = 38 g per 100 g of water
solubility at 40 °C = 14 g per 100 g of water
Part b. Find the mass of KClO₃ that crystallizes out of the solution, showing the worded equation, the substituted values, and the result with its unit.
Model: The worded equation:
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
mass that crystallizes = 38 − 14
mass that crystallizes = 24 g of KClO3
Part c. In one or two sentences, explain why the solute leaves the solution when it cools, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: When the solution cools, the cap falls with the temperature. The solute above the new, lower cap cannot stay dissolved, so 24 g of KClO3 crystallizes out of the solution.
Full model answer. (a) A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve:
solubility at 80 °C = 38 g per 100 g of water
solubility at 40 °C = 14 g per 100 g of water
(b) The worded equation:
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
mass that crystallizes = 38 − 14
mass that crystallizes = 24 g of KClO3
(c) When the solution cools, the cap falls with the temperature. The solute above the new, lower cap cannot stay dissolved, so 24 g of KClO3 crystallizes out of the solution.
Form B
SOL-014 · easy · recall · seq 1 · authored (blueprint row 1 case B — report-meaning read; instance KClO₃ 10 g at 30 °C from the unit's canonical curve data, value ≤12 and temperature on the taught gridline) · mcq4
A label reads: 'The solubility of KClO₃ at 30 °C is 10 g per 100 g of water.' Which reading of this label is correct?
At 30 °C, 100 g of water can dissolve at most 10 g of KClO₃. KEY
Any amount of water at 30 °C can dissolve up to 10 g of KClO₃.
100 g of water always holds exactly 10 g of KClO₃, whatever the temperature.
KClO₃ takes 10 minutes to dissolve in 100 g of water at 30 °C.
SOL-015 · easy · recall · seq 2 · authored (blueprint row 2 case B — pattern-statement pick; key restates the taught rule pair, distractors keyed to no-exceptions, flipped-direction, and speed-only errors) · mcq4
Which statement correctly describes how temperature affects the solubility of solid solutes in water?
Most solid solutes dissolve more at higher temperature; a few exceptions dissolve less. KEY
Most solid solutes dissolve less at higher temperature; a few exceptions dissolve more.
Each solid keeps one fixed maximum, whatever the water's temperature.
Higher temperature raises how much can dissolve for the exception solids only; typical solids hold the same amount warm or cold.
SOL-016 · easy · recall · seq 3 · authored (blueprint row 2 case B — pattern-statement pick; the no-exceptions distractor is keyed to the taught contrast with the solid pattern) · mcq4
Which statement gives the correct direction for gases dissolved in water?
A gas's solubility decreases as the water gets warmer. KEY
A gas's solubility increases as the water gets warmer.
A gas's solubility stays about the same whatever the water temperature.
Gases split the way solids do: some dissolve more when warm and some dissolve less.
SOL-019 · easy · recall · seq 4 · authored (blueprint row 4 case B — definition pick; key restates the taught definition, distractors are the saturated, unsaturated, and visible-solid confusions) · mcq4
Which description fits a supersaturated solution?
One temporarily holding more dissolved solute than its solubility allows. KEY
One holding exactly as much dissolved solute as its solubility allows.
One that can still dissolve more solute at its temperature.
One with extra solid solute resting on the bottom of its container.
A student adds baking soda to 100 g of water at 20 °C one small spoonful (about 3 g) at a time, stirring each in. The first three spoonfuls dissolve completely; part of the fourth stays solid no matter how long she stirs. When did the solution become saturated?
During the fourth spoonful — the solid that would not dissolve shows the cap was reached. KEY
After the first spoonful — dissolving any amount of solute is what saturates the water.
Never — with enough stirring, the rest of the fourth spoonful would dissolve.
Before any was added — pure water is already saturated.
A fish hauler wants the water in a transport tank to hold as much dissolved oxygen as possible during a long drive. Which choice does the particle-level reasoning support?
Keep the tank water cold, so the dissolved gas particles move more slowly and escape less easily. KEY
Keep the tank water warm, so the gas particles move fast and stir the oxygen deeper into the water.
Keep the water at any temperature, because how much oxygen stays dissolved depends only on how much was added.
Keep the tank tightly covered, because a lid stops the particles from speeding up.
SOL-021 · medium · application · seq 7 · seed:U10-G2/pool[23] (= SOL-021 gateItems[2]) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · numeric
The graph shows the solubility curves of KNO₃, KBr, and NaCl on one grid. Read the solubility of KNO₃ at 70 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "KNO₃", "points": [[10, 21], [20, 32], [30, 46], [40, 64], [50, 85], [60, 110], [70, 138], [80, 169]]}, {"label": "KBr", "points": [[10, 60], [20,… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 90,
"misconception": "you read the KBr curve instead of the KNO₃ curve",
"feedback": "This answer is wrong — you read the KBr curve instead of the KNO₃ curve."
},
{
"value": 64,
"misconception": "you read the KNO₃ curve at 40 °C instead of 70 °C",
"feedback": "This answer is wrong — you read the KNO₃ curve at 40 °C instead of 70 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled KNO₃ curve — the steeply climbing one — then read it: find 70 °C on the bottom axis, go straight up to the KNO₃ curve, then straight across. The side axis reads 138, so the solubility of KNO₃ at 70 °C is 138 g per 100 g of water."
}
SOL-022 · medium · application · seq 8 · seed:U10-G2/pool[24] (= SOL-022 gateItems[0]) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · mcq4
The graph shows the KNO₃ solubility curve and two plotted KNO₃ solutions: point A at 30 °C holding 85 g per 100 g of water, and point B at 50 °C holding 85 g. Classify the two solutions.
FIGURE (solubility-curve · ) — series: [{"label": "KNO₃", "points": [[10, 21], [20, 32], [30, 46], [40, 64], [50, 85], [60, 110], [70, 138], [80, 169]]}] · markedPoints: [{"label": "A", "x": 30, "y": 85, "guides": "dashed lines to both axes"}, {"label": "B", "x": 50, "y": 85, "guides": "dashed lines to both axes"}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, curve and points labeled in-diagram"raw spec
{
"figureSpec": {
"figureType": "solubility-curve",
"series": [
{
"label": "KNO₃",
"points": [
[
10,
21
],
[
20,
32
],
[
30,
46
],
[
40,
64
],
[
50,
85
],
[
60,
110
],
[
70,
138
],
[
80,
169
]
]
}
],
"markedPoints": [
{
"label": "A",
"x": 30,
"y": 85,
"guides": "dashed lines to both axes"
},
{
"label": "B",
"x": 50,
"y": 85,
"guides": "dashed lines to both axes"
}
],
"xAxis": "temperature (°C), 0 to 90, gridlines every 10",
"yAxis": "solubility (g per 100 g of water), 0 to 180, gridlines every 20",
"style": "clean gridded solubility graph, curve and points labeled in-diagram"
}
}
A is supersaturated and B is saturated. KEY
A is saturated and B is supersaturated.
A is unsaturated and B is saturated.
Both points are saturated, because they hold the same 85 g.
SOL-023 · medium · application · seq 9 · authored (blueprint row 9 case B — most-to-least ranking at 20 °C; fresh substance-set-plus-temperature combination; permutational options + needs-water-mass distractor) · mcq4
The graph shows the solubility curves of NaNO₃, KBr, and KClO₃. Which order ranks them from most to least soluble at 20 °C?
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}, {"label": "KBr", "points": [[10, 60], [… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
The graph cannot rank them without the mass of water.
SOL-020 · medium · application · seq 10 · authored (blueprint row 10 case B — dissolving-crystal outcome on a fresh non-curve solute (magnesium sulfate); bare category options plus a misconception-keyed refusal distractor) · mcq4
A small crystal of magnesium sulfate is dropped into a magnesium sulfate solution at constant temperature. Over ten minutes it grows smaller and smaller until it is gone. What was the solution?
Unsaturated KEY
Saturated
Supersaturated
Impossible to tell — one crystal is too small a probe to reveal any state.
SOL-024 · medium · application · seq 11 · seed:U10-G2/pool[31] (= SOL-024 gateItems[1], also demo_curated) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · numeric
A saturated NaNO₃ solution in 100 g of water is warmed from 10 °C to 60 °C. Using the NaNO₃ curve on the graph, find the extra mass of NaNO₃ that can now dissolve, in grams.
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
[
{
"value": 204,
"misconception": "you added the two curve readings instead of subtracting them",
"feedback": "This answer is wrong — you added the two curve readings instead of subtracting them."
},
{
"value": 50,
"misconception": "you subtracted the temperatures instead of the solubilities",
"feedback": "This answer is wrong — you subtracted the temperatures instead of the solubilities."
},
{
"value": 124,
"misconception": "you gave the new cap instead of the extra room the warming opened up",
"feedback": "This answer is wrong — you gave the new cap instead of the extra room the warming opened up."
}
]
feedback
{
"keyExplanation": "A saturated solution holds exactly the cap at its temperature, and warming raises the cap.\nRead the two caps from the curve:\nsolubility at 60 °C = 124 g per 100 g of water\nsolubility at 10 °C = 80 g per 100 g of water\nWrite down the equation:\nextra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature\nSubstitute in the values, and calculate:\nextra mass that can dissolve = 124 − 80\nextra mass that can dissolve = 44 g of NaNO₃"
}
SOL-025 · medium · application · seq 12 · seed:U10-G2/pool[34] (= SOL-025 gateItems[1]) — FRESHENED (SEED RULING): 'aquarium salt' -> 'water-conditioner crystals' ('salt' is an own-lesson corpus material via SOL-025's salt-in-pot WE); stem/key/distractors/explanation swept for the leftover instance (QB 3.9) · mcq4
An aquarium keeper adds the same measure of water-conditioner crystals to two identical tanks of water at the same temperature. Tank A has a circulation pump running; tank B's pump is off. Why do the crystals in tank A dissolve faster?
The moving water keeps bringing fresh water into contact with the crystals' surfaces. KEY
The pump's motor gently warms tank A's water, and that extra warmth is what speeds the dissolving.
The moving water grinds the crystals into smaller pieces.
The crystals can only dissolve in moving water.
SOL-026 · medium · application · seq 13 · seed:U10-G2/pool[37] (= SOL-026 gateItems[1]) — FRESHENED (SEED RULING): 'bath salts' -> 'bath crystals' ('salt' collides within form B against the adjacent seq-12 material set and with SOL-027's bath-salt fluency instance); key and explanation swept for the leftover 'salt' (QB 3.9) · mcq4
Bath crystals dissolve within seconds in a hot bath but take minutes in a cool one. Which statement explains the difference at the particle level?
In hot water the particles move faster, striking the crystals' surfaces more often and harder. KEY
In hot water the solute's particles expand until the crystals crumble apart.
Hot water has fewer of its own particles, leaving more room for solute particles.
Hot water is thinner, so it soaks into the middle of each crystal and dissolves it from the inside out.
A whole alum crystal takes an hour to dissolve in a jar of still water. A student wants the same mass of alum dissolved in a few minutes without touching the water's temperature. Which change does the job, and why?
Grind the crystal to a powder — grinding exposes far more alum surface for the water to contact. KEY
Use a wider jar — spreading the same water thinner makes it dissolve alum faster.
Leave the crystal whole but push it down to the jar's bottom — solutes dissolve fastest at the bottom.
Wait longer — time is the only thing that changes dissolving.
SOL-028 · medium · application · seq 15 · authored (blueprint row 15 case B — shake classify on a fresh instance (drink-mix powder in a capped flask); fixed 4-category option set per R11) · mcq4
A hiker drops drink-mix powder into a capped flask of water and shakes the flask vigorously rather than waiting. What does the shaking change about the dissolving?
Only how fast the solute dissolves KEY
Only how much solute can dissolve
Both the speed and the amount
Neither the speed nor the amount
SOL-021 · medium · application · seq 16 · authored (blueprint row 16 case B — fresh forward read; (NH₄Cl, 40 °C) pair unused in scope stems; substance family differs from B's seq-7 KNO₃ read; canonical fixed curve data) · numeric
The graph shows the solubility curves of NH₄Cl and NaCl on one grid. Read the solubility of NH₄Cl at 40 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "NH₄Cl", "points": [[10, 33], [20, 37], [30, 41], [40, 46], [50, 50], [60, 55], [70, 60], [80, 66]]}, {"label": "NaCl", "points": [[10, 36], [20, … · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 80, gridlines every 10" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 37,
"misconception": "you read the NaCl curve instead of the NH₄Cl curve",
"feedback": "This answer is wrong — you read the NaCl curve instead of the NH₄Cl curve."
},
{
"value": 50,
"misconception": "you read the NH₄Cl curve at 50 °C instead of 40 °C",
"feedback": "This answer is wrong — you read the NH₄Cl curve at 50 °C instead of 40 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled NH₄Cl curve — the climbing one — then read it: find 40 °C on the bottom axis, go straight up to the NH₄Cl curve, then straight across. The side axis reads 46, so the solubility of NH₄Cl at 40 °C is 46 g per 100 g of water."
}
SOL-020 · hard · transfer · seq 17 · seed:U10-G2/pool[20] (= SOL-020 gateItems[2], also demo_curated) — verbatim (field mapping only) · mcq4
A student drops a crystal of table salt into a KNO₃ solution to test the solution's state. Why does this test fail?
The crystal must be the same substance as the dissolved solute — salt cannot report a KNO₃ solution's state. KEY
The crystal was too small — only a crystal large enough to see clearly can run the test on a solution.
The test does not fail — a crystal of any substance reveals the state of any solution it is dropped into.
The test fails only when the solution happens to be supersaturated; for the other two states any crystal works.
SOL-023 · hard · transfer · seq 18 · seed:U10-G2/pool[29] (= SOL-023 gateItems[2]) — verbatim (field mapping only; canonical fixed curve data restored to the figure series) · mcq4
Using the same KCl and KNO₃ graph, at which temperature range do the two substances have equal solubility?
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}, {"label": "KNO₃", "points": [[10, 21], [20, 32… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram, crossing visible between 20 and 30 °C"raw spec
SOL-024 · hard · transfer · seq 19 · authored (blueprint row 19 case B — temperatures-subtracted wrong path (KBr 70→20: reported 50 = 70−20; true 90−65=25); the reported value is gridline-lattice arithmetic by construction) · mcq4
A saturated KBr solution in 100 g of water cools from 70 °C to 20 °C. The KBr curve on the graph gives the caps at both temperatures. A student reports that 50 g of KBr crystallizes out. What went wrong?
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 100, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
A saturated NH₄Cl solution in 100 g of water is warmed from 10 °C to 50 °C. Use the NH₄Cl curve on the graph.
(a) Read the solubility of NH₄Cl at each of the two temperatures, in g per 100 g of water.
(b) Find the extra mass of NH₄Cl that can now dissolve, showing the worded equation, the substituted values, and the result with its unit.
(c) In one or two sentences, explain why warming opens room for more solute, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Read the solubility of NH₄Cl at each of the two temperatures, in g per 100 g of water.
Model: A saturated solution holds exactly the cap at its temperature, and warming raises the cap. Read the two caps from the curve:
solubility at 50 °C = 50 g per 100 g of water
solubility at 10 °C = 33 g per 100 g of water
Part b. Find the extra mass of NH₄Cl that can now dissolve, showing the worded equation, the substituted values, and the result with its unit.
Model: The worded equation:
extra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
extra mass that can dissolve = 50 − 33
extra mass that can dissolve = 17 g of NH4Cl
Part c. In one or two sentences, explain why warming opens room for more solute, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: Warming raises the cap. The solution was holding exactly the old, lower cap, so the rise opens room below the new cap — 17 g more of NH4Cl can now dissolve.
Full model answer. (a) A saturated solution holds exactly the cap at its temperature, and warming raises the cap. Read the two caps from the curve:
solubility at 50 °C = 50 g per 100 g of water
solubility at 10 °C = 33 g per 100 g of water
(b) The worded equation:
extra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
extra mass that can dissolve = 50 − 33
extra mass that can dissolve = 17 g of NH4Cl
(c) Warming raises the cap. The solution was holding exactly the old, lower cap, so the rise opens room below the new cap — 17 g more of NH4Cl can now dissolve.
Form C
SOL-014 · easy · recall · seq 1 · authored (blueprint row 1 case C — can-it-all-dissolve vs a stated cap; washing soda 22 g cap, 16 g added: both values off the canonical curve lattice and outside the scope corpus) · mcq4
Washing soda's solubility at 20 °C is 22 g per 100 g of water. A student stirs 16 g of washing soda into 100 g of water at 20 °C. Can all of it dissolve?
Yes — 16 g is below the 22 g maximum. KEY
No — only part of any added amount ever dissolves.
No — 100 g of water can only dissolve amounts smaller than 10 g.
It cannot be decided without knowing how fast the student stirs.
A data sheet marks solid Z as 'one of the exception solids' for temperature and solubility. What does that label say about solid Z?
Its solubility decreases as temperature increases. KEY
Its solubility increases as temperature increases, like most solids.
Its solubility stays the same at every temperature.
It cannot dissolve in warm water at all.
SOL-016 · easy · recall · seq 3 · authored (blueprint row 3 case C — direction pick on a fresh open-water scenario; number-free, pressure-free per D2) · mcq4
A garden pond warms through a hot, sunny afternoon. What happens to the maximum amount of dissolved oxygen the pond water can hold?
It decreases. KEY
It increases.
It stays the same, because sunlight is not a chemical change.
It increases in the sunlit shallows and decreases everywhere else.
SOL-019 · easy · recall · seq 4 · authored (blueprint row 4 case C — two-step procedure pick; key is the taught procedure, distractors spoil one step each) · mcq4
A student wants to prepare a supersaturated solution of a typical solid solute. Which procedure does the job?
Saturate the solution at a high temperature, then let it cool slowly without disturbing it. KEY
Stir extra solute into a saturated solution while holding the temperature constant.
Saturate the solution at a high temperature, then shake it vigorously as it cools.
Chill the solvent first, then add the solute a little at a time.
SOL-018 · medium · application · seq 5 · authored (blueprint row 5 case C — fresh two-beaker classify; potassium sulfate 11 g cap (real value, ≤12 so the number scan ignores it), amounts 11/4) · mcq4
The solubility of potassium sulfate at 20 °C is 11 g per 100 g of water. Glass 1 holds 11 g of potassium sulfate dissolved in 100 g of water at 20 °C; glass 2 holds 4 g dissolved in the same conditions. Classify the two solutions.
Glass 1 is saturated and glass 2 is unsaturated. KEY
Glass 1 is unsaturated and glass 2 is saturated.
Both glasses are saturated, because both hold dissolved solute.
Neither can be classified without watching solid rest on the bottom.
A hot spring holds far less dissolved oxygen than the cool river beside it. A student says the spring's minerals soak up the oxygen. What is the correct explanation?
In the hot water, the dissolved oxygen particles move faster and escape the liquid more easily, so less stays dissolved. KEY
The student is right — minerals dissolved in the spring water absorb the oxygen and keep it locked out of solution.
The hot water burns the dissolved oxygen away as fuel, using it up faster than the river uses its own.
The oxygen sinks through the hot water and collects at the bottom of the spring, out of the dissolved count.
SOL-021 · medium · application · seq 7 · authored (blueprint row 7 case C — fresh forward read; (NaNO₃, 50 °C) pair unused in the scope corpus; canonical fixed curve data) · numeric
The graph shows the solubility curves of NaNO₃ and KCl on one grid. Read the solubility of NaNO₃ at 50 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}, {"label": "KCl", "points": [[10, 31], [… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 43,
"misconception": "you read the KCl curve instead of the NaNO₃ curve",
"feedback": "This answer is wrong — you read the KCl curve instead of the NaNO₃ curve."
},
{
"value": 104,
"misconception": "you read the NaNO₃ curve at 40 °C instead of 50 °C",
"feedback": "This answer is wrong — you read the NaNO₃ curve at 40 °C instead of 50 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled NaNO₃ curve — the upper one — then read it: find 50 °C on the bottom axis, go straight up to the NaNO₃ curve, then straight across. The side axis reads 114, so the solubility of NaNO₃ at 50 °C is 114 g per 100 g of water."
}
SOL-022 · medium · application · seq 8 · seed:U10-G2/pool[26] (= SOL-022 gateItems[2]) — NOTA distractor REPLACED (ANSWERS #2): 'None of the above' -> mental-plot refusal distractor keyed to the demanded-drawn-point error; recorded for DAVID_REVIEW; rest verbatim · mcq4
100 g of water at 70 °C holds 60 g of NH₄Cl dissolved, with no point drawn for you. Use the NH₄Cl curve on the graph to classify this solution.
FIGURE (solubility-curve · ) — series: [{"label": "NH₄Cl", "points": [[10, 33], [20, 37], [30, 41], [40, 46], [50, 50], [60, 55], [70, 60], [80, 66]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 80, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
SOL-020 · medium · application · seq 10 · authored (blueprint row 10 case C — rapid-growth outcome on a fresh non-curve solute (citric acid); bare category options plus a misconception-keyed chemical-change distractor) · mcq4
One small crystal of citric acid is dropped into a citric acid solution at constant temperature. Within seconds a mass of new crystals grows outward from it. What was the solution?
Supersaturated KEY
Saturated
Unsaturated
The crystal caused a chemical change, so no saturation state applies.
SOL-024 · medium · application · seq 11 · re-instanced 2026-10-13 (A1 forms pass; James ruling 2026-10-08 #19, WE-duplicate clearance): the KClO₃ 20→60 warming case was SOL-024 WE2 word for word -> KCl warmed 30→70 on the gate's canonical KCl grid (48 − 37 = 11); same skill (warming-direction difference of two curve readings), same misconception set (added readings / temperature difference / new cap alone); KCl is fresh at seq 11 across forms (A KNO₃, B NaNO₃, D NH₄Cl, E KBr) and the (KCl, 30/70) pair appears in no lesson item (fluency uses 10→70). Previous source: authored (blueprint row 11 case C — warming difference KClO₃ 20→60). · numeric
A saturated KCl solution in 100 g of water is warmed from 30 °C to 70 °C. Using the KCl curve on the graph, find the extra mass of KCl that can now dissolve, in grams.
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 60, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
[
{
"value": 85,
"misconception": "you added the two curve readings instead of subtracting them",
"feedback": "This answer is wrong — you added the two curve readings instead of subtracting them."
},
{
"value": 40,
"misconception": "you subtracted the temperatures instead of the solubilities",
"feedback": "This answer is wrong — you subtracted the temperatures instead of the solubilities."
},
{
"value": 48,
"misconception": "you gave the new cap instead of the extra room the warming opened up",
"feedback": "This answer is wrong — you gave the new cap instead of the extra room the warming opened up."
}
]
feedback
{
"keyExplanation": "A saturated solution holds exactly the cap at its temperature, and warming raises the cap.\nRead the two caps from the curve:\nsolubility at 70 °C = 48 g per 100 g of water\nsolubility at 30 °C = 37 g per 100 g of water\nWrite down the equation:\nextra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature\nSubstitute in the values, and calculate:\nextra mass that can dissolve = 48 − 37\nextra mass that can dissolve = 11 g of KCl"
}
A student gently stirring washing soda into a bucket of water with a spoon says, 'Stirring speeds this up because the spoon grinds the crystals into powder.' What is the real reason stirring speeds up the dissolving?
Stirring keeps bringing fresh water to the crystals' surfaces, where dissolving happens. KEY
The student is right — stirring works by breaking the crystals into smaller pieces.
Stirring transfers heat from the student's hand into the water.
Stirring pushes the crystals to the bottom, where water dissolves things best.
A student claims, 'Solutes dissolve faster in cold water, because cold water is thicker and grips the solute better.' Which correction fixes the claim?
Solutes dissolve faster in HOT water — its particles move faster and strike the solute surface more often and harder. KEY
The student's direction is right, but the reason is wrong — cold water's particles move faster than hot water's.
Temperature makes no difference to dissolving speed — agitation is the only factor that can change it.
The claim is fully correct as written — thicker water really does grip and dissolve solutes faster.
Equal masses of potassium permanganate go into three identical unstirred beakers of water at the same temperature: as one single crystal, as a few broken chunks, and as a fine powder. Rank the samples from fastest-dissolving to slowest.
Powder, then chunks, then the single crystal KEY
The single crystal, then chunks, then powder
Chunks, then powder, then the single crystal
All three tie, because the masses and temperatures are equal
SOL-028 · medium · application · seq 15 · authored (blueprint row 15 case C — warm-typical-solid classify on a fresh instance (borax in wash water); fixed 4-category option set per R11) · mcq4
A launderer warms the wash water before dissolving a scoop of borax, a typical solid solute. What does the warming change about the dissolving?
Both the speed and the amount KEY
Only how fast the solute dissolves
Only how much solute can dissolve
Neither the speed nor the amount
SOL-021 · medium · application · seq 16 · authored (blueprint row 16 case C — fresh forward read; (KBr, 40 °C) pair appears nowhere in scope stems; substance family differs from C's seq-7 NaNO₃ read; canonical fixed curve data) · numeric
The graph shows the solubility curves of KBr and NaCl on one grid. Read the solubility of KBr at 40 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}, {"label": "NaCl", "points": [[10, 36], [20, 36… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 100, gridlines every 10" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 37,
"misconception": "you read the NaCl curve instead of the KBr curve",
"feedback": "This answer is wrong — you read the NaCl curve instead of the KBr curve."
},
{
"value": 81,
"misconception": "you read the KBr curve at 50 °C instead of 40 °C",
"feedback": "This answer is wrong — you read the KBr curve at 50 °C instead of 40 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled KBr curve — the upper one — then read it: find 40 °C on the bottom axis, go straight up to the KBr curve, then straight across. The side axis reads 76, so the solubility of KBr at 40 °C is 76 g per 100 g of water."
}
SOL-020 · hard · transfer · seq 17 · authored (blueprint row 17 case C — rest + dissolve permutation pair; parallels the seed's two-solution shape on a different outcome pair) · mcq4
Solutions C and D contain the same solute at the same temperature. An identical small crystal of the solute is dropped into each. In C the crystal rests on the bottom unchanged; in D it dissolves away. Identify the two solutions.
C was saturated and D was unsaturated. KEY
C was unsaturated and D was saturated.
C was supersaturated and D was unsaturated.
C was saturated and D was supersaturated.
SOL-023 · hard · transfer · seq 18 · authored (blueprint row 18 case C — more-soluble below the crossing; canonical KCl/KNO₃ crossover is binding content; steepness distractor keyed to the slope-for-height error) · mcq4
The graph shows the KCl and KNO₃ solubility curves, which cross between 20 °C and 30 °C. A plant runs its dissolving step at 10 °C and must pick the substance that is more soluble there. Which pick does the graph support, and why?
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}, {"label": "KNO₃", "points": [[10, 21], [20, 32… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram, crossing visible between 20 and 30 °C"raw spec
KCl — below the crossing, its curve runs above the KNO₃ curve. KEY
KNO₃ — its curve is higher over most of the graph, so it wins everywhere.
Either one — the two are equal at 10 °C because their curves cross on this graph.
KNO₃ — the steeper curve always marks the more soluble substance.
SOL-024 · hard · transfer · seq 19 · authored (blueprint row 19 case C — final-cap wrong path (NH₄Cl 60→10: reported 33 = the cooler cap; true 55−33=22); a final-cap error report IS a curve reading by construction — pre-declared adjudication class) · mcq4
A saturated NH₄Cl solution in 100 g of water cools from 60 °C to 10 °C. The NH₄Cl curve on the graph gives the caps at both temperatures. A student reports that 33 g of NH₄Cl crystallizes out. What went wrong?
FIGURE (solubility-curve · ) — series: [{"label": "NH₄Cl", "points": [[10, 33], [20, 37], [30, 41], [40, 46], [50, 50], [60, 55], [70, 60], [80, 66]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 80, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
The student reported the cooler temperature's cap, not the difference between the caps. KEY
The student added the two curve readings together instead of subtracting the smaller from the larger.
The student subtracted the temperatures instead of the solubilities.
Nothing went wrong — 33 g is correct.
SOL-028 · hard · transfer · seq 20 · authored (blueprint row 20 case C — saturated washing-soda amount trap; parallels the seed's which-change shape with a different agitation distractor set) · mcq4
A pail holds a saturated washing-soda solution with undissolved lumps on the bottom. The launderer stirs hard and even crushes the lumps against the pail wall, but the solid will not shrink. Which change would let more washing soda dissolve?
Warming the water KEY
Stirring for much longer
Crushing the lumps still finer
Pouring the mixture back and forth between two pails
SOL-024 · medium · application · seq 21 · authored (blueprint row 21 form C — cooling FRQ, NaNO₃ 80→50 (148−114=34); rubric 30/40/30; R15 shape; no symbol algebra — the taught equation is worded) · frq
A saturated NaNO₃ solution in 100 g of water cools from 80 °C to 50 °C. Use the NaNO₃ curve on the graph.
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
A saturated NaNO₃ solution in 100 g of water cools from 80 °C to 50 °C. Use the NaNO₃ curve on the graph.
(a) Read the solubility of NaNO₃ at each of the two temperatures, in g per 100 g of water.
(b) Find the mass of NaNO₃ that crystallizes out of the solution, showing the worded equation, the substituted values, and the result with its unit.
(c) In one or two sentences, explain why the solute leaves the solution when it cools, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Read the solubility of NaNO₃ at each of the two temperatures, in g per 100 g of water.
Model: A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve:
solubility at 80 °C = 148 g per 100 g of water
solubility at 50 °C = 114 g per 100 g of water
Part b. Find the mass of NaNO₃ that crystallizes out of the solution, showing the worded equation, the substituted values, and the result with its unit.
Model: The worded equation:
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
mass that crystallizes = 148 − 114
mass that crystallizes = 34 g of NaNO3
Part c. In one or two sentences, explain why the solute leaves the solution when it cools, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: When the solution cools, the cap falls with the temperature. The solute above the new, lower cap cannot stay dissolved, so 34 g of NaNO3 crystallizes out of the solution.
Full model answer. (a) A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve:
solubility at 80 °C = 148 g per 100 g of water
solubility at 50 °C = 114 g per 100 g of water
(b) The worded equation:
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
mass that crystallizes = 148 − 114
mass that crystallizes = 34 g of NaNO3
(c) When the solution cools, the cap falls with the temperature. The solute above the new, lower cap cannot stay dissolved, so 34 g of NaNO3 crystallizes out of the solution.
Form D
SOL-014 · easy · recall · seq 1 · authored (blueprint row 1 case D — definition pick; key is the taught rule line verbatim, distractors keyed to current-contents, speed-for-amount, and solvent-for-solute swaps) · mcq4
Which quantity is a substance's solubility?
The maximum amount of it that can dissolve in a set amount of solvent at a given temperature. KEY
The amount of it that just happens to be dissolved in one particular solution at the moment.
The time it takes for a sample of it to dissolve completely.
The amount of solvent needed to cover the solute completely.
SOL-015 · easy · recall · seq 2 · authored (blueprint row 2 case D — fresh claim-correction run in the cooling direction; key applies the rule pair in reverse) · mcq4
A student says: 'Cooling water always lets it hold more dissolved solid.' Which correction makes the statement accurate?
Cooling lowers most solids' maximum — only the few exception solids can hold more when cool. KEY
No correction is needed — the statement is already accurate as written for solids.
Cooling raises most solids' maximum, while the exception solids keep their usual amounts.
Cooling mostly changes how fast solids dissolve, not how much of them can stay dissolved.
SOL-016 · easy · recall · seq 3 · authored (blueprint row 3 case D — direction pick, fresh open-container scenario; number-free, pressure-free per D2) · mcq4
A shallow birdbath warms in the afternoon sun. Compare the maximum dissolved air its water can hold in the warm afternoon with the cool morning.
Less in the warm afternoon — a gas's maximum falls as the water warms. KEY
More in the warm afternoon — warmth helps everything dissolve.
The same at both times — air is not affected by water temperature.
It depends on how much air the wind stirs into the water.
SOL-019 · easy · recall · seq 4 · authored (blueprint row 4 case D — definition-to-label read; live numbers 118/82 chosen off the unit's canonical curve lattice and outside the scope corpus) · mcq4
A concentrated glucose stock solution holds 118 g of glucose dissolved in 100 g of water, at a temperature where glucose's solubility is 82 g per 100 g of water. Classify the solution.
Supersaturated KEY
Saturated
Unsaturated
Impossible — no solution can hold more than its solubility allows.
SOL-018 · medium · application · seq 5 · authored (blueprint row 5 case D — added-solute-dissolves diagnosis; cream of tartar keeps the instance off the corpus and off the curve set) · mcq4
A student adds a few small crystals of cream of tartar to a cream-of-tartar solution held at constant temperature, and the added crystals vanish completely as she stirs. What was the solution before the addition?
Unsaturated — it still had room to dissolve more at that temperature. KEY
Saturated — dissolving the addition proves the solution was at its cap.
Saturated — any solution that already holds solute is at its cap.
It cannot be decided — adding solute changes the temperature and spoils the test.
SOL-017 · medium · application · seq 6 · authored (blueprint row 6 case D — warm-bay-vs-cold-depth O₂; key carries the taught mechanism, distractors keyed to pressure, chemical-change, and wrong-direction errors) · mcq4
By late summer the sun has warmed a shallow bay, and its water holds less dissolved oxygen than the same water held in early spring, when it was cold. Why is the warm water's maximum lower?
The warm water's dissolved oxygen particles move faster and escape the liquid more easily, so less stays dissolved. KEY
Back when the bay was cold it was also denser, and denser water automatically locks in extra oxygen.
Sunlight bleaches the dissolved oxygen out of the warm water chemically, destroying it.
The oxygen particles settle to the bay's floor as the water warms, so they stop counting as dissolved.
SOL-021 · medium · application · seq 7 · authored (blueprint row 7 case D — fresh forward read; (KCl, 40 °C) pair unused in the scope corpus stems; canonical fixed curve data) · numeric
The graph shows the solubility curves of KCl and KClO₃ on one grid. Read the solubility of KCl at 40 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}, {"label": "KClO₃", "points": [[10, 5], [20, 7]… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 60, gridlines every 10" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 14,
"misconception": "you read the KClO₃ curve instead of the KCl curve",
"feedback": "This answer is wrong — you read the KClO₃ curve instead of the KCl curve."
},
{
"value": 43,
"misconception": "you read the KCl curve at 50 °C instead of 40 °C",
"feedback": "This answer is wrong — you read the KCl curve at 50 °C instead of 40 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled KCl curve — the upper one — then read it: find 40 °C on the bottom axis, go straight up to the KCl curve, then straight across. The side axis reads 40, so the solubility of KCl at 40 °C is 40 g per 100 g of water."
}
SOL-022 · medium · application · seq 8 · re-instanced 2026-10-13 (A1 forms pass; James ruling 2026-10-08 #19): the NaNO₃ above-the-curve point repeated SOL-022 WE2's salt and answer -> KClO₃ point P (40 °C, 20 g) above its curve (reads 14 g); same skill (plot the point, read below/on/above), same option set; KClO₃ is fresh for SOL-022 at seq 8 across forms (A KCl, B KNO₃, C NH₄Cl, E KBr) and the (KClO₃, 40 °C, 20 g) instance appears in no lesson item (fluency-2 uses 50 °C / 19 g, saturated). KClO₃ genuinely supersaturates on slow cooling (course constraint kept). Held mass off the canonical lattice. Previous source: authored (blueprint row 8 case D — plotted point ABOVE the curve; NaNO₃). · mcq4
The graph shows the KClO₃ solubility curve and one plotted KClO₃ solution, point P at 40 °C holding 20 g per 100 g of water. Classify the solution at P.
FIGURE (solubility-curve · ) — series: [{"label": "KClO₃", "points": [[10, 5], [20, 7], [30, 10], [40, 14], [50, 19], [60, 24], [70, 30], [80, 38]]}] · markedPoints: [{"label": "P", "x": 40, "y": 20, "guides": "dashed lines to both axes"}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 50, gridlines every 10" · style: "clean gridded solubility graph, curve and point labeled in-diagram"raw spec
{
"figureSpec": {
"figureType": "solubility-curve",
"series": [
{
"label": "KClO₃",
"points": [
[
10,
5
],
[
20,
7
],
[
30,
10
],
[
40,
14
],
[
50,
19
],
[
60,
24
],
[
70,
30
],
[
80,
38
]
]
}
],
"markedPoints": [
{
"label": "P",
"x": 40,
"y": 20,
"guides": "dashed lines to both axes"
}
],
"xAxis": "temperature (°C), 0 to 90, gridlines every 10",
"yAxis": "solubility (g per 100 g of water), 0 to 50, gridlines every 10",
"style": "clean gridded solubility graph, curve and point labeled in-diagram"
}
}
Supersaturated KEY
Saturated
Unsaturated
It cannot be classified without knowing the mass of water.
SOL-023 · medium · application · seq 9 · authored (blueprint row 9 case D — most-soluble pick at 20 °C; the ends-highest distractor is keyed to the steep-curve/endpoint error, plus a refusal distractor keyed to the temperature caution) · mcq4
The graph shows the solubility curves of KBr, KCl, and KNO₃. Which substance is the most soluble at 20 °C?
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}, {"label": "KCl", "points": [[10, 31], [20, 34]… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
It cannot be decided, because the ranking changes with temperature.
SOL-020 · medium · application · seq 10 · authored (blueprint row 10 case D — resting-crystal outcome on a fresh non-curve solute (borax); bare category options plus a misconception-keyed no-information distractor) · mcq4
A borax crystal dropped into a borax solution rests on the bottom, unchanged after an hour at constant temperature. What was the solution?
Saturated KEY
Unsaturated
Supersaturated
The test says nothing, because the crystal never moved.
SOL-024 · medium · application · seq 11 · authored (blueprint row 11 case D — cooling difference NH₄Cl 70→10 (60−33=27); stem carries substance/saturated/temperatures/100 g of water only per the course figure-stimulus rule) · numeric
A saturated NH₄Cl solution in 100 g of water cools from 70 °C to 10 °C. Using the NH₄Cl curve on the graph, find the mass of NH₄Cl that crystallizes out, in grams.
FIGURE (solubility-curve · ) — series: [{"label": "NH₄Cl", "points": [[10, 33], [20, 37], [30, 41], [40, 46], [50, 50], [60, 55], [70, 60], [80, 66]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 80, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
[
{
"value": 93,
"misconception": "you added the two curve readings instead of subtracting them",
"feedback": "This answer is wrong — you added the two curve readings instead of subtracting them."
},
{
"value": 60,
"misconception": "you reported 60, which matches both the temperature difference (70 − 10) and the cap at 70 °C itself — either way, the answer is the difference between the two caps: 60 − 33 = 27 g",
"feedback": "This answer is wrong — you reported 60, which matches both the temperature difference (70 − 10) and the cap at 70 °C itself — either way, the answer is the difference between the two caps: 60 − 33 = 27 g."
},
{
"value": 33,
"misconception": "you gave the cap at the cooler temperature instead of the difference between the caps",
"feedback": "This answer is wrong — you gave the cap at the cooler temperature instead of the difference between the caps."
}
]
feedback
{
"keyExplanation": "A saturated solution holds exactly the cap at its temperature.\nRead the two caps from the curve:\nsolubility at 70 °C = 60 g per 100 g of water\nsolubility at 10 °C = 33 g per 100 g of water\nWrite down the equation:\nmass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature\nSubstitute in the values, and calculate:\nmass that crystallizes = 60 − 33\nmass that crystallizes = 27 g of NH₄Cl"
}
SOL-025 · medium · application · seq 12 · authored (blueprint row 12 case D — stirred dye vat; key carries the taught fresh-solvent mechanism, distractors keyed to warming, crushing, and motion-required errors) · mcq4
A dyer adds the same measure of dye powder to two identical vats of water at the same temperature, stirring one steadily and leaving the other still. Why does the stirred vat's dye dissolve faster?
Stirring keeps bringing fresh water to the dye powder's surfaces, where dissolving happens. KEY
Stirring warms the vat's water a little as it spins, and the warming does the speeding up.
The stirring paddle mashes the dye grains into a finer powder.
Dye spreads evenly only while the water spins, so the still vat's dye never dissolves.
SOL-026 · medium · application · seq 13 · authored (blueprint row 13 case D — fresh unstirred hot-vs-cold why; key carries the taught strike clause, distractors keyed to soaking, crushing, and agitation-required errors) · mcq4
A canner dissolves identical scoops of preserving powder in two unstirred kettles of water, one hot and one cold. The hot kettle's scoop vanishes far sooner. What are the hot water's particles doing differently?
Moving faster, so they strike the powder's surfaces more often and harder. KEY
Sticking to the powder longer, giving each grain more time to soften.
Splitting the powder grains apart the way a grinder would.
Nothing different — hot water dissolves faster only when it is also stirred.
SOL-027 · medium · application · seq 14 · authored (blueprint row 14 case D — fresh crushed-vs-whole why; key carries the taught surface-exposure mechanism, distractors keyed to heating, lightness, and inside-out errors) · mcq4
A water-treatment operator crushes alkalinity chips before adding them to a holding tank, instead of dropping them in whole. Why do the crushed chips dissolve faster than whole chips of the same mass?
Crushing exposes surfaces hidden inside the chips, giving water more places to contact at once. KEY
Crushing warms the chips, and the warmth speeds the dissolving.
Crushed pieces each weigh less individually, and lighter pieces always dissolve faster than whole ones.
Crushing lets the chips hold more water inside themselves as they dissolve.
SOL-028 · medium · application · seq 15 · authored (blueprint row 15 case D — grind classify on a fresh instance (citric acid at a drink stand); fixed 4-category option set per R11) · mcq4
A drink-stand worker grinds coarse citric acid crystals to a fine powder before dissolving them in the mixing water. What does the grinding change about the dissolving?
Only how fast the solute dissolves KEY
Only how much solute can dissolve
Both the speed and the amount
Neither the speed nor the amount
SOL-021 · medium · application · seq 16 · authored (blueprint row 16 case D — fresh forward read; (KNO₃, 80 °C) pair unused anywhere in the scope; substance family differs from D's seq-7 KCl read; canonical fixed curve data) · numeric
The graph shows the solubility curves of KNO₃ and NaCl on one grid. Read the solubility of KNO₃ at 80 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "KNO₃", "points": [[10, 21], [20, 32], [30, 46], [40, 64], [50, 85], [60, 110], [70, 138], [80, 169]]}, {"label": "NaCl", "points": [[10, 36], [20… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 38,
"misconception": "you read the NaCl curve instead of the KNO₃ curve",
"feedback": "This answer is wrong — you read the NaCl curve instead of the KNO₃ curve."
},
{
"value": 138,
"misconception": "you read the KNO₃ curve at 70 °C instead of 80 °C",
"feedback": "This answer is wrong — you read the KNO₃ curve at 70 °C instead of 80 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled KNO₃ curve — the steeply climbing one — then read it: find 80 °C on the bottom axis, go straight up to the KNO₃ curve, then straight across. The side axis reads 169, so the solubility of KNO₃ at 80 °C is 169 g per 100 g of water."
}
SOL-020 · hard · transfer · seq 17 · authored (blueprint row 17 case D — fresh wrong-substance procedure-integrity case (alum probe in washing-soda solution); parallels the seed's same-substance rule) · mcq4
To test a solution of washing soda for saturation, a student drops in a small crystal of potassium sulfate and watches it rest unchanged on the bottom. Why does this reading prove nothing about the tested solution?
The crystal must match the dissolved solute — potassium sulfate cannot report washing soda's state. KEY
The reading fails only because the student used one crystal instead of several.
Nothing is wrong — the resting crystal shows this washing-soda solution is saturated.
The potassium sulfate crystal rested only because it was dropped too gently to start dissolving.
SOL-023 · hard · transfer · seq 18 · authored (blueprint row 18 case D — why rankings must name a temperature; key states the crossing consequence, distractors keyed to axis-units, gridline-only, and right-edge errors) · mcq4
Two substances' curves — KCl's and KNO₃'s — cross on the graph between 20 °C and 30 °C. Why must any solubility ranking of these two substances always name a temperature?
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}, {"label": "KNO₃", "points": [[10, 21], [20, 32… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram, crossing visible between 20 and 30 °C"raw spec
The curves cross, so the order flips from one side of the crossing to the other — a ranking holds only at its named temperature. KEY
Because the y-axis units change with temperature, and the units set the order.
Because curve readings only exist at the printed gridline temperatures.
No temperature is actually needed — the higher curve on the right side of the graph is always the more soluble substance.
SOL-024 · hard · transfer · seq 19 · authored (blueprint row 19 case D — direction-backwards-on-warming wrong path (KClO₃ 30→80: correct difference 28 g misread as crystallizing); reported value 28 is off the canonical lattice and outside the scope corpus; distractors kept number-free) · mcq4
A saturated KClO₃ solution in 100 g of water is warmed from 30 °C to 80 °C. The KClO₃ curve on the graph gives the caps at both temperatures. A student reports that 28 g of KClO₃ crystallizes out of the warming solution. What went wrong?
FIGURE (solubility-curve · ) — series: [{"label": "KClO₃", "points": [[10, 5], [20, 7], [30, 10], [40, 14], [50, 19], [60, 24], [70, 30], [80, 38]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 50, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
The student ran the routine backwards — on warming, the 28 g difference is room to dissolve more, not crystals coming out. KEY
The student misread the curve — the caps were taken at the wrong temperatures.
The student subtracted the caps when the temperatures should have been subtracted.
Nothing went wrong — a temperature change in either direction always pushes some solute out of a saturated solution.
SOL-028 · hard · transfer · seq 20 · authored (blueprint row 20 case D — crushing-raised-the-cap claim fix on a fresh instance (borax); distractors keyed to full, partial, and inverted versions of the amount error) · mcq4
A student crushes the leftover solid in a saturated borax mixture and says, 'Now the water's limit is higher, because powder fits into water better than lumps do.' Which correction fixes the claim?
Crushing changes only how fast solute dissolves — this water's limit moves only when its temperature changes. KEY
The claim is right — powdered solute genuinely raises the amount the water can hold.
The claim is wrong only about the size of the rise — powder raises the limit, just by less than the student thinks.
The correction is that crushing LOWERS the limit, because powder packs the water full of solid bits.
SOL-024 · medium · application · seq 21 · authored (blueprint row 21 form D — warming FRQ, KBr 20→60 (86−65=21); rubric 30/40/30; R15 shape; no symbol algebra — the taught equation is worded) · frq
A saturated KBr solution in 100 g of water is warmed from 20 °C to 60 °C. Use the KBr curve on the graph.
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 100, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
A saturated KBr solution in 100 g of water is warmed from 20 °C to 60 °C. Use the KBr curve on the graph.
(a) Read the solubility of KBr at each of the two temperatures, in g per 100 g of water.
(b) Find the extra mass of KBr that can now dissolve, showing the worded equation, the substituted values, and the result with its unit.
(c) In one or two sentences, explain why warming opens room for more solute, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Read the solubility of KBr at each of the two temperatures, in g per 100 g of water.
Model: A saturated solution holds exactly the cap at its temperature, and warming raises the cap. Read the two caps from the curve:
solubility at 60 °C = 86 g per 100 g of water
solubility at 20 °C = 65 g per 100 g of water
Part b. Find the extra mass of KBr that can now dissolve, showing the worded equation, the substituted values, and the result with its unit.
Model: The worded equation:
extra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
extra mass that can dissolve = 86 − 65
extra mass that can dissolve = 21 g of KBr
Part c. In one or two sentences, explain why warming opens room for more solute, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: Warming raises the cap. The solution was holding exactly the old, lower cap, so the rise opens room below the new cap — 21 g more of KBr can now dissolve.
Full model answer. (a) A saturated solution holds exactly the cap at its temperature, and warming raises the cap. Read the two caps from the curve:
solubility at 60 °C = 86 g per 100 g of water
solubility at 20 °C = 65 g per 100 g of water
(b) The worded equation:
extra mass that can dissolve = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
extra mass that can dissolve = 86 − 65
extra mass that can dissolve = 21 g of KBr
(c) Warming raises the cap. The solution was holding exactly the old, lower cap, so the rise opens room below the new cap — 21 g more of KBr can now dissolve.
Form E
SOL-014 · easy · recall · seq 1 · authored (blueprint row 1 case E — cap-enforcement read; alum ~6 g per 100 g at 20 °C (real value), all live numbers ≤12) · mcq4
The solubility of KClO₃ at 20 °C is 7 g per 100 g of water. A student stirs 9 g of KClO₃ into 100 g of water at 20 °C for a long time. What is the most KClO₃ that dissolves?
7 g KEY
9 g
2 g
0 g — KClO₃ cannot dissolve once the added amount is over the maximum
SOL-015 · easy · recall · seq 2 · authored (blueprint row 2 case E — fresh claim-correction on the same-direction overgeneralization; key preserves both taught directions) · mcq4
A lab notebook claims: 'Temperature moves every solid's solubility in the same direction.' Which correction makes the claim accurate?
Most solids' solubility rises with temperature and a few exceptions fall — no single direction fits all. KEY
The claim is accurate — all solids respond to temperature identically.
The direction is the same for all solids, but only above room temperature.
Temperature never changes any solid's solubility, so there is no direction for the claim to compare.
SOL-016 · easy · recall · seq 3 · authored (blueprint row 3 case E — direction pick along a warming gradient; number-free, pressure-free per D2) · mcq4
Spring water emerges cold from the ground, then flows down a shallow, sun-warmed channel. What happens to the maximum dissolved oxygen the water can hold as it warms along the way?
It falls steadily as the water warms. KEY
It rises steadily as the water warms.
It stays fixed at whatever the cold source water could hold.
It falls only if fish are present to breathe the oxygen.
SOL-019 · easy · recall · seq 4 · authored (blueprint row 4 case E — supersaturated-vs-unsaturated compare; parallels the seed's P-vs-Q shape on the other state pair) · mcq4
At one temperature, solution R is supersaturated and solution S is unsaturated, both of the same solute. How do the two compare against their shared solubility cap?
R temporarily holds more than the cap; S holds less and could dissolve more. KEY
R holds less than their shared cap, and S is the one holding more than the cap allows.
Both hold exactly the cap, but R's solute is warmer.
R and S cannot share a solubility cap if their states differ.
SOL-018 · medium · application · seq 5 · authored (blueprint row 5 case E — fresh two-jar classify stated in words (full cap vs half the cap), number-free by design) · mcq4
Jar 1 and jar 2 hold equal amounts of the same solvent at 30 °C. Jar 1 holds exactly as much dissolved solid Q as its solubility allows; jar 2 holds half that amount. Classify the two solutions.
Jar 1 is saturated and jar 2 is unsaturated. KEY
Jar 1 is unsaturated and jar 2 is saturated.
Both jars are saturated, because both hold dissolved solid Q.
Both jars are unsaturated, because neither shows solid on the bottom.
SOL-017 · medium · application · seq 6 · authored (blueprint row 6 case E — sun-warmed rain barrel; key carries the taught mechanism, distractors keyed to particle-resizing, chemical-change, and grip-flip errors) · mcq4
A rain barrel stands in full sun all afternoon and its water turns warm. It now keeps less dissolved air than it did in the cool of the morning. Why?
The dissolved gas particles move faster in the warm water and escape the liquid more easily, so fewer stay dissolved. KEY
The warm water's own particles swell up in the heat and squeeze the dissolved air out of the spaces between them.
The afternoon sun converts the dissolved air into water, removing it from solution.
Warm water grips gas particles more tightly, so the air that remains can never leave.
SOL-021 · medium · application · seq 7 · authored (blueprint row 7 case E — fresh forward read; (NH₄Cl, 50 °C) pair unused in the scope corpus; canonical fixed curve data) · numeric
The graph shows the solubility curves of NH₄Cl, NaCl, and KClO₃ on one grid. Read the solubility of NH₄Cl at 50 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "NH₄Cl", "points": [[10, 33], [20, 37], [30, 41], [40, 46], [50, 50], [60, 55], [70, 60], [80, 66]]}, {"label": "NaCl", "points": [[10, 36], [20, … · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 80, gridlines every 10" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 37,
"misconception": "you read the NaCl curve instead of the NH₄Cl curve",
"feedback": "This answer is wrong — you read the NaCl curve instead of the NH₄Cl curve."
},
{
"value": 55,
"misconception": "you read the NH₄Cl curve at 60 °C instead of 50 °C",
"feedback": "This answer is wrong — you read the NH₄Cl curve at 60 °C instead of 50 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled NH₄Cl curve — the middle one — then read it: find 50 °C on the bottom axis, go straight up to the NH₄Cl curve, then straight across. The side axis reads 50, so the solubility of NH₄Cl at 50 °C is 50 g per 100 g of water."
}
SOL-022 · medium · application · seq 8 · authored (blueprint row 8 case E — plotted point BELOW the curve; held mass 62 g off the canonical lattice) · mcq4
The graph shows the KBr solubility curve and one plotted KBr solution, point P at 60 °C holding 62 g per 100 g of water. Classify the solution at P.
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}] · markedPoints: [{"label": "P", "x": 60, "y": 62, "guides": "dashed lines to both axes"}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 100, gridlines every 10" · style: "clean gridded solubility graph, curve and point labeled in-diagram"raw spec
{
"figureSpec": {
"figureType": "solubility-curve",
"series": [
{
"label": "KBr",
"points": [
[
10,
60
],
[
20,
65
],
[
30,
70
],
[
40,
76
],
[
50,
81
],
[
60,
86
],
[
70,
90
],
[
80,
95
]
]
}
],
"markedPoints": [
{
"label": "P",
"x": 60,
"y": 62,
"guides": "dashed lines to both axes"
}
],
"xAxis": "temperature (°C), 0 to 90, gridlines every 10",
"yAxis": "solubility (g per 100 g of water), 0 to 100, gridlines every 10",
"style": "clean gridded solubility graph, curve and point labeled in-diagram"
}
}
Unsaturated KEY
Saturated
Supersaturated
It cannot be classified, because 62 g is not a value on the curve.
SOL-023 · medium · application · seq 9 · authored (blueprint row 9 case E — most-to-least ranking at 70 °C; fresh substance-set-plus-temperature combination; permutational options + frozen-ranking distractor) · mcq4
The graph shows the solubility curves of NaNO₃, NH₄Cl, and KClO₃. Which order ranks them from most to least soluble at 70 °C?
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}, {"label": "NH₄Cl", "points": [[10, 33],… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
SOL-020 · medium · application · seq 10 · authored (blueprint row 10 case E — dissolving-crystal outcome on a fresh non-curve solute (washing soda); bare category options plus a misconception-keyed wrong-substance distractor) · mcq4
A small washing-soda crystal dropped into a washing-soda solution at constant temperature shrinks steadily and disappears. What was the solution?
Unsaturated KEY
Saturated
Supersaturated
Impossible to tell from a single test crystal.
SOL-024 · medium · application · seq 11 · authored (blueprint row 11 case E — cooling difference KBr 60→10 (86−60=26); stem carries substance/saturated/temperatures/100 g of water only per the course figure-stimulus rule) · numeric
A saturated KBr solution in 100 g of water cools from 60 °C to 10 °C. Using the KBr curve on the graph, find the mass of KBr that crystallizes out, in grams.
FIGURE (solubility-curve · ) — series: [{"label": "KBr", "points": [[10, 60], [20, 65], [30, 70], [40, 76], [50, 81], [60, 86], [70, 90], [80, 95]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 100, gridlines every 10" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
[
{
"value": 146,
"misconception": "you added the two curve readings instead of subtracting them",
"feedback": "This answer is wrong — you added the two curve readings instead of subtracting them."
},
{
"value": 50,
"misconception": "you subtracted the temperatures instead of the solubilities",
"feedback": "This answer is wrong — you subtracted the temperatures instead of the solubilities."
},
{
"value": 60,
"misconception": "you gave the cap at the cooler temperature instead of the difference between the caps",
"feedback": "This answer is wrong — you gave the cap at the cooler temperature instead of the difference between the caps."
}
]
feedback
{
"keyExplanation": "A saturated solution holds exactly the cap at its temperature.\nRead the two caps from the curve:\nsolubility at 60 °C = 86 g per 100 g of water\nsolubility at 10 °C = 60 g per 100 g of water\nWrite down the equation:\nmass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature\nSubstitute in the values, and calculate:\nmass that crystallizes = 86 − 60\nmass that crystallizes = 26 g of KBr"
}
SOL-025 · medium · application · seq 12 · authored (blueprint row 12 case E — paddle-mixed softener pellets; key carries the taught fresh-solvent mechanism, distractors keyed to pressure, crushing, and motion-is-warmth errors) · mcq4
Water-softener pellets dissolve noticeably sooner in a tank where a slow paddle keeps the water moving, without ever touching the pellets, than in an identical still tank at the same temperature. Why do the paddled tank's pellets finish first?
The moving water keeps bringing fresh water into contact with the pellets' surfaces, where dissolving happens. KEY
The paddle's motion presses the pellets harder against the tank floor, squeezing solute out of them.
The paddle blades chip the pellets into smaller and smaller pieces as they turn.
Moving water is warmer than still water, and the warmth speeds the dissolving.
SOL-026 · medium · application · seq 13 · authored (blueprint row 13 case E — fresh hot-vs-cold why on bouillon granules; key carries the taught strike clause, distractors keyed to thinner-water, melting, and airborne-solute errors) · mcq4
Cream of tartar crystals dissolve much faster in near-boiling water than in lukewarm water, both left unstirred. Which statement gives the particle-level reason?
In the hotter water the particles move faster, so they strike the crystals' surfaces more often and harder. KEY
Near-boiling water is emptier, so the crystals' particles find open spaces more easily.
The near-boiling water partly melts each crystal before dissolving it.
Bubbles rising through the near-boiling water carry the crystals' particles up and out of the liquid, emptying them faster.
SOL-027 · medium · application · seq 14 · authored (blueprint row 14 case E — fresh ground-vs-pressed comparison; key carries the taught surface-exposure mechanism, distractors keyed to substance-change, equal-mass, and crowding errors) · mcq4
A pool manager can add pH-adjusting granules either as large pressed tablets or ground fine, equal masses into the same still water. Which prediction fits the ground form?
It dissolves faster, because grinding exposes more of the solute's surface for water to contact. KEY
It dissolves faster, because grinding turns the granules into a chemically more soluble substance.
It dissolves at the same speed, because the masses and the water are identical.
It dissolves more slowly, because small grains crowd each other and block the water.
SOL-028 · medium · application · seq 15 · authored (blueprint row 15 case E — cool-typical-solid classify on a fresh instance (plant-food crystals); fixed 4-category option set per R11; both-change key runs the downward direction) · mcq4
A greenhouse keeper chills the mixing water before dissolving plant-food crystals, a typical solid solute. Compared with unchilled water, what does the chilling change?
Both how fast the crystals dissolve and the maximum amount that can dissolve KEY
Only how fast the crystals dissolve, not the maximum amount
Only the maximum amount that can dissolve, not how fast it dissolves
Neither the speed nor the maximum amount
SOL-021 · medium · application · seq 16 · authored (blueprint row 16 case E — fresh forward read; (NaNO₃, 30 °C) pair unused in scope stems; substance family differs from E's seq-7 NH₄Cl read; canonical fixed curve data) · numeric
The graph shows the solubility curves of NaNO₃ and KCl on one grid. Read the solubility of NaNO₃ at 30 °C, in g per 100 g of water.
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}, {"label": "KCl", "points": [[10, 31], [… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram at its right end"raw spec
[
{
"value": 37,
"misconception": "you read the KCl curve instead of the NaNO₃ curve",
"feedback": "This answer is wrong — you read the KCl curve instead of the NaNO₃ curve."
},
{
"value": 104,
"misconception": "you read the NaNO₃ curve at 40 °C instead of 30 °C",
"feedback": "This answer is wrong — you read the NaNO₃ curve at 40 °C instead of 30 °C."
}
]
feedback
{
"keyExplanation": "Pick the labeled NaNO₃ curve — the upper one — then read it: find 30 °C on the bottom axis, go straight up to the NaNO₃ curve, then straight across. The side axis reads 96, so the solubility of NaNO₃ at 30 °C is 96 g per 100 g of water."
}
SOL-020 · hard · transfer · seq 17 · authored (blueprint row 17 case E — three-flask full permutation; options span the plausible permutations of the three outcomes) · mcq4
Three flasks hold solutions of the same solute at one temperature. An identical small crystal of the solute goes into each: in flask 1 it dissolves; in flask 2 it rests unchanged; in flask 3 it sets off rapid growth of new crystals. Identify the three solutions.
SOL-023 · hard · transfer · seq 18 · authored (blueprint row 18 case E — order flips across two temperatures (20/50 °C, both off the seed's 10/40 pair); canonical KCl/KNO₃ crossover is binding content) · mcq4
A student reads the KCl and KNO₃ curves twice: once at 20 °C and once at 50 °C. The curves cross between 20 °C and 30 °C. What does the pair of readings show?
FIGURE (solubility-curve · ) — series: [{"label": "KCl", "points": [[10, 31], [20, 34], [30, 37], [40, 40], [50, 43], [60, 46], [70, 48], [80, 51]]}, {"label": "KNO₃", "points": [[10, 21], [20, 32… · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, each curve labeled in-diagram, crossing visible between 20 and 30 °C"raw spec
The order flips: KCl sits higher at 20 °C, and KNO₃ sits higher at 50 °C. KEY
The same substance leads at both temperatures, because a curve that starts higher stays higher.
The two substances are equal at both temperatures, since both readings use the same graph.
The readings cannot be compared, because each temperature needs its own separate graph.
SOL-024 · hard · transfer · seq 19 · authored (blueprint row 19 case E — cannot-exceed-what-was-dissolved sanity check (NaNO₃ 70→20: 155 g reported > 134 g ever held; 155 chosen off the corpus lattice); the 134 g ceiling in the key IS a curve reading by construction — pre-declared adjudication class) · mcq4
A saturated NaNO₃ solution in 100 g of water cools from 70 °C to 20 °C. The NaNO₃ curve on the graph gives the caps at both temperatures. A student reports that 155 g of NaNO₃ crystallizes out. Before recalculating anything, which check already proves the report wrong?
FIGURE (solubility-curve · ) — series: [{"label": "NaNO₃", "points": [[10, 80], [20, 88], [30, 96], [40, 104], [50, 114], [60, 124], [70, 134], [80, 148]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 160, gridlines every 20" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
The solution only ever held 134 g dissolved, so no more than 134 g could possibly crystallize out. KEY
No check can fail the report yet — 155 g stays possible if the solution was holding extra solute.
The report must instead equal 70 − 20 = 50 g, so 155 g fails that test.
155 g is impossible only because it does not sit on a printed gridline.
SOL-028 · hard · transfer · seq 20 · authored (blueprint row 20 case E — shaking-a-saturated-solution claim fix on a fresh instance (alum); distractors keyed to cap-beating, lid-blame, and inverted-effect errors) · mcq4
A jar holds a saturated alum solution above a spoonful of undissolved alum. A student shakes the sealed jar hard and says, 'Shaking will make the water take in the rest.' Which correction fixes the claim?
Shaking changes only the speed — the solution already holds its cap; only warming can fit more in. KEY
The claim is right — shaking forces extra solute into solution beyond the solubility.
Shaking would have worked if the jar were open — the sealed lid is what blocks the extra dissolving.
The claim is backwards — shaking makes dissolved alum crystallize out instead.
SOL-024 · medium · application · seq 21 · authored (blueprint row 21 form E — cooling FRQ, KNO₃ 50→30 (85−46=39); rubric 30/40/30; R15 shape; fresh (substance, temperature-pair) vs the article's 60→30 worked example; no symbol algebra — the taught equation is worded) · frq
A saturated KNO₃ solution in 100 g of water cools from 50 °C to 30 °C. Use the KNO₃ curve on the graph.
FIGURE (solubility-curve · ) — series: [{"label": "KNO₃", "points": [[10, 21], [20, 32], [30, 46], [40, 64], [50, 85], [60, 110], [70, 138], [80, 169]]}] · xAxis: "temperature (°C), 0 to 90, gridlines every 10" · yAxis: "solubility (g per 100 g of water), 0 to 180, gridlines every 20" · style: "clean gridded solubility graph, curve labeled in-diagram"raw spec
A saturated KNO₃ solution in 100 g of water cools from 50 °C to 30 °C. Use the KNO₃ curve on the graph.
(a) Read the solubility of KNO₃ at each of the two temperatures, in g per 100 g of water.
(b) Find the mass of KNO₃ that crystallizes out of the solution, showing the worded equation, the substituted values, and the result with its unit.
(c) In one or two sentences, explain why the solute leaves the solution when it cools, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Read the solubility of KNO₃ at each of the two temperatures, in g per 100 g of water.
Model: A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve:
solubility at 50 °C = 85 g per 100 g of water
solubility at 30 °C = 46 g per 100 g of water
Part b. Find the mass of KNO₃ that crystallizes out of the solution, showing the worded equation, the substituted values, and the result with its unit.
Model: The worded equation:
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
mass that crystallizes = 85 − 46
mass that crystallizes = 39 g of KNO3
Part c. In one or two sentences, explain why the solute leaves the solution when it cools, in terms of the cap.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: When the solution cools, the cap falls with the temperature. The solute above the new, lower cap cannot stay dissolved, so 39 g of KNO3 crystallizes out of the solution.
Full model answer. (a) A saturated solution holds exactly the cap at its temperature. Read the two caps from the curve:
solubility at 50 °C = 85 g per 100 g of water
solubility at 30 °C = 46 g per 100 g of water
(b) The worded equation:
mass that crystallizes = solubility at the warmer temperature − solubility at the cooler temperature
Substituting the two readings and calculating:
mass that crystallizes = 85 − 46
mass that crystallizes = 39 g of KNO3
(c) When the solution cools, the cap falls with the temperature. The solute above the new, lower cap cannot stay dissolved, so 39 g of KNO3 crystallizes out of the solution.
End of Topic Test: Solubility rules and precipitates (u10-t3 · 5 forms × 21 items · 105 total)
David's notes for James:
FRQ COMPOSITION - frq_per_form = 1 (the baseline): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - every scope atom is qualitative (SOL-029/030/031 atomKind fact, SOL-032/033 atomKind discrimination; zero atomKind quant, zero numeric seeds, zero formula substitution anywhere in scope). U10's quantitative topics (molarity SOL-034..040, dilution/solution stoichiometry SOL-041..045) belong to U10-G4, where the 2-3-FRQ calc split applies instead. The single FRQ is UNSEEDED (your prod_seeds list is empty and the pool holds zero production items): the SOL-033 capstone chain run on one fresh mixing per form - (a) swap partners and write BOTH product formulas (30), (b) classify each product with the rules, naming the deciding rule or exception (40), (c) the chemist's outcome line with (s)/(aq) labels and what an observer sees (30). Case classes rotate across the five forms covering all three taught outcome classes (insoluble-family solid x2, exception-made solid x2, no-precipitate). NO 5-line scaffold (nothing substitutes into a formula - qualitative chain). D5 scaling evaluated and NOT triggered: min(20, 6 x 5 lessons) = 20 objective rows; your target_size=20 honored. Total 20 objective + 1 FRQ = 21 items/form.
SEED CENSUS AND FRESHENINGS UNDER THE STRICT SEED RULING (no freshness exemption; validator never weakened): all 15 distinct pool seeds are placed on form A (rows 4-15 and 17-19; your three demo_curated items are stem-normalized byte matches of pool[12]/[13]/[14] - deduped, both refs cited), zero spares; forms B-E are authored on the same blueprint rows. 12 seeds ride VERBATIM (field mapping only). 3 carried your stage-1 must_freshen_instance flags, executed with structure and misconception keys preserved and every field swept (QB 3.9) - confirm: (1) A#12 pool[8] washing-soda Na2CO3 -> Cs2CO3 (the seed's key restated SOL-031's OWN rule-line example 'Na2CO3 is a soluble carbonate, because sodium is a group 1 metal'; rescue key + family-flip + grinding-speed + formula-misread flanks kept). (2) A#17 pool[12] (= demo_curated[0]) partner salt NaCl -> NaBr (NaCl is a reactant in SOL-033's OWN article mechanism case AgNO3 + NaCl; KNO3 retained - not the colliding reactant; no-precipitate key + all three flanks kept). (3) A#18 pool[13] (= demo_curated[1]) reactant Pb(NO3)2 -> Ba(NO3)2 (Pb(NO3)2 is SOL-033's OWN WE1 reactant; Na2SO4 retained; typed answer now BaSO4; the soluble-pick / no-swap / criss-cross anticipated-wrong ledger rewritten to the new instance, feedback-leftover sweep clean).
SEED-TEMPLATE WRONG-KEY FIX AT BLIND QC - row 9 C/D forms (confirm; both blind reviewers caught it independently): the stage-2 template clause 'every common bromide compound is soluble' is FALSE under the taught rules (AgBr and PbBr2 are roster exceptions). C#9/D#9 keys now argue roster membership ('the exception is on the sulfate list only - this metal is not a bromide exception, so [salt] dissolves'); A/B/E row-9 keys keep the every-compound clause only where the family is a true guarantee (nitrate/acetate).
QUALIFIER-HINGED DISTRACTOR CLASS (sol authoritative QC, loops 1-3 - bank-wide sweep, confirm the approach): distractors whose wrongness rode ONLY on single/only/all/every around an otherwise-true norm were replaced with independently false claims (exception-roster misplacement errors; invented charge-test / vote-count / temperature / purpose / appearance mechanisms). Row-9 D1/D2 and the row-4 'outranks' distractor were IDENTICAL sentences across forms and were swept as whole rows for form-parallel consistency ('never trust a judge to re-catch').
FEEDBACK FAMILY-PROMOTION FIX (sol loop 1, 11 items + 1 class-sweep): misconception/feedback sentences that promoted bromide/sulfate NORMS to guarantees (contradicting the taught usually-vs-always distinction) were rewritten; two were stage-3 sweep leftovers. Keys were never affected.
ADJUDICATED INSTANCE FAMILIES (confirm; the gate's structural whitelists, enforced by a gate-local wrapper over the UNTOUCHED shared validator; slots enumerated from the scan): (1) AQUEOUS-SOLVENT-WATER (52 slots) - the gate's topic IS solubility in water; every scope rule line states its families' behavior 'in water' and the taught task on every row is a water verdict (U10-G1's identical family is the unit precedent); solutes stay globally unique under the home-seq ledger. (2) SOL030-EXCEPTION-ROSTER (silver/lead/calcium, 12 slots) - YOUR specNotes' own ruling: 'exception rosters ARE the content, so recall items necessarily reuse the rosters'; an item cannot test the exceptions without naming them, and SOL-032's guarantee-beats-exception counterintuition inherits the same names; barium (also roster) is not in the scan lexicon so needs no relief; fresh non-roster metals stay fully scanned. (3) GROUP1-RESCUE-SODIUM (3 slots: A#17/A#18 seed spectator prose + E#12's TSP rescue construct) - sodium is the taught group-1 exemplar of SOL-031's own rule line; it names the always-soluble spectator, never the tested outcome. Adjacency rulings recorded: A#8/A#9 share 'silver' as roster vocabulary (A#8 key = Lead, A#9 key = the nitrate guarantee - neither key leaks the other); A#17/A#18 share the sodium spectator (both are YOUR placed seeds; A#17 key = no precipitate, A#18 key = BaSO4).
GATE_CONFIG TASK-IDIOM WHITELIST (2 recorded entries, confirm) - both are YOUR SOL-033 registers, zero instance signal: 'are mixed. Which precipitate, if any, forms?' (the spec-MANDATED mixing-question register; your specNotes fix the register and your fluency + gate stems all use exactly this clause) and 'are mixed, and a white solid appears. Write the formula of the precipitate.' (your text-entry register, run at the seq-18 production row on every form by design). The runtime idiom-derivation misses both only because the clauses follow formula tokens (sentence-split lands inside the formula); every substance name, formula token, and number in the mixing stems still scans in full.
AgNO3 REUSE IN FRESH ITEMS (B#18, C#19, E#21 - disclosure, chemistry-forced): AgNO3 is the ONLY taught-soluble silver source (silver acetate is BANNED by your SOL-032 extraNotes; every other silver salt in scope is insoluble), so any fresh item needing dissolved silver must start from it. All three pairs differ from the article/WE/fluency/seed pairs.
CROSS-ITEM CUEING ADVISORY CLASS (59 round-1 blind flags - recorded test-level adjudication, no per-item fix): a single-topic RULES test necessarily restates the five taught rules across its explain-rows and apply-rows; the restated rule IS the skill under test. Blind reviewers picked every key without lesson access either way.
TEXT-ROW CASE CONTRACT (seq 18, all forms - schema-gap carrier note, same class as U10-G1's seq-7 note): the typed precipitate formulas are graded CASE-SENSITIVELY on the element symbols (source answer.caseSensitive = true - capitalization is part of the formula; 'BaSO4' vs 'baso4' must not both pass), but the section-6 answer shape has no caseSensitive lane, so the flag is carried here and restated inside each item's exported gradingNote. Unicode subscripts are NEVER required to type: the plain-digit spelling is an acceptedAlternate everywhere (your SOL-033 extraNotes' own convention). On forms B and E the correct formulas (AgI, ZnS) contain no subscripts at all, so the display and plain forms coincide - their alternate lists carry the same string as the display value by the seed convention (redundant but harmless; grading unaffected).
'No precipitate forms' IS NOT NOTA (recorded at stage 1): it is the spec-mandated substantive outcome option of your SOL-033 extraNotes register (one option per outcome; the 'Both X and Y' option covers the two-solids misconception), same for pool[1]'s 'Nothing can be concluded until the metal is identified' claim option. V-T5's literal NOTA/AOTA regex passes on all forms; zero replace_distractor flags were needed across the 15-seed pool.
LESSON TITLES (verification request): unit10_specs_merged.json carries null title fields for SOL-029..032, so the exported lessonTitle values for those four are OUR reconstructions from your rule/extraNotes content ('The always-soluble ions', 'Usually soluble, with exceptions', 'The usually-insoluble families', 'Classify any compound with the rules'); SOL-033's 'Predict the precipitate' is from your gate data's capstones entry. The student-facing test title was verified against the nested manifest (Unit 10 Topic 6 'Solubility rules and precipitates'). Please confirm or supply your canonical lesson titles.
HONEST SURVIVOR MINORS (2, recorded): C#4 dry-storage and C#5 sealed-jar condition distractors judged weak-but-functional by sol (style notes only; both are the taught conditions-override misconception family, kept).
QC PROVENANCE (no action needed): deterministic validator 11/12 shared PASS + V-T7 100% excused under the recorded adjudications (parity-guard wrapper exit 0); blind rounds 1-5 (r1=gpt-5.6-sol blind role, r2=gemini; fresh tags each re-blind) = 258/258 key picks, 0 open hard fails; keyless lanes hand-graded 10/10 match/match for both reviewers (5 seq-18 formulas typed exactly + 5 FRQs reproducing both product formulas, the deciding rules, and the (s)/(aq) verdict cold); sol authoritative loops 22 -> 5 -> 1 -> 0 FAILs, final ledger 105/105 pass.
BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings; classroom-safe instance set throughout (lab bottles, cleaners, pigments, mineral specimens, water-treatment and photography salts; your fireworks/tarnish/jeweler seed framings kept). Swept at packaging over forms_final AND this export body: ZERO hits on every banned token including 'dose'.
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u10-t3 — notes/queries from David's side (U10-G3)
1. FRQ COMPOSITION - frq_per_form = 1 (the baseline): the wave-6 calc-heavy FRQ rule was EVALUATED AND NOT TRIGGERED after an honest predicate check - every scope atom is qualitative (SOL-029/030/031 atomKind fact, SOL-032/033 atomKind discrimination; zero atomKind quant, zero numeric seeds, zero formula substitution anywhere in scope). U10's quantitative topics (molarity SOL-034..040, dilution/solution stoichiometry SOL-041..045) belong to U10-G4, where the 2-3-FRQ calc split applies instead. The single FRQ is UNSEEDED (your prod_seeds list is empty and the pool holds zero production items): the SOL-033 capstone chain run on one fresh mixing per form - (a) swap partners and write BOTH product formulas (30), (b) classify each product with the rules, naming the deciding rule or exception (40), (c) the chemist's outcome line with (s)/(aq) labels and what an observer sees (30). Case classes rotate across the five forms covering all three taught outcome classes (insoluble-family solid x2, exception-made solid x2, no-precipitate). NO 5-line scaffold (nothing substitutes into a formula - qualitative chain). D5 scaling evaluated and NOT triggered: min(20, 6 x 5 lessons) = 20 objective rows; your target_size=20 honored. Total 20 objective + 1 FRQ = 21 items/form.
2. SEED CENSUS AND FRESHENINGS UNDER THE STRICT SEED RULING (no freshness exemption; validator never weakened): all 15 distinct pool seeds are placed on form A (rows 4-15 and 17-19; your three demo_curated items are stem-normalized byte matches of pool[12]/[13]/[14] - deduped, both refs cited), zero spares; forms B-E are authored on the same blueprint rows. 12 seeds ride VERBATIM (field mapping only). 3 carried your stage-1 must_freshen_instance flags, executed with structure and misconception keys preserved and every field swept (QB 3.9) - confirm: (1) A#12 pool[8] washing-soda Na2CO3 -> Cs2CO3 (the seed's key restated SOL-031's OWN rule-line example 'Na2CO3 is a soluble carbonate, because sodium is a group 1 metal'; rescue key + family-flip + grinding-speed + formula-misread flanks kept). (2) A#17 pool[12] (= demo_curated[0]) partner salt NaCl -> NaBr (NaCl is a reactant in SOL-033's OWN article mechanism case AgNO3 + NaCl; KNO3 retained - not the colliding reactant; no-precipitate key + all three flanks kept). (3) A#18 pool[13] (= demo_curated[1]) reactant Pb(NO3)2 -> Ba(NO3)2 (Pb(NO3)2 is SOL-033's OWN WE1 reactant; Na2SO4 retained; typed answer now BaSO4; the soluble-pick / no-swap / criss-cross anticipated-wrong ledger rewritten to the new instance, feedback-leftover sweep clean).
3. SEED-TEMPLATE WRONG-KEY FIX AT BLIND QC - row 9 C/D forms (confirm; both blind reviewers caught it independently): the stage-2 template clause 'every common bromide compound is soluble' is FALSE under the taught rules (AgBr and PbBr2 are roster exceptions). C#9/D#9 keys now argue roster membership ('the exception is on the sulfate list only - this metal is not a bromide exception, so [salt] dissolves'); A/B/E row-9 keys keep the every-compound clause only where the family is a true guarantee (nitrate/acetate).
4. QUALIFIER-HINGED DISTRACTOR CLASS (sol authoritative QC, loops 1-3 - bank-wide sweep, confirm the approach): distractors whose wrongness rode ONLY on single/only/all/every around an otherwise-true norm were replaced with independently false claims (exception-roster misplacement errors; invented charge-test / vote-count / temperature / purpose / appearance mechanisms). Row-9 D1/D2 and the row-4 'outranks' distractor were IDENTICAL sentences across forms and were swept as whole rows for form-parallel consistency ('never trust a judge to re-catch').
5. FEEDBACK FAMILY-PROMOTION FIX (sol loop 1, 11 items + 1 class-sweep): misconception/feedback sentences that promoted bromide/sulfate NORMS to guarantees (contradicting the taught usually-vs-always distinction) were rewritten; two were stage-3 sweep leftovers. Keys were never affected.
6. ADJUDICATED INSTANCE FAMILIES (confirm; the gate's structural whitelists, enforced by a gate-local wrapper over the UNTOUCHED shared validator; slots enumerated from the scan): (1) AQUEOUS-SOLVENT-WATER (52 slots) - the gate's topic IS solubility in water; every scope rule line states its families' behavior 'in water' and the taught task on every row is a water verdict (U10-G1's identical family is the unit precedent); solutes stay globally unique under the home-seq ledger. (2) SOL030-EXCEPTION-ROSTER (silver/lead/calcium, 12 slots) - YOUR specNotes' own ruling: 'exception rosters ARE the content, so recall items necessarily reuse the rosters'; an item cannot test the exceptions without naming them, and SOL-032's guarantee-beats-exception counterintuition inherits the same names; barium (also roster) is not in the scan lexicon so needs no relief; fresh non-roster metals stay fully scanned. (3) GROUP1-RESCUE-SODIUM (3 slots: A#17/A#18 seed spectator prose + E#12's TSP rescue construct) - sodium is the taught group-1 exemplar of SOL-031's own rule line; it names the always-soluble spectator, never the tested outcome. Adjacency rulings recorded: A#8/A#9 share 'silver' as roster vocabulary (A#8 key = Lead, A#9 key = the nitrate guarantee - neither key leaks the other); A#17/A#18 share the sodium spectator (both are YOUR placed seeds; A#17 key = no precipitate, A#18 key = BaSO4).
7. GATE_CONFIG TASK-IDIOM WHITELIST (2 recorded entries, confirm) - both are YOUR SOL-033 registers, zero instance signal: 'are mixed. Which precipitate, if any, forms?' (the spec-MANDATED mixing-question register; your specNotes fix the register and your fluency + gate stems all use exactly this clause) and 'are mixed, and a white solid appears. Write the formula of the precipitate.' (your text-entry register, run at the seq-18 production row on every form by design). The runtime idiom-derivation misses both only because the clauses follow formula tokens (sentence-split lands inside the formula); every substance name, formula token, and number in the mixing stems still scans in full.
8. AgNO3 REUSE IN FRESH ITEMS (B#18, C#19, E#21 - disclosure, chemistry-forced): AgNO3 is the ONLY taught-soluble silver source (silver acetate is BANNED by your SOL-032 extraNotes; every other silver salt in scope is insoluble), so any fresh item needing dissolved silver must start from it. All three pairs differ from the article/WE/fluency/seed pairs.
9. CROSS-ITEM CUEING ADVISORY CLASS (59 round-1 blind flags - recorded test-level adjudication, no per-item fix): a single-topic RULES test necessarily restates the five taught rules across its explain-rows and apply-rows; the restated rule IS the skill under test. Blind reviewers picked every key without lesson access either way.
10. TEXT-ROW CASE CONTRACT (seq 18, all forms - schema-gap carrier note, same class as U10-G1's seq-7 note): the typed precipitate formulas are graded CASE-SENSITIVELY on the element symbols (source answer.caseSensitive = true - capitalization is part of the formula; 'BaSO4' vs 'baso4' must not both pass), but the section-6 answer shape has no caseSensitive lane, so the flag is carried here and restated inside each item's exported gradingNote. Unicode subscripts are NEVER required to type: the plain-digit spelling is an acceptedAlternate everywhere (your SOL-033 extraNotes' own convention). On forms B and E the correct formulas (AgI, ZnS) contain no subscripts at all, so the display and plain forms coincide - their alternate lists carry the same string as the display value by the seed convention (redundant but harmless; grading unaffected).
11. 'No precipitate forms' IS NOT NOTA (recorded at stage 1): it is the spec-mandated substantive outcome option of your SOL-033 extraNotes register (one option per outcome; the 'Both X and Y' option covers the two-solids misconception), same for pool[1]'s 'Nothing can be concluded until the metal is identified' claim option. V-T5's literal NOTA/AOTA regex passes on all forms; zero replace_distractor flags were needed across the 15-seed pool.
12. LESSON TITLES (verification request): unit10_specs_merged.json carries null title fields for SOL-029..032, so the exported lessonTitle values for those four are OUR reconstructions from your rule/extraNotes content ('The always-soluble ions', 'Usually soluble, with exceptions', 'The usually-insoluble families', 'Classify any compound with the rules'); SOL-033's 'Predict the precipitate' is from your gate data's capstones entry. The student-facing test title was verified against the nested manifest (Unit 10 Topic 6 'Solubility rules and precipitates'). Please confirm or supply your canonical lesson titles.
13. HONEST SURVIVOR MINORS (2, recorded): C#4 dry-storage and C#5 sealed-jar condition distractors judged weak-but-functional by sol (style notes only; both are the taught conditions-override misconception family, kept).
14. QC PROVENANCE (no action needed): deterministic validator 11/12 shared PASS + V-T7 100% excused under the recorded adjudications (parity-guard wrapper exit 0); blind rounds 1-5 (r1=gpt-5.6-sol blind role, r2=gemini; fresh tags each re-blind) = 258/258 key picks, 0 open hard fails; keyless lanes hand-graded 10/10 match/match for both reviewers (5 seq-18 formulas typed exactly + 5 FRQs reproducing both product formulas, the deciding rules, and the (s)/(aq) verdict cold); sol authoritative loops 22 -> 5 -> 1 -> 0 FAILs, final ledger 105/105 pass.
15. BINDING INSTANCE RULE (course-wide, wave 6): zero fissile/weapons/dose framings; classroom-safe instance set throughout (lab bottles, cleaners, pigments, mineral specimens, water-treatment and photography salts; your fireworks/tarnish/jeweler seed framings kept). Swept at packaging over forms_final AND this export body: ZERO hits on every banned token including 'dose'.
Form A
SOL-029 · easy · recall · seq 1 · authored (blueprint row 1 — recall case rotation, form A) · mcq4
At a review session, four students each state a guarantee they say comes from the solubility rules. Which student states a real guarantee?
“Any common compound containing the ammonium ion is soluble in water.” KEY
“Any common compound containing the iodide ion is soluble in water.”
“Any common compound containing the sulfate ion is soluble in water.”
“Any common compound containing the hydroxide ion is soluble in water.”
SOL-030 · easy · recall · seq 2 · authored (blueprint row 2 — recall case rotation, form A) · mcq4
What do the solubility rules say about chloride compounds as a family?
Most of them dissolve in water — a few named metals supply the exceptions. KEY
Most of them are insoluble in water; only compounds of group 1 metals dissolve.
Every one of them is soluble — the family carries a no-exceptions guarantee.
The rules do not cover this family at all.
SOL-031 · easy · recall · seq 3 · authored (blueprint row 3 — recall case rotation, form A) · mcq4
Which negative-ion family is usually INSOLUBLE in water?
A student classifies Pb(NO₃)₂ as insoluble 'because lead is an exception.' What did the student get wrong?
The routine's order — nitrate settles the compound as soluble before lead's exceptions ever apply. KEY
Nothing — lead compounds are insoluble no matter the other ion.
The metal — lead is not an exception metal at all, so the compound is soluble simply because no exception applies.
Only the charge — the student should have written the compound as PbNO₃ first.
SOL-032 · medium · application · seq 16 · authored (blueprint row 16 — application case rotation, form A) · mcq4
The compound Rb₂CO₃ is run through the classification routine. At which point does the routine settle its verdict?
At the positive-ion check — the positive ion is one of the guarantee ions. KEY
At the negative-ion check — the negative ion is nitrate or acetate.
At the family norm — and the exception check leaves the usual answer standing.
At the exception check — a named exception flips the family's usual answer.
SOL-033 · medium · application · seq 17 · seed:U10-G3.pool[12] (= demo_curated[0], James) — partner salt freshened per must_freshen_instance: NaCl (SOL-033 article mechanism reactant) → NaBr; no-precipitate key and all three misconception flanks preserved · mcq4
Solutions of potassium nitrate, KNO₃, and sodium bromide, NaBr, are mixed. Which precipitate, if any, forms?
No precipitate forms KEY
KBr
NaNO₃
Both KBr and NaNO₃
SOL-033 · hard · transfer · seq 18 · seed:U10-G3.pool[13] (= demo_curated[1], James) — reactant freshened per must_freshen_instance: Pb(NO₃)₂ (SOL-033 WE1 reactant) → Ba(NO₃)₂; produce-the-insoluble-sulfate skill + soluble-pick/no-swap/criss-cross wrong-ledger preserved · text
Solutions of barium nitrate, Ba(NO₃)₂, and sodium sulfate, Na₂SO₄, are mixed and a white solid appears. Write the formula of the precipitate.
Answer: BaSO₄ KEY
also accepted: BaSO4
wrongValueDiagnostics
[
{
"value": "NaNO₃",
"misconception": "you picked the soluble product — sodium and nitrate both guarantee NaNO₃ stays dissolved; the solid is the barium sulfate",
"feedback": "This answer is wrong — NaNO₃ is the soluble product: sodium and nitrate both guarantee it stays dissolved. The solid is the barium sulfate, BaSO₄."
},
{
"value": "Ba(NO₃)₂",
"misconception": "you kept the original partners instead of swapping — Ba(NO₃)₂ was already dissolved; the new, insoluble pairing is BaSO₄",
"feedback": "This answer is wrong — Ba(NO₃)₂ is a reactant, already dissolved. Swap partners: the new pairing BaSO₄ is the insoluble product."
},
{
"value": "Ba₂(SO₄)₂",
"misconception": "you criss-crossed charges that already balance — Ba²⁺ and SO₄²⁻ pair one-to-one, so the formula is BaSO₄",
"feedback": "This answer is wrong — Ba²⁺ and SO₄²⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is BaSO₄."
}
]
feedback
{
"keyExplanation": "Swap partners: the products are BaSO₄ and NaNO₃. Barium is one of the three sulfate exceptions, so BaSO₄ is insoluble and forms the white solid — BaSO₄(s). NaNO₃ stays dissolved as NaNO₃(aq).",
"anticipatedWrong": [
{
"value": "NaNO₃",
"misconception": "you picked the soluble product — sodium and nitrate both guarantee NaNO₃ stays dissolved; the solid is the barium sulfate",
"feedback": "This answer is wrong — NaNO₃ is the soluble product: sodium and nitrate both guarantee it stays dissolved. The solid is the barium sulfate, BaSO₄."
},
{
"value": "Ba(NO₃)₂",
"misconception": "you kept the original partners instead of swapping — Ba(NO₃)₂ was already dissolved; the new, insoluble pairing is BaSO₄",
"feedback": "This answer is wrong — Ba(NO₃)₂ is a reactant, already dissolved. Swap partners: the new pairing BaSO₄ is the insoluble product."
},
{
"value": "Ba₂(SO₄)₂",
"misconception": "you criss-crossed charges that already balance — Ba²⁺ and SO₄²⁻ pair one-to-one, so the formula is BaSO₄",
"feedback": "This answer is wrong — Ba²⁺ and SO₄²⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is BaSO₄."
}
]
}
SOL-033 · hard · transfer · seq 19 · seed:U10-G3.pool[14] (= demo_curated[2], James) — verbatim (field mapping only) · mcq4
A jeweler needs a mixture that produces a precipitate. Which pair of solutions does it?
AgNO₃ and KBr KEY
NaCl and Ca(NO₃)₂
K₂SO₄ and MgCl₂
NaNO₃ and K₂CO₃
SOL-033 · hard · transfer · seq 20 · authored (blueprint row 20 — transfer case rotation, form A) · mcq4
Solutions of Co(NO₃)₂ and LiOH are mixed. Which statement describes the outcome the way a chemist would write it?
Co(OH)₂(s) forms as a solid, and the other product stays dissolved as LiNO₃(aq). KEY
LiNO₃(s) forms as a solid, and the other product stays dissolved as Co(OH)₂(aq).
Both products are written (aq) — any solid that appears dissolves back within a few moments.
Both products are written (s) — mixing two solutions always solidifies both new pairings.
SOL-033 · medium · application · seq 21 · authored (blueprint row 21 — UNSEEDED capstone-chain FRQ; insoluble-family case (carbonate norm, no rescue ion); rubric 30/40/30; no 5-line scaffold — no formula substitution anywhere) · frq
Solutions of FeCl₂ and K₂CO₃ are mixed.
Solutions of FeCl₂ and K₂CO₃ are mixed.
(a) Swap the partners and write correct formulas for BOTH candidate products.
(b) Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
(c) Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Swap the partners and write correct formulas for BOTH candidate products.
Model: Swap partners: the iron pairs with carbonate and the potassium pairs with chloride. The candidate products are FeCO3 and KCl. The iron and carbonate ions carry equal-and-opposite 2+ and 2− charges, so they pair one-to-one; potassium and chloride also pair one-to-one.
Part b. Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
Model: FeCO3 — carbonates are usually insoluble, and iron is not a group 1 metal or ammonium, so no rescue applies: insoluble. KCl — potassium is a group 1 metal, so the always-soluble guarantee settles it: soluble.
Part c. Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: FeCO3(s) forms as the precipitate, and KCl stays dissolved as KCl(aq). An observer sees a solid appear and cloud the mixture the moment the solutions meet, then settle.
Full model answer. (a) Swap partners: the iron pairs with carbonate and the potassium pairs with chloride. The candidate products are FeCO3 and KCl. The iron and carbonate ions carry equal-and-opposite 2+ and 2− charges, so they pair one-to-one; potassium and chloride also pair one-to-one.
(b) FeCO3 — carbonates are usually insoluble, and iron is not a group 1 metal or ammonium, so no rescue applies: insoluble. KCl — potassium is a group 1 metal, so the always-soluble guarantee settles it: soluble.
(c) FeCO3(s) forms as the precipitate, and KCl stays dissolved as KCl(aq). An observer sees a solid appear and cloud the mixture the moment the solutions meet, then settle.
Form B
SOL-029 · easy · recall · seq 1 · authored (blueprint row 1 — recall case rotation, form B) · mcq4
Four flashcards each claim that one ion makes every common compound containing it soluble in water. Which flashcard is correct?
The acetate ion. KEY
The chloride ion.
The carbonate ion.
The zinc ion.
SOL-030 · easy · recall · seq 2 · authored (blueprint row 2 — recall case rotation, form B) · mcq4
What do the solubility rules say about iodide compounds as a family?
Most of them dissolve in water, and a short list of named metals forms the insoluble exceptions. KEY
Most of them stay undissolved in water; only group 1 compounds are exceptions.
They all dissolve because iodide is one of the ions on the always-soluble list.
The rules are silent on this family.
SOL-031 · easy · recall · seq 3 · authored (blueprint row 3 — recall case rotation, form B) · mcq4
Most carbonate compounds do not dissolve in water. Which positive ions turn a compound from one of the insoluble families into a soluble exception?
A group 1 metal ion or the ammonium ion. KEY
The silver or lead ion.
The barium or calcium ion.
No positive ion can do that — solubility is decided by the negative ion alone.
SOL-029 · medium · application · seq 4 · authored (blueprint row 4 — application case rotation, form B) · mcq4
A prep-room bottle holds rubidium acetate, RbC₂H₃O₂. What do the solubility rules say about it?
It is soluble — each of its ions carries the always-soluble guarantee on its own. KEY
It is insoluble — two guaranteed ions in one formula cancel each other out.
It is insoluble — rubidium is one of the named exception metals in the rules.
The rules cannot decide without a beaker test.
SOL-029 · medium · application · seq 5 · authored (blueprint row 5 — application case rotation, form B) · mcq4
A mineral-supplement label has been water-damaged: all that is readable is that the active compound is “a nitrate of some metal.” What can the solubility rules conclude?
The compound is soluble in water, whichever metal it is. KEY
Nothing can be concluded until the metal is identified.
The compound is insoluble, because nitrate sits in one of the usually-insoluble families.
The rules say it dissolves in warm water but stays a solid in cold water.
SOL-029 · medium · application · seq 6 · authored (blueprint row 6 — application case rotation, form B) · mcq4
The always-soluble rule can settle the solubility of exactly one of these compounds. Which one?
CsCl KEY
FePO₄
NiSO₄
CuBr₂
SOL-030 · medium · application · seq 7 · authored (blueprint row 7 — application case rotation, form B) · mcq4
Which chloride compound is insoluble in water?
AgCl KEY
BaCl₂
LiCl
ZnCl₂
SOL-030 · medium · application · seq 8 · authored (blueprint row 8 — application case rotation, form B) · mcq4
Which metal breaks ONLY the sulfate rule — its sulfate is insoluble, but its chloride, bromide, and iodide all dissolve?
Barium KEY
Lead
Silver
Zinc
SOL-030 · medium · application · seq 9 · authored (blueprint row 9 — application case rotation, form B) · mcq4
PbSO₄ is insoluble in water, yet Pb(NO₃)₂ dissolves freely. Which statement explains why Pb(NO₃)₂ dissolves?
The exception applies to the sulfate family, not to nitrate — and every common nitrate compound is soluble. KEY
The rules track the metal, not the negative ion — and this metal is on the rules' list of soluble metals.
Pb(NO₃)₂ dissolves because its two nitrate ions outnumber the metal ion and outvote its exception.
Pb(NO₃)₂ dissolves only because it is usually stored as a fine powder.
SOL-031 · medium · application · seq 10 · authored (blueprint row 10 — application case rotation, form B) · mcq4
Which hydroxide compound is insoluble in water?
Fe(OH)₃ KEY
KOH
RbOH
NaOH
SOL-031 · medium · application · seq 11 · authored (blueprint row 11 — application case rotation, form B) · mcq4
Zinc sulfide, ZnS, is the white solid in some luminous pigments. What do the solubility rules say about ZnS in water?
It is insoluble. KEY
It is soluble.
It is soluble because pigment powders are milled extremely fine.
Factory-made compounds sit outside what the solubility rules can classify.
SOL-031 · medium · application · seq 12 · authored (blueprint row 12 — application case rotation, form B) · mcq4
A leavening powder once used by bakers is (NH₄)₂CO₃ — and it dissolves freely, even though most carbonates do not. Why?
Ammonium is a rescue ion — ammonium compounds of the insoluble families are the soluble exceptions. KEY
Carbonates are actually a usually-soluble family, so the powder needs no special explanation to dissolve.
Its formula holds two positive ions against one negative ion, and the majority ion type decides solubility.
Its formula contains no carbonate ion, so the carbonate rule does not apply.
SOL-032 · medium · application · seq 13 · authored (blueprint row 13 — application case rotation, form B) · mcq4
Apply the solubility rules: which one of these compounds dissolves in water?
LiNO₃ KEY
PbBr₂
NiS
Cu(OH)₂
SOL-032 · medium · application · seq 14 · authored (blueprint row 14 — application case rotation, form B) · mcq4
Use the solubility rules: which of these zinc compounds is soluble in water?
ZnSO₄ KEY
ZnS
ZnCO₃
Zn(OH)₂
SOL-032 · medium · application · seq 15 · authored (blueprint row 15 — application case rotation, form B) · mcq4
A student classifies K₂CO₃ as insoluble “because carbonates are insoluble.” What did the student get wrong?
The routine's order — potassium's group 1 guarantee settles K₂CO₃ as soluble before the carbonate norm is ever consulted. KEY
Nothing — the carbonate norm outranks the guarantee checks whenever the two disagree.
Only the family — carbonates are actually usually soluble, so the compound dissolves for a different reason.
Only the formula — the compound should have been written KCO₃ before classifying it.
SOL-032 · medium · application · seq 16 · authored (blueprint row 16 — application case rotation, form B) · mcq4
The compound Zn(C₂H₃O₂)₂ is run through the classification routine. At which point does the routine settle its verdict?
At the negative-ion check — the negative ion is nitrate or acetate. KEY
At the positive-ion check — the positive ion is one of the guarantee ions.
At the family norm — and the exception check leaves the usual answer standing.
At the exception check — a named exception flips the family's usual answer.
SOL-033 · medium · application · seq 17 · authored (blueprint row 17 — application case rotation, form B) · mcq4
Solutions of LiI and RbNO₃ are mixed. Which precipitate, if any, forms?
No precipitate forms KEY
RbI
LiNO₃
Both RbI and LiNO₃
SOL-033 · hard · transfer · seq 18 · authored (blueprint row 18 — fresh precipitate-production pair, form B) · text
Solutions of AgNO₃ and NaI are mixed, and a pale-yellow solid appears. Write the formula of the precipitate.
Answer: AgI KEY
also accepted: AgI
wrongValueDiagnostics
[
{
"value": "NaNO₃",
"misconception": "you picked the soluble product — sodium and nitrate both guarantee NaNO₃ stays dissolved; the solid is the silver iodide",
"feedback": "This answer is wrong — NaNO₃ is the soluble product: sodium and nitrate both guarantee it. The solid is the silver iodide, AgI."
},
{
"value": "AgNO₃",
"misconception": "you kept the original partners instead of swapping — AgNO₃ was already dissolved; the new, insoluble pairing is AgI",
"feedback": "This answer is wrong — AgNO₃ is a reactant, already dissolved. Swap partners: the new pairing AgI is the insoluble product."
},
{
"value": "Ag₂I",
"misconception": "you criss-crossed charges that already balance — Ag⁺ and I⁻ pair one-to-one, so the formula is AgI",
"feedback": "This answer is wrong — Ag⁺ and I⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is AgI."
}
]
feedback
{
"keyExplanation": "Swap partners: the products are AgI and NaNO₃. Silver is a halide exception, so AgI is insoluble and forms the pale-yellow solid — AgI(s). NaNO₃ stays dissolved as NaNO₃(aq).",
"anticipatedWrong": [
{
"value": "NaNO₃",
"misconception": "you picked the soluble product — sodium and nitrate both guarantee NaNO₃ stays dissolved; the solid is the silver iodide",
"feedback": "This answer is wrong — NaNO₃ is the soluble product: sodium and nitrate both guarantee it. The solid is the silver iodide, AgI."
},
{
"value": "AgNO₃",
"misconception": "you kept the original partners instead of swapping — AgNO₃ was already dissolved; the new, insoluble pairing is AgI",
"feedback": "This answer is wrong — AgNO₃ is a reactant, already dissolved. Swap partners: the new pairing AgI is the insoluble product."
},
{
"value": "Ag₂I",
"misconception": "you criss-crossed charges that already balance — Ag⁺ and I⁻ pair one-to-one, so the formula is AgI",
"feedback": "This answer is wrong — Ag⁺ and I⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is AgI."
}
]
}
SOL-033 · hard · transfer · seq 19 · authored (blueprint row 19 — transfer case rotation, form B) · mcq4
A water-treatment bench needs a mixture that traps a dissolved metal as a solid. Which pair of solutions produces a precipitate when mixed?
Ca(NO₃)₂ and K₂CO₃ KEY
LiCl and KNO₃
NaI and Mg(NO₃)₂
RbBr and LiNO₃
SOL-033 · hard · transfer · seq 20 · authored (blueprint row 20 — transfer case rotation, form B) · mcq4
Solutions of FeSO₄ and Rb₂CO₃ are mixed. Which statement describes the outcome the way a chemist would write it?
FeCO₃(s) forms as a solid, and the other product stays dissolved as Rb₂SO₄(aq). KEY
Rb₂SO₄(s) forms as a solid, and the other product stays dissolved as FeCO₃(aq).
Both products are written (aq) — the carbonate solid breaks back up as soon as it forms.
Both products are written (s) — the mixing turns the whole beaker's contents solid.
SOL-033 · medium · application · seq 21 · authored (blueprint row 21 — UNSEEDED capstone-chain FRQ; exception-made case (sulfate roster: calcium); rubric 30/40/30; no 5-line scaffold — no formula substitution anywhere) · frq
Solutions of CaBr₂ and Cs₂SO₄ are mixed.
Solutions of CaBr₂ and Cs₂SO₄ are mixed.
(a) Swap the partners and write correct formulas for BOTH candidate products.
(b) Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
(c) Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Swap the partners and write correct formulas for BOTH candidate products.
Model: Swap partners: the calcium pairs with sulfate and the cesium pairs with bromide. The candidate products are CaSO4 and CsBr. The calcium and sulfate ions carry equal-and-opposite 2+ and 2− charges, so they pair one-to-one; cesium and bromide also pair one-to-one.
Part b. Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
Model: CaSO4 — sulfates are usually soluble, BUT calcium is one of the three named sulfate exceptions, so the exception flips it: insoluble. CsBr — cesium is a group 1 metal, so the always-soluble guarantee settles it: soluble.
Part c. Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: CaSO4(s) forms as the precipitate, and CsBr stays dissolved as CsBr(aq). An observer sees the clear mixture turn cloudy with white solid, which then settles.
Full model answer. (a) Swap partners: the calcium pairs with sulfate and the cesium pairs with bromide. The candidate products are CaSO4 and CsBr. The calcium and sulfate ions carry equal-and-opposite 2+ and 2− charges, so they pair one-to-one; cesium and bromide also pair one-to-one.
(b) CaSO4 — sulfates are usually soluble, BUT calcium is one of the three named sulfate exceptions, so the exception flips it: insoluble. CsBr — cesium is a group 1 metal, so the always-soluble guarantee settles it: soluble.
(c) CaSO4(s) forms as the precipitate, and CsBr stays dissolved as CsBr(aq). An observer sees the clear mixture turn cloudy with white solid, which then settles.
Form C
SOL-029 · easy · recall · seq 1 · authored (blueprint row 1 — recall case rotation, form C) · mcq4
A lab poster lists four ions, each with the caption “every common compound of this ion dissolves in water.” For which ion is the caption actually true?
The cesium ion, Cs⁺. KEY
The bromide ion, Br⁻.
The sulfide ion, S²⁻.
The iron(III) ion, Fe³⁺.
SOL-030 · easy · recall · seq 2 · authored (blueprint row 2 — recall case rotation, form C) · mcq4
What do the solubility rules say about sulfate compounds as a family?
Most of them are soluble in water, with a short roster of named metals as the exceptions. KEY
Most of them are insoluble in water unless a group 1 metal or the ammonium ion rescues them.
Every sulfate compound is soluble — the family has no exceptions.
The rules do not classify this family either way.
SOL-031 · easy · recall · seq 3 · authored (blueprint row 3 — recall case rotation, form C) · mcq4
Which of these negative-ion families follows the norm “usually insoluble in water”?
Sulfide. KEY
Chloride.
Nitrate.
Sulfate.
SOL-029 · medium · application · seq 4 · authored (blueprint row 4 — application case rotation, form C) · mcq4
Lithium bromide, LiBr, is used in some building air-conditioning systems. What do the solubility rules say about it?
It is soluble — lithium is a group 1 metal ion, and one guaranteed ion settles the compound. KEY
It is insoluble — bromide is not on the always-soluble list, and an off-list ion blocks the guarantee.
Only a beaker test could answer this — the list of rules says nothing here.
It is soluble only while the sample is kept dry beforehand.
SOL-029 · medium · application · seq 5 · authored (blueprint row 5 — application case rotation, form C) · mcq4
A reagent jar is labeled “ammonium …” but the negative ion's name has rubbed off. What can the solubility rules conclude about the compound?
It is soluble in water, whatever the missing negative ion is. KEY
Nothing — the negative ion's family decides, so the label is useless as it stands.
It is soluble only if the missing ion is nitrate or acetate.
It is soluble only if the jar has been kept tightly sealed.
SOL-029 · medium · application · seq 6 · authored (blueprint row 6 — application case rotation, form C) · mcq4
For exactly one of these compounds, the always-soluble rule alone settles the answer. Which one is it?
Rb₂S KEY
AlPO₄
ZnBr₂
MgI₂
SOL-030 · medium · application · seq 7 · authored (blueprint row 7 — application case rotation, form C) · mcq4
Which bromide compound is insoluble in water?
PbBr₂ KEY
CaBr₂
CsBr
ZnBr₂
SOL-030 · medium · application · seq 8 · authored (blueprint row 8 — application case rotation, form C) · mcq4
Which metal breaks ONLY the halide rule — its chloride, bromide, and iodide are insoluble, but its sulfate is not a named exception?
Silver KEY
Lead
Calcium
Barium
SOL-030 · medium · application · seq 9 · authored (blueprint row 9 — application case rotation, form C) · mcq4
BaSO₄ is insoluble in water, yet BaBr₂ dissolves freely. Which statement explains why BaBr₂ dissolves?
The exception is on the sulfate list only — this metal is not a bromide exception, so BaBr₂ dissolves. KEY
The rules track the metal, not the negative ion — and this metal is on the rules' list of soluble metals.
BaBr₂ dissolves because its ions' charges are small, and small charges cannot hold a solid together.
BaBr₂ dissolves only while the mixture is kept stirred.
SOL-031 · medium · application · seq 10 · authored (blueprint row 10 — application case rotation, form C) · mcq4
Which sulfide compound is insoluble in water?
ZnS KEY
Na₂S
Li₂S
Cs₂S
SOL-031 · medium · application · seq 11 · authored (blueprint row 11 — real-world insoluble-family case, form C; stage-3 instance refresh, see fix_log P8) · mcq4
Cobalt(II) sulfide, CoS, gives some mineral ores a dark metallic sheen. What do the solubility rules say about CoS in water?
It is insoluble. KEY
It is soluble.
It is soluble because crushing the ore into a powder makes it dissolve.
Minerals dug from the ground fall outside what the rules can classify.
SOL-031 · medium · application · seq 12 · authored (blueprint row 12 — application case rotation, form C) · mcq4
K₂S dissolves completely in water even though most sulfides stay solid. Which statement gives the reason?
Potassium is a group 1 metal, and group 1 compounds of the insoluble families are the soluble exceptions. KEY
Sulfides are actually a usually-soluble family with a short exception roster.
Lab-grade K₂S is unusually pure, and pure compounds always dissolve.
The compound contains two potassium ions, and two of any metal ion force a compound to dissolve.
SOL-032 · medium · application · seq 13 · authored (blueprint row 13 — application case rotation, form C) · mcq4
According to the solubility rules, which of these compounds is soluble in water?
RbNO₃ KEY
AgBr
FeCO₃
Ni(OH)₂
SOL-032 · medium · application · seq 14 · authored (blueprint row 14 — application case rotation, form C) · mcq4
Use the solubility rules: which of these manganese compounds is soluble in water?
MnSO₄ KEY
MnCO₃
Mn(OH)₂
Mn₃(PO₄)₂
SOL-032 · medium · application · seq 15 · authored (blueprint row 15 — application case rotation, form C) · mcq4
A student classifies RbOH as insoluble “because hydroxides are insoluble.” What did the student get wrong?
The routine's order — rubidium's group 1 guarantee settles RbOH as soluble before the hydroxide norm is ever consulted. KEY
Nothing — the hydroxide norm applies to every hydroxide, whatever the positive ion is.
Only the family — hydroxides are actually usually soluble, so the compound dissolves for a different reason.
Only the charge — the student should first rewrite the compound as Rb(OH)₂.
SOL-032 · medium · application · seq 16 · authored (blueprint row 16 — application case rotation, form C) · mcq4
The compound NiCl₂ is run through the classification routine. At which point does the routine settle its verdict?
At the family norm — and the exception check leaves the usual answer standing. KEY
At the positive-ion check — the positive ion is one of the guarantee ions.
At the negative-ion check — the negative ion is nitrate or acetate.
At the exception check — a named exception flips the family's usual answer.
SOL-033 · medium · application · seq 17 · authored (blueprint row 17 — application case rotation, form C) · mcq4
Solutions of LiC₂H₃O₂ and MgBr₂ are mixed. Which precipitate, if any, forms?
No precipitate forms KEY
LiBr
Mg(C₂H₃O₂)₂
Both LiBr and Mg(C₂H₃O₂)₂
SOL-033 · hard · transfer · seq 18 · authored (blueprint row 18 — fresh precipitate-production pair, form C) · text
Solutions of Co(NO₃)₂ and Na₂CO₃ are mixed, and a pink solid appears. Write the formula of the precipitate.
Answer: CoCO₃ KEY
also accepted: CoCO3
wrongValueDiagnostics
[
{
"value": "NaNO₃",
"misconception": "you picked the soluble product — sodium and nitrate both guarantee NaNO₃ stays dissolved; the solid is the cobalt carbonate",
"feedback": "This answer is wrong — NaNO₃ is the soluble product: sodium and nitrate both guarantee it. The solid is the cobalt carbonate, CoCO₃."
},
{
"value": "Co(NO₃)₂",
"misconception": "you kept the original partners instead of swapping — Co(NO₃)₂ was already dissolved; the new, insoluble pairing is CoCO₃",
"feedback": "This answer is wrong — Co(NO₃)₂ is a reactant, already dissolved. Swap partners: the new pairing CoCO₃ is the insoluble product."
},
{
"value": "Co₂(CO₃)₂",
"misconception": "you criss-crossed charges that already balance — Co²⁺ and CO₃²⁻ pair one-to-one, so the formula is CoCO₃",
"feedback": "This answer is wrong — Co²⁺ and CO₃²⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is CoCO₃."
}
]
feedback
{
"keyExplanation": "Swap partners: the products are CoCO₃ and NaNO₃. Carbonates are usually insoluble and cobalt is not a rescue ion, so CoCO₃ forms the pink solid — CoCO₃(s). NaNO₃ stays dissolved as NaNO₃(aq).",
"anticipatedWrong": [
{
"value": "NaNO₃",
"misconception": "you picked the soluble product — sodium and nitrate both guarantee NaNO₃ stays dissolved; the solid is the cobalt carbonate",
"feedback": "This answer is wrong — NaNO₃ is the soluble product: sodium and nitrate both guarantee it. The solid is the cobalt carbonate, CoCO₃."
},
{
"value": "Co(NO₃)₂",
"misconception": "you kept the original partners instead of swapping — Co(NO₃)₂ was already dissolved; the new, insoluble pairing is CoCO₃",
"feedback": "This answer is wrong — Co(NO₃)₂ is a reactant, already dissolved. Swap partners: the new pairing CoCO₃ is the insoluble product."
},
{
"value": "Co₂(CO₃)₂",
"misconception": "you criss-crossed charges that already balance — Co²⁺ and CO₃²⁻ pair one-to-one, so the formula is CoCO₃",
"feedback": "This answer is wrong — Co²⁺ and CO₃²⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is CoCO₃."
}
]
}
SOL-033 · hard · transfer · seq 19 · authored (blueprint row 19 — transfer case rotation, form C) · mcq4
A prep bench needs a mixture that yields a solid product. Which pair of solutions produces a precipitate when mixed?
AgNO₃ and CsI KEY
KNO₃ and Na₂SO₄
RbCl and Li₂SO₄
NH₄I and CsNO₃
SOL-033 · hard · transfer · seq 20 · authored (blueprint row 20 — transfer case rotation, form C) · mcq4
Solutions of CsI and LiNO₃ are mixed. Which statement describes the outcome the way a chemist would write it?
Nothing solid forms — the products stay dissolved as CsNO₃(aq) and LiI(aq), and the mixture stays clear. KEY
CsNO₃(s) forms as a solid, and the other product stays dissolved as LiI(aq).
LiI(s) forms as a solid, and the other product stays dissolved as CsNO₃(aq).
Both products form as solids at first, then slowly redissolve while the mixture is being stirred.
SOL-033 · medium · application · seq 21 · authored (blueprint row 21 — UNSEEDED capstone-chain FRQ; insoluble-family case (hydroxide norm, no rescue ion; 1:2 formula-writing challenge); rubric 30/40/30; no 5-line scaffold — no formula substitution anywhere) · frq
Solutions of Zn(NO₃)₂ and CsOH are mixed.
Solutions of Zn(NO₃)₂ and CsOH are mixed.
(a) Swap the partners and write correct formulas for BOTH candidate products.
(b) Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
(c) Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Swap the partners and write correct formulas for BOTH candidate products.
Model: Swap partners: the zinc pairs with hydroxide and the cesium pairs with nitrate. The candidate products are Zn(OH)2 and CsNO3. The zinc ion's 2+ charge needs two of the 1− hydroxide ions, so the zinc product takes the (OH)2 group; cesium and nitrate pair one-to-one.
Part b. Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
Model: Zn(OH)2 — hydroxides are usually insoluble, and zinc is not a group 1 metal or ammonium, so no rescue applies: insoluble. CsNO3 — cesium is a group 1 metal AND nitrate is an always-soluble family; either guarantee alone settles it: soluble.
Part c. Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: Zn(OH)2(s) forms as the precipitate, and CsNO3 stays dissolved as CsNO3(aq). An observer sees white solid cloud the mixture as the solutions meet, then settle.
Full model answer. (a) Swap partners: the zinc pairs with hydroxide and the cesium pairs with nitrate. The candidate products are Zn(OH)2 and CsNO3. The zinc ion's 2+ charge needs two of the 1− hydroxide ions, so the zinc product takes the (OH)2 group; cesium and nitrate pair one-to-one.
(b) Zn(OH)2 — hydroxides are usually insoluble, and zinc is not a group 1 metal or ammonium, so no rescue applies: insoluble. CsNO3 — cesium is a group 1 metal AND nitrate is an always-soluble family; either guarantee alone settles it: soluble.
(c) Zn(OH)2(s) forms as the precipitate, and CsNO3 stays dissolved as CsNO3(aq). An observer sees white solid cloud the mixture as the solutions meet, then settle.
A compound from one of the usually-insoluble families turns out to dissolve freely. Which positive ion could it contain?
The rubidium ion — a group 1 metal ion. KEY
The silver ion.
Any metal ion with a 2+ charge.
None — a compound from an insoluble family can never dissolve.
SOL-029 · medium · application · seq 4 · authored (blueprint row 4 — application case rotation, form D) · mcq4
Ammonium bromide, NH₄Br, was once a standard chemical in photographic darkrooms. What do the solubility rules say about it?
It is soluble — the ammonium ion guarantees it, whatever the other ion is. KEY
It is insoluble — bromide is not on the always-soluble list, and that decides the compound.
It is soluble only in warm water.
No verdict is possible until a sample has been stirred into water and watched.
SOL-029 · medium · application · seq 5 · authored (blueprint row 5 — application case rotation, form D) · mcq4
Half a label survives on an old stock bottle: the compound is “potassium something” — the negative ion's name is torn away. What can the solubility rules conclude?
The compound is soluble in water no matter what the torn-off ion is. KEY
Nothing can be concluded until the torn half of the label is found.
The compound is soluble only if the missing ion is also on the always-soluble list.
The compound is insoluble, because most negative ions form insoluble potassium compounds.
SOL-029 · medium · application · seq 6 · authored (blueprint row 6 — application case rotation, form D) · mcq4
Which one of these compounds can the always-soluble rule settle by itself?
Li₂SO₄ KEY
CuCO₃
NiBr₂
ZnI₂
SOL-030 · medium · application · seq 7 · authored (blueprint row 7 — application case rotation, form D) · mcq4
Which of these iodide compounds does NOT dissolve in water?
AgI KEY
BaI₂
NaI
NiI₂
SOL-030 · medium · application · seq 8 · authored (blueprint row 8 — application case rotation, form D) · mcq4
A student needs a sulfate that stays dissolved in water. Of these metals, whose sulfate should be chosen?
Potassium KEY
Barium
Lead
Calcium
SOL-030 · medium · application · seq 9 · authored (blueprint row 9 — application case rotation, form D) · mcq4
CaSO₄ is insoluble in water, yet CaBr₂ dissolves freely. Which statement explains why CaBr₂ dissolves?
The exception is on the sulfate list only — this metal is not a bromide exception, so CaBr₂ dissolves. KEY
The rules track the metal, not the negative ion — and this metal is on the rules' list of soluble metals.
CaBr₂ dissolves because splitting the metal's 2+ charge between two bromide ions leaves the solid too weak to hold together.
CaBr₂ dissolves only because it is usually stored as a fine powder.
SOL-031 · medium · application · seq 10 · authored (blueprint row 10 — application case rotation, form D) · mcq4
Which phosphate compound is insoluble in water?
Mg₃(PO₄)₂ KEY
Na₃PO₄
Rb₃PO₄
(NH₄)₃PO₄
SOL-031 · medium · application · seq 11 · authored (blueprint row 11 — real-world insoluble-family case, form D; stage-3 instance refresh, see fix_log P9) · mcq4
A pottery-glaze powder contains cobalt carbonate, CoCO₃. What do the solubility rules say about CoCO₃ in water?
It is insoluble. KEY
It is soluble.
It is soluble because the glaze is stirred into water before firing.
A manufactured glaze powder falls outside the solubility rules.
SOL-031 · medium · application · seq 12 · authored (blueprint row 12 — application case rotation, form D) · mcq4
CsOH, used in some specialty batteries, dissolves freely in water even though most hydroxides do not. Why?
Cesium is a group 1 metal, and group 1 compounds of the insoluble families are the soluble exceptions. KEY
Hydroxides are actually a usually-soluble family with named metal exceptions, just like the sulfates.
Battery chemicals are made unusually pure, and pure compounds always dissolve.
CsOH contains no hydroxide ion, so the hydroxide rule never applied to it.
SOL-032 · medium · application · seq 13 · authored (blueprint row 13 — application case rotation, form D) · mcq4
Exactly one of these compounds is classified as soluble by the solubility rules. Which one is it?
(NH₄)₂SO₄ KEY
CaSO₄
ZnCO₃
FePO₄
SOL-032 · medium · application · seq 14 · authored (blueprint row 14 — application case rotation, form D) · mcq4
Use the solubility rules: which of these lead compounds is soluble in water?
Pb(C₂H₃O₂)₂ KEY
PbBr₂
PbS
Pb₃(PO₄)₂
SOL-032 · medium · application · seq 15 · authored (blueprint row 15 — application case rotation, form D) · mcq4
A student classifies calcium acetate, Ca(C₂H₃O₂)₂, as insoluble “because calcium is an exception.” What did the student get wrong?
Two things at once — acetate's guarantee settles the compound as soluble first, and calcium's exception lives on the sulfate rule anyway. KEY
Nothing — an exception metal makes all of its compounds insoluble.
The metal — calcium is not an exception metal at all, so the compound is soluble simply because no exception applies.
Only the formula — the compound should be written CaC₂H₃O₂ before classifying it.
SOL-032 · medium · application · seq 16 · authored (blueprint row 16 — application case rotation, form D) · mcq4
The compound AgBr is run through the classification routine. At which point does the routine settle its verdict?
At the exception check — a named exception flips the family's usual answer. KEY
At the positive-ion check — the positive ion is one of the guarantee ions.
At the negative-ion check — the negative ion is nitrate or acetate.
At the family norm — and the exception check leaves the usual answer standing.
SOL-033 · medium · application · seq 17 · authored (blueprint row 17 — application case rotation, form D) · mcq4
Solutions of ZnSO₄ and LiCl are mixed. Which precipitate, if any, forms?
No precipitate forms KEY
ZnCl₂
Li₂SO₄
Both ZnCl₂ and Li₂SO₄
SOL-033 · hard · transfer · seq 18 · authored (blueprint row 18 — fresh precipitate-production pair, form D) · text
Solutions of FeCl₃ and KOH are mixed, and a rust-orange solid appears. Write the formula of the precipitate.
Answer: Fe(OH)₃ KEY
also accepted: Fe(OH)3
wrongValueDiagnostics
[
{
"value": "KCl",
"misconception": "you picked the soluble product — potassium guarantees KCl stays dissolved; the solid is the iron(III) hydroxide",
"feedback": "This answer is wrong — KCl is the soluble product: potassium guarantees it. The solid is the iron(III) hydroxide, Fe(OH)₃."
},
{
"value": "FeOH",
"misconception": "you paired the ions one-to-one — iron here is 3+ and hydroxide is 1−, so three hydroxides are needed: Fe(OH)₃",
"feedback": "This answer is wrong — the charges do not balance one-to-one: Fe³⁺ needs three OH⁻ ions, so the formula is Fe(OH)₃."
},
{
"value": "Fe(OH)₂",
"misconception": "you used iron(II)'s charge — the reactant FeCl₃ shows iron carrying 3+, so the hydroxide is Fe(OH)₃",
"feedback": "This answer is wrong — FeCl₃ shows this iron is 3+, so it pairs with three hydroxide ions: Fe(OH)₃."
}
]
feedback
{
"keyExplanation": "Swap partners: the products are Fe(OH)₃ and KCl. Hydroxides are usually insoluble and iron(III) is not a rescue ion, so Fe(OH)₃ forms the rust-orange solid — Fe(OH)₃(s). KCl stays dissolved as KCl(aq). The 3+ iron pairs with THREE hydroxide ions, so the formula needs the (OH)₃ group.",
"anticipatedWrong": [
{
"value": "KCl",
"misconception": "you picked the soluble product — potassium guarantees KCl stays dissolved; the solid is the iron(III) hydroxide",
"feedback": "This answer is wrong — KCl is the soluble product: potassium guarantees it. The solid is the iron(III) hydroxide, Fe(OH)₃."
},
{
"value": "FeOH",
"misconception": "you paired the ions one-to-one — iron here is 3+ and hydroxide is 1−, so three hydroxides are needed: Fe(OH)₃",
"feedback": "This answer is wrong — the charges do not balance one-to-one: Fe³⁺ needs three OH⁻ ions, so the formula is Fe(OH)₃."
},
{
"value": "Fe(OH)₂",
"misconception": "you used iron(II)'s charge — the reactant FeCl₃ shows iron carrying 3+, so the hydroxide is Fe(OH)₃",
"feedback": "This answer is wrong — FeCl₃ shows this iron is 3+, so it pairs with three hydroxide ions: Fe(OH)₃."
}
]
}
SOL-033 · hard · transfer · seq 19 · authored (blueprint row 19 — transfer case rotation, form D) · mcq4
A pigment workshop needs a mixture that drops a colored solid out of solution. Which pair of solutions produces a precipitate when mixed?
Cu(NO₃)₂ and RbOH KEY
ZnCl₂ and KNO₃
Li₂SO₄ and NaBr
NH₄I and Cs₂SO₄
SOL-033 · hard · transfer · seq 20 · authored (blueprint row 20 — transfer case rotation, form D) · mcq4
Solutions of Na₂S and NiSO₄ are mixed. Which statement describes the outcome the way a chemist would write it?
NiS(s) forms as a solid, and the other product stays dissolved as Na₂SO₄(aq). KEY
Na₂SO₄(s) forms as a solid, and the other product stays dissolved as NiS(aq).
Both products are written (aq) — the sulfide solid dissolves back almost immediately.
Both products are written (s) — swapping partners always produces two solids.
SOL-033 · medium · application · seq 21 · authored (blueprint row 21 — UNSEEDED capstone-chain FRQ; no-precipitate case (all products guarantee-settled); rubric 30/40/30; no 5-line scaffold — no formula substitution anywhere) · frq
Solutions of CsC₂H₃O₂ and LiBr are mixed.
Solutions of CsC₂H₃O₂ and LiBr are mixed.
(a) Swap the partners and write correct formulas for BOTH candidate products.
(b) Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
(c) Give the chemist's verdict — if any product forms a solid, name it with its (s) label; if none does, write both products with their (aq) labels and explain why the mixture stays clear.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Swap the partners and write correct formulas for BOTH candidate products.
Model: Swap partners: the cesium pairs with bromide and the lithium pairs with acetate. The candidate products are CsBr and LiC2H3O2. Cesium and bromide pair one-to-one; lithium and acetate also pair one-to-one.
Part b. Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
Model: CsBr — cesium is a group 1 metal, so the always-soluble guarantee settles it: soluble. LiC2H3O2 — lithium is a group 1 metal and acetate is an always-soluble family; either guarantee alone settles it: soluble.
Part c. Give the chemist's verdict — if any product forms a solid, name it with its (s) label; if none does, write both products with their (aq) labels and explain why the mixture stays clear.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: No precipitate forms. Both products stay dissolved — CsBr(aq) and LiC2H3O2(aq) — because both are soluble, and if both products are soluble no solid can form. An observer sees the mixture stay completely clear.
Full model answer. (a) Swap partners: the cesium pairs with bromide and the lithium pairs with acetate. The candidate products are CsBr and LiC2H3O2. Cesium and bromide pair one-to-one; lithium and acetate also pair one-to-one.
(b) CsBr — cesium is a group 1 metal, so the always-soluble guarantee settles it: soluble. LiC2H3O2 — lithium is a group 1 metal and acetate is an always-soluble family; either guarantee alone settles it: soluble.
(c) No precipitate forms. Both products stay dissolved — CsBr(aq) and LiC2H3O2(aq) — because both are soluble, and if both products are soluble no solid can form. An observer sees the mixture stay completely clear.
Which negative-ion family has “insoluble” as its usual answer in the solubility rules?
Carbonate. KEY
Acetate.
Chloride.
Nitrate.
SOL-029 · medium · application · seq 4 · authored (blueprint row 4 — application case rotation, form E) · mcq4
Potassium acetate, KC₂H₃O₂, is used at airports as a runway de-icer. What do the solubility rules say about it?
It is soluble — both of its ions carry the always-soluble guarantee. KEY
It is insoluble — when both ions carry a guarantee, the two guarantees cancel.
The rules give no verdict here, because its two guarantees point to conflicting answers.
De-icers are designed to melt ice rather than dissolve, so the rules class it as insoluble.
SOL-029 · medium · application · seq 5 · authored (blueprint row 5 — application case rotation, form E) · mcq4
A lab notebook entry reads only “rubidium compound of an unrecorded negative ion — white crystals.” What can the solubility rules conclude?
The crystals are soluble in water, whatever the unrecorded ion is. KEY
Nothing — without the negative ion, the rules cannot be applied at all.
Nothing is settled — the unrecorded ion might be one of rubidium's named exceptions.
White crystals signal an insoluble compound — color is how the rules sort solids.
SOL-029 · medium · application · seq 6 · authored (blueprint row 6 — application case rotation, form E) · mcq4
Exactly one compound below is settled by the always-soluble rule, with no other rule needed. Which one?
NH₄C₂H₃O₂ KEY
FeBr₃
Cu(OH)₂
MgBr₂
SOL-030 · medium · application · seq 7 · authored (blueprint row 7 — application case rotation, form E) · mcq4
One of these chloride compounds stays undissolved in water. Which one?
PbCl₂ KEY
MgCl₂
RbCl
NiCl₂
SOL-030 · medium · application · seq 8 · authored (blueprint row 8 — application case rotation, form E) · mcq4
Which statement about the two exception rosters — the halide exceptions and the sulfate exceptions — is correct?
Lead is on both rosters; silver is only on the halide roster. KEY
Silver is on both rosters; lead is only on the halide roster.
Barium is on the halide roster as well as the sulfate roster.
Calcium's chloride is one of the named insoluble exceptions.
SOL-030 · medium · application · seq 9 · authored (blueprint row 9 — application case rotation, form E) · mcq4
PbBr₂ is insoluble in water, yet Pb(C₂H₃O₂)₂ dissolves freely. Which statement explains why Pb(C₂H₃O₂)₂ dissolves?
The exception applies to the bromide family, not to acetate — and every common acetate compound is soluble. KEY
The rules track the metal, not the negative ion — and this metal is on the rules' list of soluble metals.
Pb(C₂H₃O₂)₂ dissolves because its two acetate ions outnumber the metal ion and pull it into solution.
Pb(C₂H₃O₂)₂ dissolves only because it is usually stored as a fine powder.
SOL-031 · medium · application · seq 10 · authored (blueprint row 10 — application case rotation, form E) · mcq4
One of these carbonate compounds is insoluble in water. Which one?
NiCO₃ KEY
Rb₂CO₃
Li₂CO₃
K₂CO₃
SOL-031 · medium · application · seq 11 · authored (blueprint row 11 — real-world insoluble-family case, form E; stage-3 instance refresh, see fix_log P10) · mcq4
Chromium(III) hydroxide, Cr(OH)₃, is the gray-green solid that settles out in some leather-tanning tanks. What do the solubility rules say about Cr(OH)₃ in water?
It is insoluble. KEY
It is soluble.
It is soluble because tanning tanks hold warm liquid.
Industrial by-products fall outside what the solubility rules cover.
SOL-031 · medium · application · seq 12 · authored (blueprint row 12 — application case rotation, form E) · mcq4
The degreasing cleaner TSP is Na₃PO₄ — and it dissolves freely, even though most phosphates do not. Which statement gives the reason?
Sodium is a group 1 metal, and group 1 compounds of the insoluble families are the soluble exceptions. KEY
Phosphates are actually a usually-soluble family with named metal exceptions.
Cleaning products contain added dissolving agents that force any solid to dissolve.
Na₃PO₄ holds three metal ions, and three positive ions always outvote a single negative ion in solubility.
SOL-032 · medium · application · seq 13 · authored (blueprint row 13 — application case rotation, form E) · mcq4
Which one of these compounds do the solubility rules classify as soluble in water?
Na₂S KEY
PbSO₄
CuCO₃
Zn(OH)₂
SOL-032 · medium · application · seq 14 · authored (blueprint row 14 — application case rotation, form E) · mcq4
Use the solubility rules: which of these nickel compounds is soluble in water?
Ni(NO₃)₂ KEY
NiS
NiCO₃
Ni(OH)₂
SOL-032 · medium · application · seq 15 · authored (blueprint row 15 — application case rotation, form E) · mcq4
A student refuses to classify (NH₄)₂SO₄ without a lab test, “because some sulfates are exceptions.” What did the student get wrong?
The routine's order — the ammonium guarantee settles the compound as soluble before the sulfate rule is consulted. KEY
Nothing — a family that carries named exceptions really does need a beaker test for every single one of its compounds.
Only the roster — ammonium is one of the named sulfate exceptions, so the compound is insoluble.
Only the formula — the compound should have been written NH₄SO₄ before classifying it.
SOL-032 · medium · application · seq 16 · authored (blueprint row 16 — application case rotation, form E) · mcq4
The compound (NH₄)₃PO₄ is run through the classification routine. At which point does the routine settle its verdict?
At the positive-ion check — the positive ion is one of the guarantee ions. KEY
At the negative-ion check — the negative ion is nitrate or acetate.
At the family norm — and the exception check leaves the usual answer standing.
At the exception check — a named exception flips the family's usual answer.
SOL-033 · medium · application · seq 17 · authored (blueprint row 17 — application case rotation, form E) · mcq4
Solutions of NH₄Br and LiC₂H₃O₂ are mixed. Which precipitate, if any, forms?
No precipitate forms KEY
LiBr
NH₄C₂H₃O₂
Both LiBr and NH₄C₂H₃O₂
SOL-033 · hard · transfer · seq 18 · authored (blueprint row 18 — fresh precipitate-production pair, form E) · text
Solutions of ZnBr₂ and Li₂S are mixed, and a white solid appears. Write the formula of the precipitate.
Answer: ZnS KEY
also accepted: ZnS
wrongValueDiagnostics
[
{
"value": "LiBr",
"misconception": "you picked the soluble product — lithium's group 1 guarantee keeps LiBr dissolved (and lithium is no bromide exception); the solid is the zinc sulfide",
"feedback": "This answer is wrong — LiBr is the soluble product: lithium's group 1 guarantee keeps it dissolved. The solid is the zinc sulfide, ZnS."
},
{
"value": "ZnBr₂",
"misconception": "you kept the original partners instead of swapping — ZnBr₂ was already dissolved; the new, insoluble pairing is ZnS",
"feedback": "This answer is wrong — ZnBr₂ is a reactant, already dissolved. Swap partners: the new pairing ZnS is the insoluble product."
},
{
"value": "Zn₂S₂",
"misconception": "you criss-crossed charges that already balance — Zn²⁺ and S²⁻ pair one-to-one, so the formula is ZnS",
"feedback": "This answer is wrong — Zn²⁺ and S²⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is ZnS."
}
]
feedback
{
"keyExplanation": "Swap partners: the products are ZnS and LiBr. Sulfides are usually insoluble and zinc is not a rescue ion, so ZnS forms the white solid — ZnS(s). LiBr stays dissolved as LiBr(aq).",
"anticipatedWrong": [
{
"value": "LiBr",
"misconception": "you picked the soluble product — lithium's group 1 guarantee keeps LiBr dissolved (and lithium is no bromide exception); the solid is the zinc sulfide",
"feedback": "This answer is wrong — LiBr is the soluble product: lithium's group 1 guarantee keeps it dissolved. The solid is the zinc sulfide, ZnS."
},
{
"value": "ZnBr₂",
"misconception": "you kept the original partners instead of swapping — ZnBr₂ was already dissolved; the new, insoluble pairing is ZnS",
"feedback": "This answer is wrong — ZnBr₂ is a reactant, already dissolved. Swap partners: the new pairing ZnS is the insoluble product."
},
{
"value": "Zn₂S₂",
"misconception": "you criss-crossed charges that already balance — Zn²⁺ and S²⁻ pair one-to-one, so the formula is ZnS",
"feedback": "This answer is wrong — Zn²⁺ and S²⁻ have equal-and-opposite charges, so they pair one-to-one. The formula is ZnS."
}
]
}
SOL-033 · hard · transfer · seq 19 · authored (blueprint row 19 — transfer case rotation, form E) · mcq4
A filtration demonstration needs a mixture that visibly clouds with solid the moment it is stirred. Which pair of solutions produces a precipitate when mixed?
Ba(NO₃)₂ and Li₂SO₄ KEY
CsCl and LiNO₃
Mg(NO₃)₂ and Cs₂SO₄
NaBr and KC₂H₃O₂
SOL-033 · hard · transfer · seq 20 · authored (blueprint row 20 — transfer case rotation, form E) · mcq4
Solutions of MnSO₄ and Cs₂CO₃ are mixed. Which statement describes the outcome the way a chemist would write it?
MnCO₃(s) forms as a solid, and the other product stays dissolved as Cs₂SO₄(aq). KEY
Cs₂SO₄(s) forms as a solid, and the other product stays dissolved as MnCO₃(aq).
Both products are written (aq) — the carbonate solid redissolves while the beaker is stirred.
Both products are written (s) — the swap locks both new pairings into solids.
SOL-033 · medium · application · seq 21 · authored (blueprint row 21 — UNSEEDED capstone-chain FRQ; exception-made case (halide roster: silver); rubric 30/40/30; no 5-line scaffold — no formula substitution anywhere) · frq
Solutions of AgNO₃ and RbI are mixed.
Solutions of AgNO₃ and RbI are mixed.
(a) Swap the partners and write correct formulas for BOTH candidate products.
(b) Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
(c) Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Part a. Swap the partners and write correct formulas for BOTH candidate products.
Model: Swap partners: the silver pairs with iodide and the rubidium pairs with nitrate. The candidate products are AgI and RbNO3. Silver and iodide pair one-to-one; rubidium and nitrate also pair one-to-one.
Part b. Use the solubility rules to classify each product as soluble or insoluble, and name the rule or exception that decides each one.
Model: AgI — iodides are usually soluble, BUT silver is one of the two named halide exceptions, so the exception flips it: insoluble. RbNO3 — rubidium is a group 1 metal and nitrate is an always-soluble family; either guarantee alone settles it: soluble.
Part c. Give the chemist's verdict: name the precipitate with its (s) label and the dissolved product with its (aq) label, and say what an observer at the bench would see.
Write your answers in plain letters and digits — subscripts are not required, and plain-digit formulas are accepted.
Model: AgI(s) forms as the precipitate, and RbNO3 stays dissolved as RbNO3(aq). An observer sees a pale-yellow solid cloud the mixture the instant the solutions meet, then settle.
Full model answer. (a) Swap partners: the silver pairs with iodide and the rubidium pairs with nitrate. The candidate products are AgI and RbNO3. Silver and iodide pair one-to-one; rubidium and nitrate also pair one-to-one.
(b) AgI — iodides are usually soluble, BUT silver is one of the two named halide exceptions, so the exception flips it: insoluble. RbNO3 — rubidium is a group 1 metal and nitrate is an always-soluble family; either guarantee alone settles it: soluble.
(c) AgI(s) forms as the precipitate, and RbNO3 stays dissolved as RbNO3(aq). An observer sees a pale-yellow solid cloud the mixture the instant the solutions meet, then settle.
End of Topic Test: Molarity & Dilution and solution stoichiometry (u10-t4 · 5 forms × 23 items · 115 total)
David's notes for James:
CALC-HEAVY FRQ SPLIT - frq_per_form = 3, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): QB-v2 sec.4 directs 2-3 FRQs for calculation-heavy / multi-step-procedure topics, and this gate's three NAMED capstones are exactly three distinct multi-step procedure families, each needing its own end-to-end production evidence: FRQ-1 at seq 21 = the SOL-040 grams -> molarity two-stage chain (Topic 7 capstone), FRQ-2 at seq 22 = the SOL-044 stock-volume rearrangement + finish-the-preparation bookkeeping (solvent to add = V2 - V1; Topic 8 dilution capstone), FRQ-3 at seq 23 = the SOL-045 solution-stoichiometry chain (n = M x V -> mole ratio -> moles of product; unit synthesis capstone). The single-substitution routines SOL-038/039/043 do NOT justify FRQs (one equation, one substitution - the numeric-entry objective rows evidence them fully; same logic U9-G3 used to reject a third FRQ). Total 20 objective + 3 FRQ = 23 items/form. D5 scaling evaluated and NOT triggered: min(20, 6 x 12 lessons = 72) = 20; your target_size=20 concurs.
SEED USAGE (confirm the dispositions): pool 36 = the 12 scope lessons' three gate items each; your 9 demo_curated items are byte-dups of pool[18,19,20,30,31,32,33,34,35] (the three capstone lessons' gate items - verified programmatically, deduped, both refs cited on the placements). 19 of the 20 objective rows are seeded on form A (forms B-E authored fresh); seq 20 is authored on all forms because no pool seed matches the combined-load transfer case (pool[34]/pool[35] are 1:1-ratio single-load chains - held as spares); FRQs 21-23 are unseeded (prod_seeds = []; in-lesson fluency prods are student-seen = spares-only). Face check: 27/36 seeds carried V-T7 freshness collisions on their face (deterministic scan with the shared validator's own corpus machinery; report _seed_face_check_report.json), 7 clean, 2 warn-only. Dominant collision classes: (1) own-lesson material words - nearly every seed says 'dissolved in enough water' / 'diluted with water' / 'add water' and 'water' is spec text in six of the twelve lessons -> freshened to 'made up to a total volume of' / 'diluted to' / 'add solvent' (the course's own vocabulary); (2) spec-burned numbers -> fresh value grids; (3) single-lesson gram joins -> fresh substances/clauses or the two recorded A2 adjudications below; (4) corpus formula 4-grams: Na2SO4/Na2CO3/K2CO3/K2SO4/Pb(NO3)2/Ba(NO3)2 are banned bank-wide, and every precipitate your spec corpus shows in an '(aq) -> X(s)' equation (CaCO3, Mg(OH)2, MgCO3, PbI2, BaSO4) is unusable in ANY authored equation - the corpus-fresh precipitates AgCl, PbSO4, Cu(OH)2, Fe(OH)3 carry the SOL-045 columns instead. Freshenings preserve each seed's skill, case shape, scaffold, and misconception set (SEED RULING strict - the validator was never weakened).
SIG-FIGS CONTRACT (the headline adjudication of this gate, two linked rulings - confirm): (a) NO ROUNDING TAILS: the U9-G4 stem tail 'Round your answer to N significant figures.' could not be carried here - 'round' is not in this gate's corpus lexicon, so the tail grams collide cross-form under V-T7; the course's own SOL-038/039/040/043/044 fluency items carry NO rounding instruction, so removing the tails is course-style faithful. (b) LIMITING-SF KEY (sol QC authoritative overturn, supersedes an earlier alternates-only approach): on 14 numeric items whose exact engineered value overflows the limiting significant figures of the givens, the marked key is the limiting-sig-fig HALF-UP report (e.g. D#11: 0.52 M x 0.3750 L = 0.195 exact -> key 0.20 mol at 2 sf); the exact unrounded value is admitted by the tolerance band, keyExplanations show the rounding line, and answer.gradingNote records the contract on each item. After a further sol ruling, explicit exact-value alternate STRINGS were dropped on the wide-band items (internally inconsistent with the declared half-up rule) - the tolerance band still admits the unrounded value, so students typing the exact result are never penalized.
FRQ SEQ-22 REPORTING CONVENTION (same class, confirm): the model answers report parts (b)/(c) at the limiting 2 significant figures (half-up) and the rubrics name the exact accepted values (e.g. form A: V1 = 76.5 mL accept 76.5 or 77 mL; solvent to add 773.5 accept 773.5/774/773/770 mL). FRQ 21/23 finals are rounding-clean by construction (engineered exact) and carry no such note.
SEQ-7 SEED LICENSE ANALYSIS (confirm - the one place a must-freshen changed what the KEY may claim): pool[10]'s sugar + water instance licensed the course's taught direction claim ('dissolving sugar into 1.0 L of water leaves slightly more than 1.0 L of solution'); our freshened generic-solvent + ionic-salt instances LOSE that license (dissolution does not universally increase volume, and the finished volume is unmeasured). Blind round 1 hard-failed four forms on exactly this rigor gap. The key was rewritten to the PRIMARY SOL-037 rule with no direction claim: 'The NNN mL measures the solvent poured in - molarity divides by the volume of the finished solution.'; feedback and the paired distractor were swept of the direction rider on all five forms.
REALISM SWAPS from sol QC (confirm; full-field sweeps, values preserved): A#17 asked for 5.0 M KNO3 stock - exceeds room-temperature solubility -> NaNO3; C#16 asked for a 4.8 M NaF solution - physically impossible -> CsCl.
DISTRACTOR REWRITES from QC (confirm): E#3 'wrong only about speed' was genuinely defensible against E's stirring claim (unstirred dilution still weakens, slowly) -> replaced with a stirring-removes-solute misconception; seq-4 distractors that paired a correct conclusion with a factually TRUE premise (not cleanly wrong) -> replaced with cleanly-false-premise settling/timing misconceptions on four forms; seq-14's 'solvent added = V2 - V1' distractor asserted volume additivity AND arithmetic outside SOL-042's scope -> replaced with the value-slot-swap error, and the seq-18 feedback line quoting V2 - V1 was dropped; seq-18 keys trimmed to '...add solvent up to a NNN.N mL total.' (longest-option cue); B#4/D#4/E#4 stems now state the solute is fully dissolved.
GATE_CONFIG ADJUDICATIONS - two A2 task-idiom families (recorded, confirm; the shared validator was NOT changed): (1) 'stock solution is needed to prepare' - SOL-044's fluency items all ask the stock-volume question with this exact clause (it IS the lesson's taught task wording), but the runtime A2 derivation misses it because decimal molarities like '2.0 M' break the question-clause extractor's sentence split; (2) 'reacts completely with excess' - SOL-045's extraNotes make the excess statement a BINDING lesson contract (the dissolved reactant is always the limiting one, stated as 'reacts completely with excess X'), and it sits in the setup clause outside the question-clause derivation lane. Both entries whitelist ONLY the task phrasing - every substance name, formula token run, and number in those clauses still scans. No extra_materials, no figure relief this gate.
FRQ-3 STOPS AT MOLES (judgment call, confirm): the optional mass finish is NOT in FRQ3 - each real molar-mass constant may appear only once bank-wide (V-T7 number freshness) and the corpus-fresh precipitates' molar masses are spent on the seq-20 objective column, where the mass finish IS evidenced end-to-end on forms A (PbSO4 303.3) and B (AgCl 143.3); FRQ3 evidences the SOL-045 chain (n = M x V -> ratio -> moles) end-to-end, matching the lesson's own gate-item pattern. Related style call: the seq-18 finish-the-preparation options name NO substance (generic 'stock solution') - the procedure is substance-free in the lesson's own gate item, and it keeps the transfer row free of instance load.
NOTA/AOTA: zero literal hits in your pool; zero replace_distractor flags. V-T5 PASS.
BINDING INSTANCE RULE (course-wide, wave 6): zero uranium/plutonium/radium/actinide/reactor/weapon/dose framings in pool + demo, and the packaging re-sweep over forms_final + this export body is clean; all instances are classroom-safe drink/lab-salt framings.
ACCEPTED ADVISORY CLASSES (recorded, no fix): (a) cross-item cueing - entry rows (seqs 14/17/18) state the routines the FRQ capstones then apply; inherent to the easy->hard blueprint (numbers differ; knowing the rule still requires the computation); (b) definition-echo recall keys on claim-correction items (keyed on the taught rule verbatim by design; V-T4 margins pass); (c) key-position pattern flags = per-reviewer shuffle artifacts (keys are authored at 'a'; student forms get independent export shuffles); (d) 30 sol minors on file, all in three adjudicated classes: stem-does-not-state-precision (the rounding-tail adjudication above - gradingNote + tolerance carry the contract), plainly-wrong-but-fanciful distractor style notes (misconception-keyed by design, sol pass), and guard-digit intermediate notes (sol itself rules them acceptable).
FRQ delivery: your `prod` draw-on-paper class is delivered as a typed diagram-PLAN task (state kinds/recipes/counts/spacing; rubric grades the same observables). Confirm this mapping (pending from the U01-G1 round-trip).
full notes file
# u10-t4 — notes/queries from David's side (U10-G4)
1. CALC-HEAVY FRQ SPLIT - frq_per_form = 3, OVERRIDING your gate data's frq_per_form=1 baseline (confirm the composition): QB-v2 sec.4 directs 2-3 FRQs for calculation-heavy / multi-step-procedure topics, and this gate's three NAMED capstones are exactly three distinct multi-step procedure families, each needing its own end-to-end production evidence: FRQ-1 at seq 21 = the SOL-040 grams -> molarity two-stage chain (Topic 7 capstone), FRQ-2 at seq 22 = the SOL-044 stock-volume rearrangement + finish-the-preparation bookkeeping (solvent to add = V2 - V1; Topic 8 dilution capstone), FRQ-3 at seq 23 = the SOL-045 solution-stoichiometry chain (n = M x V -> mole ratio -> moles of product; unit synthesis capstone). The single-substitution routines SOL-038/039/043 do NOT justify FRQs (one equation, one substitution - the numeric-entry objective rows evidence them fully; same logic U9-G3 used to reject a third FRQ). Total 20 objective + 3 FRQ = 23 items/form. D5 scaling evaluated and NOT triggered: min(20, 6 x 12 lessons = 72) = 20; your target_size=20 concurs.
2. SEED USAGE (confirm the dispositions): pool 36 = the 12 scope lessons' three gate items each; your 9 demo_curated items are byte-dups of pool[18,19,20,30,31,32,33,34,35] (the three capstone lessons' gate items - verified programmatically, deduped, both refs cited on the placements). 19 of the 20 objective rows are seeded on form A (forms B-E authored fresh); seq 20 is authored on all forms because no pool seed matches the combined-load transfer case (pool[34]/pool[35] are 1:1-ratio single-load chains - held as spares); FRQs 21-23 are unseeded (prod_seeds = []; in-lesson fluency prods are student-seen = spares-only). Face check: 27/36 seeds carried V-T7 freshness collisions on their face (deterministic scan with the shared validator's own corpus machinery; report _seed_face_check_report.json), 7 clean, 2 warn-only. Dominant collision classes: (1) own-lesson material words - nearly every seed says 'dissolved in enough water' / 'diluted with water' / 'add water' and 'water' is spec text in six of the twelve lessons -> freshened to 'made up to a total volume of' / 'diluted to' / 'add solvent' (the course's own vocabulary); (2) spec-burned numbers -> fresh value grids; (3) single-lesson gram joins -> fresh substances/clauses or the two recorded A2 adjudications below; (4) corpus formula 4-grams: Na2SO4/Na2CO3/K2CO3/K2SO4/Pb(NO3)2/Ba(NO3)2 are banned bank-wide, and every precipitate your spec corpus shows in an '(aq) -> X(s)' equation (CaCO3, Mg(OH)2, MgCO3, PbI2, BaSO4) is unusable in ANY authored equation - the corpus-fresh precipitates AgCl, PbSO4, Cu(OH)2, Fe(OH)3 carry the SOL-045 columns instead. Freshenings preserve each seed's skill, case shape, scaffold, and misconception set (SEED RULING strict - the validator was never weakened).
3. SIG-FIGS CONTRACT (the headline adjudication of this gate, two linked rulings - confirm): (a) NO ROUNDING TAILS: the U9-G4 stem tail 'Round your answer to N significant figures.' could not be carried here - 'round' is not in this gate's corpus lexicon, so the tail grams collide cross-form under V-T7; the course's own SOL-038/039/040/043/044 fluency items carry NO rounding instruction, so removing the tails is course-style faithful. (b) LIMITING-SF KEY (sol QC authoritative overturn, supersedes an earlier alternates-only approach): on 14 numeric items whose exact engineered value overflows the limiting significant figures of the givens, the marked key is the limiting-sig-fig HALF-UP report (e.g. D#11: 0.52 M x 0.3750 L = 0.195 exact -> key 0.20 mol at 2 sf); the exact unrounded value is admitted by the tolerance band, keyExplanations show the rounding line, and answer.gradingNote records the contract on each item. After a further sol ruling, explicit exact-value alternate STRINGS were dropped on the wide-band items (internally inconsistent with the declared half-up rule) - the tolerance band still admits the unrounded value, so students typing the exact result are never penalized.
4. FRQ SEQ-22 REPORTING CONVENTION (same class, confirm): the model answers report parts (b)/(c) at the limiting 2 significant figures (half-up) and the rubrics name the exact accepted values (e.g. form A: V1 = 76.5 mL accept 76.5 or 77 mL; solvent to add 773.5 accept 773.5/774/773/770 mL). FRQ 21/23 finals are rounding-clean by construction (engineered exact) and carry no such note.
5. SEQ-7 SEED LICENSE ANALYSIS (confirm - the one place a must-freshen changed what the KEY may claim): pool[10]'s sugar + water instance licensed the course's taught direction claim ('dissolving sugar into 1.0 L of water leaves slightly more than 1.0 L of solution'); our freshened generic-solvent + ionic-salt instances LOSE that license (dissolution does not universally increase volume, and the finished volume is unmeasured). Blind round 1 hard-failed four forms on exactly this rigor gap. The key was rewritten to the PRIMARY SOL-037 rule with no direction claim: 'The NNN mL measures the solvent poured in - molarity divides by the volume of the finished solution.'; feedback and the paired distractor were swept of the direction rider on all five forms.
6. REALISM SWAPS from sol QC (confirm; full-field sweeps, values preserved): A#17 asked for 5.0 M KNO3 stock - exceeds room-temperature solubility -> NaNO3; C#16 asked for a 4.8 M NaF solution - physically impossible -> CsCl.
7. DISTRACTOR REWRITES from QC (confirm): E#3 'wrong only about speed' was genuinely defensible against E's stirring claim (unstirred dilution still weakens, slowly) -> replaced with a stirring-removes-solute misconception; seq-4 distractors that paired a correct conclusion with a factually TRUE premise (not cleanly wrong) -> replaced with cleanly-false-premise settling/timing misconceptions on four forms; seq-14's 'solvent added = V2 - V1' distractor asserted volume additivity AND arithmetic outside SOL-042's scope -> replaced with the value-slot-swap error, and the seq-18 feedback line quoting V2 - V1 was dropped; seq-18 keys trimmed to '...add solvent up to a NNN.N mL total.' (longest-option cue); B#4/D#4/E#4 stems now state the solute is fully dissolved.
8. GATE_CONFIG ADJUDICATIONS - two A2 task-idiom families (recorded, confirm; the shared validator was NOT changed): (1) 'stock solution is needed to prepare' - SOL-044's fluency items all ask the stock-volume question with this exact clause (it IS the lesson's taught task wording), but the runtime A2 derivation misses it because decimal molarities like '2.0 M' break the question-clause extractor's sentence split; (2) 'reacts completely with excess' - SOL-045's extraNotes make the excess statement a BINDING lesson contract (the dissolved reactant is always the limiting one, stated as 'reacts completely with excess X'), and it sits in the setup clause outside the question-clause derivation lane. Both entries whitelist ONLY the task phrasing - every substance name, formula token run, and number in those clauses still scans. No extra_materials, no figure relief this gate.
9. FRQ-3 STOPS AT MOLES (judgment call, confirm): the optional mass finish is NOT in FRQ3 - each real molar-mass constant may appear only once bank-wide (V-T7 number freshness) and the corpus-fresh precipitates' molar masses are spent on the seq-20 objective column, where the mass finish IS evidenced end-to-end on forms A (PbSO4 303.3) and B (AgCl 143.3); FRQ3 evidences the SOL-045 chain (n = M x V -> ratio -> moles) end-to-end, matching the lesson's own gate-item pattern. Related style call: the seq-18 finish-the-preparation options name NO substance (generic 'stock solution') - the procedure is substance-free in the lesson's own gate item, and it keeps the transfer row free of instance load.
10. NOTA/AOTA: zero literal hits in your pool; zero replace_distractor flags. V-T5 PASS.
11. BINDING INSTANCE RULE (course-wide, wave 6): zero uranium/plutonium/radium/actinide/reactor/weapon/dose framings in pool + demo, and the packaging re-sweep over forms_final + this export body is clean; all instances are classroom-safe drink/lab-salt framings.
12. ACCEPTED ADVISORY CLASSES (recorded, no fix): (a) cross-item cueing - entry rows (seqs 14/17/18) state the routines the FRQ capstones then apply; inherent to the easy->hard blueprint (numbers differ; knowing the rule still requires the computation); (b) definition-echo recall keys on claim-correction items (keyed on the taught rule verbatim by design; V-T4 margins pass); (c) key-position pattern flags = per-reviewer shuffle artifacts (keys are authored at 'a'; student forms get independent export shuffles); (d) 30 sol minors on file, all in three adjudicated classes: stem-does-not-state-precision (the rounding-tail adjudication above - gradingNote + tolerance carry the contract), plainly-wrong-but-fanciful distractor style notes (misconception-keyed by design, sol pass), and guard-digit intermediate notes (sol itself rules them acceptable).
A hospital stocks two 350 mL bags of dextrose drip solution. Bag 1's label reads '6.5 g of dextrose'; bag 2's reads '13.5 g of dextrose'. A nurse needs the more dilute bag. Which bag is it, and why?
Bag 1, because it holds less dextrose in the same amount of solution. KEY
Bag 2, because more dextrose makes a solution more dilute.
Neither — equal volumes always mean equal concentration.
Bag 1, because it has sat in storage longer, and stored solutions settle weaker.
SOL-035 · medium · application · seq 5 · seed:pool[4] (SOL-035 gate[1]) — freshened per must_freshen_instance: salt→NH₄Cl; misconception set kept · mcq4
A beaker of NH₄Cl solution is already saturated with NH₄Cl, and a layer of undissolved NH₄Cl sits on the bottom. A student stirs in another spoonful of NH₄Cl, and it too settles undissolved. What happens to the solution's concentration?
It stays the same, because only dissolved solute counts. KEY
It increases, because the container now holds more NH₄Cl.
It decreases, because the settled layer pulls dissolved solute down with it.
It increases slowly as the settled layer dissolves over the next few days.
A 200 mL glass of KCl solution has more solvent added until the total volume is 800 mL. No KCl is added or removed. What happens to the solution's concentration?
It decreases. KEY
It increases.
It stays the same.
It cannot be predicted without knowing the mass of KCl.
SOL-037 · medium · application · seq 7 · seed:pool[10] (SOL-037 gate[1]) — freshened per must_freshen_instance: water/sugar→solvent/KI, 1.0 L/0.50 mol→600 mL/0.33 mol; trap + misconception set kept · mcq4
A student pours exactly 600 mL of solvent into a flask, dissolves 0.33 mol of KI in it, and labels the flask 0.55 M. Why is the label not exactly right?
The 600 mL measures the solvent poured in — molarity divides by the volume of the finished solution. KEY
Nothing is wrong — 0.33 mol dissolved in 600 mL of solvent puts exactly 0.55 mol in each liter of solution.
The label should state grams — molarity counts the grams of KI dissolved in each liter, not the moles.
The flask should be labeled with its volume instead, because molarity and volume are the same reading.
SOL-038 · medium · application · seq 8 · seed:pool[12] (SOL-038 gate[0]) — freshened per must_freshen_instance: Na₂SO₄ (banned corpus 4-gram)→LiBr, 0.75/2.5→0.84/2.8; scaffold + misconception space kept · numeric
A solution contains 0.84 mol of LiBr in 2.8 L of solution. What is its molarity, in mol/L?
Answer: 0.3 mol/L KEY
also accepted: 0.30 mol/L
wrongValueDiagnostics
[
{
"value": 3.333,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.84 ÷ 2.8",
"feedback": null
},
{
"value": 2.352,
"misconception": "you multiplied n by V — molarity DIVIDES the moles by the volume: 0.84 ÷ 2.8",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.84 mol\nV = 2.8 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.84 ÷ 2.8\nM = 0.30 M"
}
A beaker holds 0.040 mol of AgNO₃ dissolved in 200.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.2 mol/L KEY
also accepted: 0.20 mol/L
wrongValueDiagnostics
[
{
"value": 0.0002,
"misconception": "you divided by the milliliters — convert first: 200.0 mL = 0.2000 L, then divide",
"feedback": null
},
{
"value": 5.0,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.040 ÷ 0.2000",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.040 mol\nV = 200.0 mL = 0.2000 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.040 ÷ 0.2000\nM = 0.20 M"
}
SOL-039 · medium · application · seq 10 · seed:pool[15] (SOL-039 gate[0]) — freshened per must_freshen_instance: Na₂SO₄ (banned gram) + burned 0.40/1.25 → KIO₃ 0.42 M/2.0 L; 'Make n the subject' scaffold + misconception space kept · numeric
A storage bottle holds 2.0 L of a 0.42 M KIO₃ solution. What amount of KIO₃, in moles, does it hold?
Answer: 0.84 mol KEY
also accepted: 0.84 mol
wrongValueDiagnostics
[
{
"value": 0.21,
"misconception": "you divided M by V — make n the subject of M = n ÷ V and MULTIPLY: 0.42 × 2.0",
"feedback": null
},
{
"value": 4.7619,
"misconception": "you divided V by M — make n the subject of M = n ÷ V and MULTIPLY: 0.42 × 2.0",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.42 M\nV = 2.0 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.42 × 2.0\nn = 0.84 mol"
}
SOL-039 · medium · application · seq 11 · seed:pool[16] (SOL-039 gate[1]) — freshened per must_freshen_instance: burned 0.20 → 0.24; NH₄Cl column-locked to seq 5 → NH₄NO₃; 750.0 kept, ledgered to seq 11 · numeric
A procedure uses 750.0 mL of a 0.24 M NH₄NO₃ solution. How many moles of NH₄NO₃ does that volume deliver?
Answer: 0.18 mol KEY
also accepted: 0.18 mol
wrongValueDiagnostics
[
{
"value": 180.0,
"misconception": "you multiplied by the milliliters — convert first: 750.0 mL = 0.7500 L, then multiply",
"feedback": null
},
{
"value": 0.32,
"misconception": "you divided instead of multiplying — make n the subject of M = n ÷ V: n = M × V = 0.24 × 0.7500",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.24 M\nV = 750.0 mL = 0.7500 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.24 × 0.7500\nn = 0.18 mol"
}
SOL-040 · medium · application · seq 12 · seed:pool[18] = demo_curated[0] (SOL-040 gate[0]) — freshened per must_freshen_instance: 'dissolved in enough water to make'→'made up to a total volume of', 29.25 g/1.25 L→46.8 g/2.50 L; NaCl + 58.5 kept (ledgered once bank-wide) · numeric
46.8 g of NaCl (molar mass of NaCl = 58.5 g/mol) is dissolved and the solution is made up to a total volume of 2.50 L. What is the solution's molarity, in mol/L?
Answer: 0.32 mol/L KEY
also accepted: 0.320 mol/L
wrongValueDiagnostics
[
{
"value": 18.72,
"misconception": "you divided grams by volume — the mass must first become moles: 46.8 ÷ 58.5 = 0.800 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.8,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.800 ÷ 2.50",
"feedback": null
},
{
"value": 0.5,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 46.8 ÷ 58.5",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 46.8 g\nmolar mass = 58.5 g/mol\nV = 2.50 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 46.8 ÷ 58.5\nn = 0.800 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.800 ÷ 2.50\nM = 0.320 M"
}
SOL-040 · medium · application · seq 13 · seed:pool[19] = demo_curated[1] (SOL-040 gate[1]) — freshened per must_freshen_instance: 'enough water'→'made up to', 11.10 g/200.0 mL→13.32 g/480.0 mL (200.0 ledger-locked to seq 9); CaCl₂ + 111.0 kept (ledgered once) · numeric
13.32 g of CaCl₂ (molar mass of CaCl₂ = 111.0 g/mol) is dissolved and the solution is made up to a total volume of 480.0 mL. What is the solution's molarity, in mol/L?
Answer: 0.25 mol/L KEY
also accepted: 0.250 mol/L
wrongValueDiagnostics
[
{
"value": 27.75,
"misconception": "you divided grams by volume — the mass must first become moles: 13.32 ÷ 111.0 = 0.1200 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.12,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.1200 ÷ 0.4800",
"feedback": null
},
{
"value": 17.3611,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 13.32 ÷ 111.0",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 13.32 g\nmolar mass = 111.0 g/mol\nV = 480.0 mL = 0.4800 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 13.32 ÷ 111.0\nn = 0.1200 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.1200 ÷ 0.4800\nM = 0.250 M"
}
A student dilutes 15.0 mL of a 3.5 M RbF solution to a total volume of 165.0 mL. Which assignment of the dilution-equation symbols is correct?
M₁ = 3.5 M, V₁ = 15.0 mL, V₂ = 165.0 mL, and M₂ is the unknown. KEY
M₁ = 3.5 M, V₁ = 165.0 mL, V₂ = 15.0 mL, and M₂ is the unknown.
M₂ = 3.5 M, V₂ = 15.0 mL, V₁ = 165.0 mL, and M₁ is the unknown.
M₁ = 15.0 M, V₁ = 3.5 mL, V₂ = 165.0 mL, and M₂ is the unknown.
SOL-043 · medium · application · seq 15 · seed:pool[28] (SOL-043 gate[1]) — freshened per must_freshen_instance: AgNO₃ (column-locked to seq 9)→LiCl, burned 0.80/25.0/500.0→0.90/35.0/630.0; 'Make M₂ the subject' scaffold + swapped-ratio misconception kept · numeric
35.0 mL of a 0.90 M LiCl solution is diluted to a total volume of 630.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.05 mol/L KEY
also accepted: 0.050 mol/L
wrongValueDiagnostics
[
{
"value": 16.2,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 0.90 M",
"feedback": null
},
{
"value": 0.9,
"misconception": "you carried the starting molarity across — dilution lowers the molarity; run M₂ = M₁V₁ ÷ V₂",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 0.90 M\nV₁ = 35.0 mL\nV₂ = 630.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (0.90 × 35.0) ÷ 630.0\nM₂ = 0.050 M\nSanity check: 0.050 M is smaller than the starting 0.90 M, as a dilution demands."
}
SOL-043 · medium · application · seq 16 · seed:pool[27] (SOL-043 gate[0]) — freshened per must_freshen_instance: 'water'→'pure solvent', KCl (column-locked to seq 6)→NaBr, 150.0/50.0→190.0/60.0; additivity assumption + V₂-total scaffold line kept · numeric
A chemist adds 190.0 mL of pure solvent to 60.0 mL of a 2.0 M NaBr solution. Assuming the volumes simply add, what is the molarity of the diluted solution, in mol/L?
Answer: 0.48 mol/L KEY
also accepted: 0.48 mol/L
wrongValueDiagnostics
[
{
"value": 0.6316,
"misconception": "you divided by the solvent added — V₂ is the TOTAL volume: 60.0 + 190.0 = 250.0 mL",
"feedback": null
},
{
"value": 8.3333,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 2.0 M",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 2.0 M\nV₁ = 60.0 mL\nV₂ = 60.0 + 190.0 = 250.0 mL — the TOTAL volume, not the solvent added\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (2.0 × 60.0) ÷ 250.0\nM₂ = 0.48 M\nSanity check: 0.48 M is smaller than the starting 2.0 M, as a dilution demands."
}
SOL-044 · medium · application · seq 17 · seed:pool[30] = demo_curated[3] (SOL-044 gate[0]) — freshened per must_freshen_instance: NH₄Cl (column-locked to seq 5)→KNO₃ stock ('KNO₃ solution at' dodges the corpus 'm kno 3 solution' gram), burned 4.5/0.30/900.0→5.0/0.45/640.0; task idiom rides the recorded A2 gate_config adjudication · numeric
What volume of a 5.0 M NaNO₃ stock solution is needed to prepare 640.0 mL of NaNO₃ solution at 0.325 M? Give your answer in mL.
Answer: 42 mL KEY
also accepted: 42 mL
wrongValueDiagnostics
[
{
"value": 9846.2,
"misconception": "you swapped the molarities — and the sanity check flags it: the stock is stronger than the target, so V₁ must come out SMALLER than 640.0 mL",
"feedback": null
},
{
"value": 208.0,
"misconception": "you forgot to divide by the stock's molarity M₁ = 5.0 — V₁ = M₂V₂ ÷ M₁",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 5.0 M\nM₂ = 0.325 M\nV₂ = 640.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake V₁ the subject:\nV₁ = M₂V₂ ÷ M₁\nSubstitute in the values, and calculate:\nV₁ = (0.325 × 640.0) ÷ 5.0\nV₁ = 41.6 mL ≈ 42 mL (2 significant figures — set by the 5.0 M given)\nSanity check: the stock is stronger than the target, so V₁ (41.6 mL) must come out smaller than V₂ (640.0 mL) — and it does."
}
SOL-044 · hard · transfer · seq 18 · seed:pool[32] (SOL-044 gate[2]) — freshened per must_freshen_instance: 'add water'→'add solvent' in every option, 20.0/1.5/300.0/0.10→34.0/3.8/680.0/0.19; option grid kept · mcq4
A student calculates that 34.0 mL of a 3.8 M stock solution is needed to prepare 680.0 mL of a 0.19 M solution. What is the correct way to finish the preparation?
Measure 34.0 mL of stock, then add solvent up to a 680.0 mL total. KEY
Measure 34.0 mL of stock, then add 680.0 mL of solvent to it.
Measure 34.0 mL of stock, then add 34.0 mL of solvent.
Add solvent to the whole stock bottle until its label value falls to 0.19 M.
SOL-045 · hard · transfer · seq 19 · seed:pool[33] = demo_curated[6] (SOL-045 gate[0]) — freshened per must_freshen_instance: CuSO₄→CuCl₂ (seq-19 Cu(OH)₂ column), burned 0.150/0.40→0.35/0.70; excess contract rides the recorded A2 adjudication; ignored-ratio/inverted-ratio wrong values kept · numeric
0.35 L of a 0.70 M CuCl₂ solution reacts completely with excess NaOH: CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq). How many moles of NaOH are used up?
Answer: 0.49 mol KEY
also accepted: 0.49 mol
wrongValueDiagnostics
[
{
"value": 0.245,
"misconception": "you ignored the mole ratio — the equation uses 2 mol of NaOH for every 1 mol of CuCl₂, so multiply by 2",
"feedback": null
},
{
"value": 0.1225,
"misconception": "you applied the ratio upside down — NaOH carries the 2 and CuCl₂ the 1, so MULTIPLY the CuCl₂ moles by 2",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.70 M\nV = 0.35 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(CuCl₂) = 0.70 × 0.35\nn(CuCl₂) = 0.245 mol\nUse the mole ratio from the balanced equation:\n1 mol CuCl₂ → 2 mol NaOH\nn(NaOH) = 0.245 × 2\nn(NaOH) = 0.49 mol"
}
SOL-045 · hard · transfer · seq 20 · authored on all forms per the blueprint (no pool seed matches the combined-load case): lead(II) acetate mass finish; PbSO₄ 303.3 used once bank-wide; 200.0 reused within form A only (seed A#9 holds it) — recorded · numeric
200.0 mL of a 1.0 M lead(II) acetate solution reacts completely with excess ZnSO₄: Pb(C₂H₃O₂)₂(aq) + ZnSO₄(aq) → PbSO₄(s) + Zn(C₂H₃O₂)₂(aq). What mass of PbSO₄ forms, in grams? (Molar mass of PbSO₄ = 303.3 g/mol.)
Answer: 61 g KEY
also accepted: 61 g
wrongValueDiagnostics
[
{
"value": 0.2,
"misconception": "you stopped at the moles — the question asks for a mass: finish with mass = moles × molar mass = 0.200 × 303.3",
"feedback": null
},
{
"value": 1516.5,
"misconception": "you divided by the moles instead of multiplying — mass = moles × molar mass = 0.200 × 303.3",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 1.0 M\nV = 200.0 mL = 0.2000 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(lead(II) acetate) = 1.0 × 0.2000\nn(lead(II) acetate) = 0.200 mol\nUse the mole ratio from the balanced equation:\n1 mol lead(II) acetate → 1 mol PbSO₄, so n(PbSO₄) = 0.20 mol\nWrite down the equation for the mass:\nmass = moles × molar mass\nSubstitute in the values, and calculate:\nmass = 0.20 × 303.3\nmass = 60.66 g ≈ 61 g (2 significant figures — set by the 1.0 M given)"
}
23.7 g of KMnO₄ (molar mass of KMnO₄ = 158.0 g/mol) is dissolved and the solution is made up to a total volume of 0.750 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
23.7 g of KMnO₄ (molar mass of KMnO₄ = 158.0 g/mol) is dissolved and the solution is made up to a total volume of 0.750 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Part a. 23.7 g of KMnO₄ (molar mass of KMnO₄ = 158.0 g/mol) is dissolved and the solution is made up to a total volume of 0.750 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Model: (a) n = m ÷ molar mass
n = 23.7 ÷ 158.0
n = 0.150 mol
(b) M = n ÷ V
M = 0.150 ÷ 0.750
(c) M = 0.200 M
Full model answer. (a) n = m ÷ molar mass
n = 23.7 ÷ 158.0
n = 0.150 mol
(b) M = n ÷ V
M = 0.150 ÷ 0.750
(c) M = 0.200 M
SOL-044 · medium · application · seq 22 · authored (FRQ2, Topic 8 dilution capstone + finish bookkeeping; the pool[31] water-to-add skill is part (c), seed held as grading reference) · frq
A technician must prepare 850.0 mL of a 0.54 M solution from a 6.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
A technician must prepare 850.0 mL of a 0.54 M solution from a 6.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Part a. A technician must prepare 850.0 mL of a 0.54 M solution from a 6.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Model: (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.54 x 850.0) / 6.0
(b) V1 = 76.5 mL, reported to the limiting 2 significant figures as 77 mL (smaller than V2 = 850.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 850.0 - 76.5 = 773.5 mL, about 770 mL at 2 significant figures
Full model answer. (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.54 x 850.0) / 6.0
(b) V1 = 76.5 mL, reported to the limiting 2 significant figures as 77 mL (smaller than V2 = 850.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 850.0 - 76.5 = 773.5 mL, about 770 mL at 2 significant figures
SOL-045 · medium · application · seq 23 · authored (FRQ3, unit synthesis capstone: n = M × V -> ratio -> moles; mass finish excluded, flagged judgment call) · frq
175.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
175.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Part a. 175.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Model: (a) M = n / V
Make n the subject:
n = M x V
V = 175.0 mL = 0.1750 L
n(FeCl3) = 2.0 x 0.1750
n(FeCl3) = 0.35 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.35 mol
Full model answer. (a) M = n / V
Make n the subject:
n = M x V
V = 175.0 mL = 0.1750 L
n(FeCl3) = 2.0 x 0.1750
n(FeCl3) = 0.35 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.35 mol
Form B
SOL-037 · easy · recall · seq 1 · authored (label-meaning recall; fresh label substance + value per the blueprint case rotation) · mcq4
A stockroom shelf holds a bottle whose label reads 1.8 M RbBr. Which statement says what that label promises?
Each liter of the solution holds 1.8 mol of dissolved RbBr. KEY
Each liter of solvent poured in holds 1.8 mol of dissolved RbBr.
The whole bottle holds 1.8 mol of RbBr, however large the bottle is.
Each liter of the solution holds 1.8 g of dissolved RbBr.
A student says: 'Diluting weakens a solution because some solute sticks to the incoming solvent and is carried out of the mixture.' Which correction is right?
Diluting removes no solute — the same amount of solute is spread through a larger volume. KEY
No correction is needed — incoming solvent really does carry solute out as it mixes in.
The student has the direction wrong — diluting makes a solution stronger, not weaker.
The student is wrong only about speed — the change happens, just slowly.
Two 275 mL cups of rehydration drink are mixed from the same fully dissolved powder: cup 1 with 18 g of powder, cup 2 with 36 g. Which cup holds the more concentrated drink, and why?
Cup 2, because it holds more powder in the same amount of drink. KEY
Cup 1, because less powder makes a drink more concentrated.
Neither — the drinks match because the volumes match.
Cup 2, because powder stirred in more vigorously makes a stronger drink.
A flask of KClO₃ solution is already saturated with KClO₃, and a layer of undissolved KClO₃ sits on the bottom. A student stirs in another spoonful of KClO₃, and it too settles undissolved. What happens to the solution's concentration?
It stays the same, because only dissolved solute counts. KEY
It increases, because the container now holds more KClO₃.
It decreases, because the settled layer pulls dissolved solute down with it.
It increases slowly as the settled layer dissolves over the next few days.
A 140 mL beaker of MgBr₂ solution has more solvent added until the total volume is 700 mL. No MgBr₂ is added or removed. What happens to the solution's concentration?
It decreases. KEY
It increases.
It stays the same.
It cannot be predicted without knowing the mass of MgBr₂.
SOL-037 · medium · application · seq 7 · authored (solution-vs-solvent trap, parallel slot; MgI₂ reused within form B only — recorded) · mcq4
A student pours exactly 575 mL of solvent into a flask, dissolves 0.92 mol of MgI₂ in it, and labels the flask 1.6 M. Why is the label not exactly right?
The 575 mL measures the solvent poured in — molarity divides by the volume of the finished solution. KEY
Nothing is wrong — 0.92 mol dissolved in 575 mL of solvent puts exactly 1.6 mol in each liter of solution.
The label should state grams — molarity counts the grams of MgI₂ dissolved in each liter, not the moles.
The flask should be labeled with its volume instead, because molarity and volume are the same reading.
SOL-038 · medium · application · seq 8 · authored (M = n ÷ V forward in liters, parallel slot; fresh substance + values) · numeric
A solution contains 0.66 mol of RbCl in 1.1 L of solution. What is its molarity, in mol/L?
Answer: 0.6 mol/L KEY
also accepted: 0.60 mol/L
wrongValueDiagnostics
[
{
"value": 1.6667,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.66 ÷ 1.1",
"feedback": null
},
{
"value": 0.726,
"misconception": "you multiplied n by V — molarity DIVIDES the moles by the volume: 0.66 ÷ 1.1",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.66 mol\nV = 1.1 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.66 ÷ 1.1\nM = 0.60 M"
}
A beaker holds 0.090 mol of SrBr₂ dissolved in 450.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.2 mol/L KEY
also accepted: 0.20 mol/L
wrongValueDiagnostics
[
{
"value": 0.0002,
"misconception": "you divided by the milliliters — convert first: 450.0 mL = 0.4500 L, then divide",
"feedback": null
},
{
"value": 5.0,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.090 ÷ 0.4500",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.090 mol\nV = 450.0 mL = 0.4500 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.090 ÷ 0.4500\nM = 0.20 M"
}
SOL-039 · medium · application · seq 10 · authored (n = M × V in liters, parallel slot; fresh substance + values) · numeric
A storage bottle holds 2.0 L of a 1.15 M CsBr solution. What amount of CsBr, in moles, does it hold?
Answer: 2.3 mol KEY
also accepted: 2.3 mol
wrongValueDiagnostics
[
{
"value": 0.575,
"misconception": "you divided M by V — make n the subject of M = n ÷ V and MULTIPLY: 1.15 × 2.0",
"feedback": null
},
{
"value": 1.7391,
"misconception": "you divided V by M — make n the subject of M = n ÷ V and MULTIPLY: 1.15 × 2.0",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 1.15 M\nV = 2.0 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 1.15 × 2.0\nn = 2.3 mol"
}
SOL-039 · medium · application · seq 11 · authored (n = M × V with mL conversion, parallel slot; fresh substance + values) · numeric
A procedure uses 125.0 mL of a 0.32 M Al(NO₃)₃ solution. How many moles of Al(NO₃)₃ does that volume deliver?
Answer: 0.04 mol KEY
also accepted: 0.040 mol
wrongValueDiagnostics
[
{
"value": 40.0,
"misconception": "you multiplied by the milliliters — convert first: 125.0 mL = 0.1250 L, then multiply",
"feedback": null
},
{
"value": 2.56,
"misconception": "you divided instead of multiplying — make n the subject of M = n ÷ V: n = M × V = 0.32 × 0.1250",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.32 M\nV = 125.0 mL = 0.1250 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.32 × 0.1250\nn = 0.040 mol"
}
SOL-040 · medium · application · seq 12 · authored (grams→molarity chain in liters, parallel slot; KOH 56.1 held from spares note) · numeric
16.83 g of KOH (molar mass of KOH = 56.1 g/mol) is dissolved and the solution is made up to a total volume of 1.50 L. What is the solution's molarity, in mol/L?
Answer: 0.2 mol/L KEY
also accepted: 0.200 mol/L
wrongValueDiagnostics
[
{
"value": 11.22,
"misconception": "you divided grams by volume — the mass must first become moles: 16.83 ÷ 56.1 = 0.300 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.3,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.300 ÷ 1.50",
"feedback": null
},
{
"value": 2.2222,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 16.83 ÷ 56.1",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 16.83 g\nmolar mass = 56.1 g/mol\nV = 1.50 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 16.83 ÷ 56.1\nn = 0.300 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.300 ÷ 1.50\nM = 0.200 M"
}
SOL-040 · medium · application · seq 13 · authored (grams→molarity chain with mL conversion, parallel slot; fresh substance + values) · numeric
29.98 g of NaI (molar mass of NaI = 149.9 g/mol) is dissolved and the solution is made up to a total volume of 625.0 mL. What is the solution's molarity, in mol/L?
Answer: 0.32 mol/L KEY
also accepted: 0.320 mol/L
wrongValueDiagnostics
[
{
"value": 47.968,
"misconception": "you divided grams by volume — the mass must first become moles: 29.98 ÷ 149.9 = 0.2000 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.2,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.2000 ÷ 0.6250",
"feedback": null
},
{
"value": 8.0,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 29.98 ÷ 149.9",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 29.98 g\nmolar mass = 149.9 g/mol\nV = 625.0 mL = 0.6250 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 29.98 ÷ 149.9\nn = 0.2000 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.2000 ÷ 0.6250\nM = 0.320 M"
}
44.0 mL of a 3.0 M MgI₂ solution is diluted to a total volume of 440.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.3 mol/L KEY
also accepted: 0.30 mol/L
wrongValueDiagnostics
[
{
"value": 30.0,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 3.0 M",
"feedback": null
},
{
"value": 3.0,
"misconception": "you carried the starting molarity across — dilution lowers the molarity; run M₂ = M₁V₁ ÷ V₂",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 3.0 M\nV₁ = 44.0 mL\nV₂ = 440.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (3.0 × 44.0) ÷ 440.0\nM₂ = 0.30 M\nSanity check: 0.30 M is smaller than the starting 3.0 M, as a dilution demands."
}
A chemist adds 425.0 mL of pure solvent to 75.0 mL of a 4.0 M LiBr solution. Assuming the volumes simply add, what is the molarity of the diluted solution, in mol/L?
Answer: 0.6 mol/L KEY
also accepted: 0.60 mol/L
wrongValueDiagnostics
[
{
"value": 0.7059,
"misconception": "you divided by the solvent added — V₂ is the TOTAL volume: 75.0 + 425.0 = 500.0 mL",
"feedback": null
},
{
"value": 26.6667,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 4.0 M",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 4.0 M\nV₁ = 75.0 mL\nV₂ = 75.0 + 425.0 = 500.0 mL — the TOTAL volume, not the solvent added\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (4.0 × 75.0) ÷ 500.0\nM₂ = 0.60 M\nSanity check: 0.60 M is smaller than the starting 4.0 M, as a dilution demands."
}
SOL-044 · medium · application · seq 17 · authored (stock-volume capstone, parallel slot; NaI reused within form B only — recorded) · numeric
What volume of a 7.0 M NaI stock solution is needed to prepare 560.0 mL of NaI solution at 0.66 M? Give your answer in mL.
Answer: 53 mL KEY
also accepted: 53 mL
wrongValueDiagnostics
[
{
"value": 5939.4,
"misconception": "you swapped the molarities — and the sanity check flags it: the stock is stronger than the target, so V₁ must come out SMALLER than 560.0 mL",
"feedback": null
},
{
"value": 369.6,
"misconception": "you forgot to divide by the stock's molarity M₁ = 7.0 — V₁ = M₂V₂ ÷ M₁",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 7.0 M\nM₂ = 0.66 M\nV₂ = 560.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake V₁ the subject:\nV₁ = M₂V₂ ÷ M₁\nSubstitute in the values, and calculate:\nV₁ = (0.66 × 560.0) ÷ 7.0\nV₁ = 52.8 mL ≈ 53 mL (2 significant figures — set by the 7.0 M given)\nSanity check: the stock is stronger than the target, so V₁ (52.8 mL) must come out smaller than V₂ (560.0 mL) — and it does."
}
SOL-044 · hard · transfer · seq 18 · authored (finish-the-preparation transfer, parallel slot; fresh values) · mcq4
A student calculates that 43.0 mL of a 4.2 M stock solution is needed to prepare 860.0 mL of a 0.21 M solution. What is the correct way to finish the preparation?
Measure 43.0 mL of stock, then add solvent up to an 860.0 mL total. KEY
Measure 43.0 mL of stock, then add 860.0 mL of solvent to it.
Measure 43.0 mL of stock, then add 43.0 mL of solvent.
Add solvent to the whole stock bottle until its label value falls to 0.21 M.
140.0 mL of a 2.0 M Cu(NO₃)₂ solution reacts completely with excess NaOH: Cu(NO₃)₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaNO₃(aq). How many moles of NaOH are used up?
Answer: 0.56 mol KEY
also accepted: 0.56 mol
wrongValueDiagnostics
[
{
"value": 0.28,
"misconception": "you ignored the mole ratio — the equation uses 2 mol of NaOH for every 1 mol of Cu(NO₃)₂, so multiply by 2",
"feedback": null
},
{
"value": 0.14,
"misconception": "you applied the ratio upside down — NaOH carries the 2 and Cu(NO₃)₂ the 1, so MULTIPLY the Cu(NO₃)₂ moles by 2",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 2.0 M\nV = 140.0 mL = 0.1400 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(Cu(NO₃)₂) = 2.0 × 0.1400\nn(Cu(NO₃)₂) = 0.280 mol\nUse the mole ratio from the balanced equation:\n1 mol Cu(NO₃)₂ → 2 mol NaOH\nn(NaOH) = 0.280 × 2\nn(NaOH) = 0.56 mol"
}
SOL-045 · hard · transfer · seq 20 · authored (combined-load mass finish; AgCl 143.3 used once bank-wide; register AgNO₃ + NaCl equation rides the whitelisted rule-line gram) · numeric
80.0 mL of a 5.0 M AgNO₃ solution reacts completely with excess NaCl: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). What mass of AgCl forms, in grams? (Molar mass of AgCl = 143.3 g/mol.)
Answer: 57 g KEY
also accepted: 57 g
wrongValueDiagnostics
[
{
"value": 0.4,
"misconception": "you stopped at the moles — the question asks for a mass: finish with mass = moles × molar mass = 0.400 × 143.3",
"feedback": null
},
{
"value": 358.25,
"misconception": "you divided by the moles instead of multiplying — mass = moles × molar mass = 0.400 × 143.3",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 5.0 M\nV = 80.0 mL = 0.0800 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(AgNO₃) = 5.0 × 0.0800\nn(AgNO₃) = 0.400 mol\nUse the mole ratio from the balanced equation:\n1 mol AgNO₃ → 1 mol AgCl, so n(AgCl) = 0.40 mol\nWrite down the equation for the mass:\nmass = moles × molar mass\nSubstitute in the values, and calculate:\nmass = 0.40 × 143.3\nmass = 57.32 g ≈ 57 g (2 significant figures — set by the 5.0 M given)"
}
SOL-040 · medium · application · seq 21 · authored (FRQ1 parallel slot; LiCl reused within-gate at seq 15 form A — different form, gram-distinct contexts, recorded) · frq
8.48 g of LiCl (molar mass of LiCl = 42.4 g/mol) is dissolved and the solution is made up to a total volume of 0.625 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
8.48 g of LiCl (molar mass of LiCl = 42.4 g/mol) is dissolved and the solution is made up to a total volume of 0.625 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Part a. 8.48 g of LiCl (molar mass of LiCl = 42.4 g/mol) is dissolved and the solution is made up to a total volume of 0.625 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Model: (a) n = m ÷ molar mass
n = 8.48 ÷ 42.4
n = 0.200 mol
(b) M = n ÷ V
M = 0.200 ÷ 0.625
(c) M = 0.320 M
Full model answer. (a) n = m ÷ molar mass
n = 8.48 ÷ 42.4
n = 0.200 mol
(b) M = n ÷ V
M = 0.200 ÷ 0.625
(c) M = 0.320 M
A technician must prepare 725.0 mL of a 0.64 M solution from a 5.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
A technician must prepare 725.0 mL of a 0.64 M solution from a 5.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Part a. A technician must prepare 725.0 mL of a 0.64 M solution from a 5.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Model: (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.64 x 725.0) / 5.0
(b) V1 = 92.8 mL, reported to the limiting 2 significant figures as 93 mL (smaller than V2 = 725.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 725.0 - 92.8 = 632.2 mL, about 630 mL at 2 significant figures
Full model answer. (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.64 x 725.0) / 5.0
(b) V1 = 92.8 mL, reported to the limiting 2 significant figures as 93 mL (smaller than V2 = 725.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 725.0 - 92.8 = 632.2 mL, about 630 mL at 2 significant figures
235.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
235.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Part a. 235.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Model: (a) M = n / V
Make n the subject:
n = M x V
V = 235.0 mL = 0.2350 L
n(FeCl3) = 2.0 x 0.2350
n(FeCl3) = 0.47 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.47 mol
Full model answer. (a) M = n / V
Make n the subject:
n = M x V
V = 235.0 mL = 0.2350 L
n(FeCl3) = 2.0 x 0.2350
n(FeCl3) = 0.47 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.47 mol
Form C
SOL-037 · easy · recall · seq 1 · authored (definition recall — the two quantities of M = n ÷ V; no instance words) · mcq4
Molarity puts a number on concentration. Which two measurements does it divide?
The moles of dissolved solute, divided by the volume of the solution in liters. KEY
The grams of dissolved solute, divided by the volume of the solution in liters.
The moles of dissolved solute, divided by the volume of solvent poured in.
The volume of the solution in liters, divided by the moles of dissolved solute.
A gardener mixes two 430 mL batches of liquid plant food from the same concentrate: batch 1 with 7.5 g of concentrate, batch 2 with 24 g. Which batch is more dilute, and why?
Batch 1, because it holds less concentrate in the same amount of solution. KEY
Batch 2, because more concentrate makes a batch more dilute.
Neither — matching volumes make the batches equally strong.
Batch 1, because it was mixed first and early batches settle weaker.
A beaker of Na₂S₂O₃ solution is already saturated with Na₂S₂O₃, and a layer of undissolved Na₂S₂O₃ sits on the bottom. A student stirs in another spoonful of Na₂S₂O₃, and it too settles undissolved. What happens to the solution's concentration?
It stays the same, because only dissolved solute counts. KEY
It increases, because the container now holds more Na₂S₂O₃.
It decreases, because the settled layer pulls dissolved solute down with it.
It increases slowly as the settled layer dissolves over the next few days.
A 260 mL flask of RbI solution has more solvent added until the total volume is 780 mL. No RbI is added or removed. What happens to the solution's concentration?
It decreases. KEY
It increases.
It stays the same.
It cannot be predicted without knowing the mass of RbI.
A student pours exactly 650 mL of solvent into a flask, dissolves 0.91 mol of LiI in it, and labels the flask 1.4 M. Why is the label not exactly right?
The 650 mL measures the solvent poured in — molarity divides by the volume of the finished solution. KEY
Nothing is wrong — 0.91 mol dissolved in 650 mL of solvent puts exactly 1.4 mol in each liter of solution.
The label should state grams — molarity counts the grams of LiI dissolved in each liter, not the moles.
The flask should be labeled with its volume instead, because molarity and volume are the same reading.
SOL-038 · medium · application · seq 8 · authored (M = n ÷ V forward in liters, parallel slot; fresh substance + values) · numeric
A solution contains 0.98 mol of Mg(NO₃)₂ in 1.4 L of solution. What is its molarity, in mol/L?
Answer: 0.7 mol/L KEY
also accepted: 0.70 mol/L
wrongValueDiagnostics
[
{
"value": 1.4286,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.98 ÷ 1.4",
"feedback": null
},
{
"value": 1.372,
"misconception": "you multiplied n by V — molarity DIVIDES the moles by the volume: 0.98 ÷ 1.4",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.98 mol\nV = 1.4 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.98 ÷ 1.4\nM = 0.70 M"
}
A beaker holds 0.070 mol of Ca(NO₃)₂ dissolved in 280.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.25 mol/L KEY
also accepted: 0.25 mol/L
wrongValueDiagnostics
[
{
"value": 0.00025,
"misconception": "you divided by the milliliters — convert first: 280.0 mL = 0.2800 L, then divide",
"feedback": null
},
{
"value": 4.0,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.070 ÷ 0.2800",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.070 mol\nV = 280.0 mL = 0.2800 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.070 ÷ 0.2800\nM = 0.25 M"
}
SOL-039 · medium · application · seq 10 · authored (n = M × V in liters, parallel slot; fresh substance + values) · numeric
A storage bottle holds 4.0 L of a 0.85 M Rb₂SO₄ solution. What amount of Rb₂SO₄, in moles, does it hold?
Answer: 3.4 mol KEY
also accepted: 3.4 mol
wrongValueDiagnostics
[
{
"value": 0.2125,
"misconception": "you divided M by V — make n the subject of M = n ÷ V and MULTIPLY: 0.85 × 4.0",
"feedback": null
},
{
"value": 4.7059,
"misconception": "you divided V by M — make n the subject of M = n ÷ V and MULTIPLY: 0.85 × 4.0",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.85 M\nV = 4.0 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.85 × 4.0\nn = 3.4 mol"
}
SOL-039 · medium · application · seq 11 · authored (n = M × V with mL conversion, parallel slot; fresh substance + values) · numeric
A procedure uses 620.0 mL of a 0.45 M Na₂SO₃ solution. How many moles of Na₂SO₃ does that volume deliver?
Answer: 0.28 mol KEY
also accepted: 0.28 mol
wrongValueDiagnostics
[
{
"value": 279.0,
"misconception": "you multiplied by the milliliters — convert first: 620.0 mL = 0.6200 L, then multiply",
"feedback": null
},
{
"value": 0.7258,
"misconception": "you divided instead of multiplying — make n the subject of M = n ÷ V: n = M × V = 0.45 × 0.6200",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.45 M\nV = 620.0 mL = 0.6200 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.45 × 0.6200\nn = 0.279 mol ≈ 0.28 mol (2 significant figures — set by the 0.45 M given)"
}
42.32 g of Sr(NO₃)₂ (molar mass of Sr(NO₃)₂ = 211.6 g/mol) is dissolved and the solution is made up to a total volume of 1.60 L. What is the solution's molarity, in mol/L?
Answer: 0.125 mol/L KEY
also accepted: 0.125 mol/L
wrongValueDiagnostics
[
{
"value": 26.45,
"misconception": "you divided grams by volume — the mass must first become moles: 42.32 ÷ 211.6 = 0.2000 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.2,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.2000 ÷ 1.60",
"feedback": null
},
{
"value": 3.125,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 42.32 ÷ 211.6",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 42.32 g\nmolar mass = 211.6 g/mol\nV = 1.60 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 42.32 ÷ 211.6\nn = 0.2000 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.2000 ÷ 1.60\nM = 0.125 M"
}
SOL-040 · medium · application · seq 13 · authored (grams→molarity chain with mL conversion, parallel slot; fresh substance + values) · numeric
13.33 g of AlCl₃ (molar mass of AlCl₃ = 133.3 g/mol) is dissolved and the solution is made up to a total volume of 160.0 mL. What is the solution's molarity, in mol/L?
Answer: 0.625 mol/L KEY
also accepted: 0.625 mol/L
wrongValueDiagnostics
[
{
"value": 83.3125,
"misconception": "you divided grams by volume — the mass must first become moles: 13.33 ÷ 133.3 = 0.1000 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.1,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.1000 ÷ 0.1600",
"feedback": null
},
{
"value": 62.5,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 13.33 ÷ 133.3",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 13.33 g\nmolar mass = 133.3 g/mol\nV = 160.0 mL = 0.1600 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 13.33 ÷ 133.3\nn = 0.1000 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.1000 ÷ 0.1600\nM = 0.625 M"
}
52.0 mL of a 3.0 M NaH₂PO₄ solution is diluted to a total volume of 390.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.4 mol/L KEY
also accepted: 0.40 mol/L
wrongValueDiagnostics
[
{
"value": 22.5,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 3.0 M",
"feedback": null
},
{
"value": 3.0,
"misconception": "you carried the starting molarity across — dilution lowers the molarity; run M₂ = M₁V₁ ÷ V₂",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 3.0 M\nV₁ = 52.0 mL\nV₂ = 390.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (3.0 × 52.0) ÷ 390.0\nM₂ = 0.40 M\nSanity check: 0.40 M is smaller than the starting 3.0 M, as a dilution demands."
}
A chemist adds 605.0 mL of pure solvent to 55.0 mL of a 4.8 M CsCl solution. Assuming the volumes simply add, what is the molarity of the diluted solution, in mol/L?
Answer: 0.4 mol/L KEY
also accepted: 0.40 mol/L
wrongValueDiagnostics
[
{
"value": 0.4364,
"misconception": "you divided by the solvent added — V₂ is the TOTAL volume: 55.0 + 605.0 = 660.0 mL",
"feedback": null
},
{
"value": 57.6,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 4.8 M",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 4.8 M\nV₁ = 55.0 mL\nV₂ = 55.0 + 605.0 = 660.0 mL — the TOTAL volume, not the solvent added\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (4.8 × 55.0) ÷ 660.0\nM₂ = 0.40 M\nSanity check: 0.40 M is smaller than the starting 4.8 M, as a dilution demands."
}
SOL-044 · medium · application · seq 17 · authored (stock-volume capstone, parallel slot; LiI reused within form C only — recorded) · numeric
What volume of a 4.0 M LiI stock solution is needed to prepare 380.0 mL of LiI solution at 0.34 M? Give your answer in mL.
Answer: 32 mL KEY
also accepted: 32 mL
wrongValueDiagnostics
[
{
"value": 4470.6,
"misconception": "you swapped the molarities — and the sanity check flags it: the stock is stronger than the target, so V₁ must come out SMALLER than 380.0 mL",
"feedback": null
},
{
"value": 129.2,
"misconception": "you forgot to divide by the stock's molarity M₁ = 4.0 — V₁ = M₂V₂ ÷ M₁",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 4.0 M\nM₂ = 0.34 M\nV₂ = 380.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake V₁ the subject:\nV₁ = M₂V₂ ÷ M₁\nSubstitute in the values, and calculate:\nV₁ = (0.34 × 380.0) ÷ 4.0\nV₁ = 32.3 mL ≈ 32 mL (2 significant figures — set by the 4.0 M and 0.34 M givens)\nSanity check: the stock is stronger than the target, so V₁ (32.3 mL) must come out smaller than V₂ (380.0 mL) — and it does."
}
SOL-044 · hard · transfer · seq 18 · authored (finish-the-preparation transfer, parallel slot; fresh values) · mcq4
A student calculates that 37.0 mL of a 4.6 M stock solution is needed to prepare 740.0 mL of a 0.23 M solution. What is the correct way to finish the preparation?
Measure 37.0 mL of stock, then add solvent up to a 740.0 mL total. KEY
Measure 37.0 mL of stock, then add 740.0 mL of solvent to it.
Measure 37.0 mL of stock, then add 37.0 mL of solvent.
Add solvent to the whole stock bottle until its label value falls to 0.23 M.
230.0 mL of a 1.0 M CuBr₂ solution reacts completely with excess NaOH: CuBr₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaBr(aq). How many moles of NaOH are used up?
Answer: 0.46 mol KEY
also accepted: 0.46 mol
wrongValueDiagnostics
[
{
"value": 0.23,
"misconception": "you ignored the mole ratio — the equation uses 2 mol of NaOH for every 1 mol of CuBr₂, so multiply by 2",
"feedback": null
},
{
"value": 0.115,
"misconception": "you applied the ratio upside down — NaOH carries the 2 and CuBr₂ the 1, so MULTIPLY the CuBr₂ moles by 2",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 1.0 M\nV = 230.0 mL = 0.2300 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(CuBr₂) = 1.0 × 0.2300\nn(CuBr₂) = 0.230 mol\nUse the mole ratio from the balanced equation:\n1 mol CuBr₂ → 2 mol NaOH\nn(NaOH) = 0.230 × 2\nn(NaOH) = 0.46 mol"
}
SOL-045 · hard · transfer · seq 20 · authored (combined-load moles chain: ratio-2 + mL conversion; parallel slot) · numeric
340.0 mL of a MgCl₂ solution at 1.0 M reacts completely with excess AgNO₃: MgCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s) + Mg(NO₃)₂(aq). How many moles of AgCl form?
Answer: 0.68 mol KEY
also accepted: 0.68 mol
wrongValueDiagnostics
[
{
"value": 0.34,
"misconception": "you ignored the mole ratio — the equation makes 2 mol of AgCl for every 1 mol of MgCl₂, so multiply by 2",
"feedback": null
},
{
"value": 680.0,
"misconception": "you skipped the mL → L conversion — divide the milliliters by 1000 first: 340.0 mL = 0.3400 L",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 1.0 M\nV = 340.0 mL = 0.3400 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(MgCl₂) = 1.0 × 0.3400\nn(MgCl₂) = 0.340 mol\nUse the mole ratio from the balanced equation:\n1 mol MgCl₂ → 2 mol AgCl\nn(AgCl) = 0.340 × 2\nn(AgCl) = 0.68 mol"
}
SOL-040 · medium · application · seq 21 · authored (FRQ1 parallel slot; KF reused from seq 2 form A — different form, gram-distinct contexts, recorded) · frq
11.62 g of KF (molar mass of KF = 58.1 g/mol) is dissolved and the solution is made up to a total volume of 4.00 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
11.62 g of KF (molar mass of KF = 58.1 g/mol) is dissolved and the solution is made up to a total volume of 4.00 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Part a. 11.62 g of KF (molar mass of KF = 58.1 g/mol) is dissolved and the solution is made up to a total volume of 4.00 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Model: (a) n = m ÷ molar mass
n = 11.62 ÷ 58.1
n = 0.200 mol
(b) M = n ÷ V
M = 0.200 ÷ 4.00
(c) M = 0.0500 M
Full model answer. (a) n = m ÷ molar mass
n = 11.62 ÷ 58.1
n = 0.200 mol
(b) M = n ÷ V
M = 0.200 ÷ 4.00
(c) M = 0.0500 M
A technician must prepare 650.0 mL of a 0.88 M solution from a 8.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
A technician must prepare 650.0 mL of a 0.88 M solution from a 8.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Part a. A technician must prepare 650.0 mL of a 0.88 M solution from a 8.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Model: (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.88 x 650.0) / 8.0
(b) V1 = 71.5 mL, reported to the limiting 2 significant figures as 72 mL (smaller than V2 = 650.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 650.0 - 71.5 = 578.5 mL, about 580 mL at 2 significant figures
Full model answer. (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.88 x 650.0) / 8.0
(b) V1 = 71.5 mL, reported to the limiting 2 significant figures as 72 mL (smaller than V2 = 650.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 650.0 - 71.5 = 578.5 mL, about 580 mL at 2 significant figures
325.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
325.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Part a. 325.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Model: (a) M = n / V
Make n the subject:
n = M x V
V = 325.0 mL = 0.3250 L
n(FeCl3) = 2.0 x 0.3250
n(FeCl3) = 0.65 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.65 mol
Full model answer. (a) M = n / V
Make n the subject:
n = M x V
V = 325.0 mL = 0.3250 L
n(FeCl3) = 2.0 x 0.3250
n(FeCl3) = 0.65 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.65 mol
A student says: 'Diluting weakens a solution because the solute particles themselves get smaller as solvent surrounds them.' Which correction is right?
Diluting removes no solute — the same amount of solute is spread through a larger volume. KEY
No correction is needed — solute particles really do shrink as more solvent surrounds them.
The student has the direction wrong — diluting makes a solution stronger, not weaker.
The student is wrong only about speed — the change happens, just slowly.
Two 320 mL jars hold fabric-dye baths mixed from one fully dissolved dye powder: jar 1 with 5.2 g of powder, jar 2 with 15.6 g. Which jar holds the more concentrated bath, and why?
Jar 2, because it holds more dye powder in the same amount of solution. KEY
Jar 1, because a smaller amount of powder colors each liter more strongly.
Neither — the baths match because each jar holds 320 mL.
Jar 2, because dye baths grow stronger the longer they stand.
A jar of NH₄Br solution is already saturated with NH₄Br, and a layer of undissolved NH₄Br sits on the bottom. A student stirs in another spoonful of NH₄Br, and it too settles undissolved. What happens to the solution's concentration?
It stays the same, because only dissolved solute counts. KEY
It increases, because the container now holds more NH₄Br.
It decreases, because the settled layer pulls dissolved solute down with it.
It increases slowly as the settled layer dissolves over the next few days.
A 210 mL beaker of NaHCO₃ solution has more solvent added until the total volume is 840 mL. No NaHCO₃ is added or removed. What happens to the solution's concentration?
It decreases. KEY
It increases.
It stays the same.
It cannot be predicted without knowing the mass of NaHCO₃.
A student pours exactly 240 mL of solvent into a flask, dissolves 0.156 mol of Zn(NO₃)₂ in it, and labels the flask 0.65 M. Why is the label not exactly right?
The 240 mL measures the solvent poured in — molarity divides by the volume of the finished solution. KEY
Nothing is wrong — 0.156 mol dissolved in 240 mL of solvent puts exactly 0.65 mol in each liter of solution.
The label should state grams — molarity counts the grams of Zn(NO₃)₂ dissolved in each liter, not the moles.
The flask should be labeled with its volume instead, because molarity and volume are the same reading.
SOL-038 · medium · application · seq 8 · authored (M = n ÷ V forward in liters, parallel slot; fresh substance + values) · numeric
A solution contains 0.72 mol of CsNO₃ in 0.96 L of solution. What is its molarity, in mol/L?
Answer: 0.75 mol/L KEY
also accepted: 0.75 mol/L
wrongValueDiagnostics
[
{
"value": 1.3333,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.72 ÷ 0.96",
"feedback": null
},
{
"value": 0.6912,
"misconception": "you multiplied n by V — molarity DIVIDES the moles by the volume: 0.72 ÷ 0.96",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.72 mol\nV = 0.96 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.72 ÷ 0.96\nM = 0.75 M"
}
A beaker holds 0.144 mol of ZnCl₂ dissolved in 360.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.4 mol/L KEY
also accepted: 0.400 mol/L
wrongValueDiagnostics
[
{
"value": 0.0004,
"misconception": "you divided by the milliliters — convert first: 360.0 mL = 0.3600 L, then divide",
"feedback": null
},
{
"value": 2.5,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.144 ÷ 0.3600",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.144 mol\nV = 360.0 mL = 0.3600 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.144 ÷ 0.3600\nM = 0.400 M (3 significant figures — set by the 0.144 mol given)"
}
SOL-039 · medium · application · seq 10 · authored (n = M × V in liters, parallel slot; fresh substance + values) · numeric
A storage bottle holds 2.0 L of a 0.95 M NH₄HCO₃ solution. What amount of NH₄HCO₃, in moles, does it hold?
Answer: 1.9 mol KEY
also accepted: 1.9 mol
wrongValueDiagnostics
[
{
"value": 0.475,
"misconception": "you divided M by V — make n the subject of M = n ÷ V and MULTIPLY: 0.95 × 2.0",
"feedback": null
},
{
"value": 2.1053,
"misconception": "you divided V by M — make n the subject of M = n ÷ V and MULTIPLY: 0.95 × 2.0",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.95 M\nV = 2.0 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.95 × 2.0\nn = 1.9 mol"
}
SOL-039 · medium · application · seq 11 · authored (n = M × V with mL conversion, parallel slot; fresh substance + values) · numeric
A procedure uses 375.0 mL of a 0.52 M CaBr₂ solution. How many moles of CaBr₂ does that volume deliver?
Answer: 0.2 mol KEY
also accepted: 0.20 mol
wrongValueDiagnostics
[
{
"value": 195.0,
"misconception": "you multiplied by the milliliters — convert first: 375.0 mL = 0.3750 L, then multiply",
"feedback": null
},
{
"value": 1.3867,
"misconception": "you divided instead of multiplying — make n the subject of M = n ÷ V: n = M × V = 0.52 × 0.3750",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.52 M\nV = 375.0 mL = 0.3750 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.52 × 0.3750\nn = 0.195 mol ≈ 0.20 mol (2 significant figures — set by the 0.52 M given)"
}
21.98 g of Li₂SO₄ (molar mass of Li₂SO₄ = 109.9 g/mol) is dissolved and the solution is made up to a total volume of 0.800 L. What is the solution's molarity, in mol/L?
Answer: 0.25 mol/L KEY
also accepted: 0.250 mol/L
wrongValueDiagnostics
[
{
"value": 27.475,
"misconception": "you divided grams by volume — the mass must first become moles: 21.98 ÷ 109.9 = 0.2000 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.2,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.2000 ÷ 0.800",
"feedback": null
},
{
"value": 6.25,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 21.98 ÷ 109.9",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 21.98 g\nmolar mass = 109.9 g/mol\nV = 0.800 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 21.98 ÷ 109.9\nn = 0.2000 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.2000 ÷ 0.800\nM = 0.250 M"
}
SOL-040 · medium · application · seq 13 · authored (grams→molarity chain with mL conversion, parallel slot; fresh substance + values) · numeric
20.67 g of LiNO₃ (molar mass of LiNO₃ = 68.9 g/mol) is dissolved and the solution is made up to a total volume of 240.0 mL. What is the solution's molarity, in mol/L?
Answer: 1.25 mol/L KEY
also accepted: 1.25 mol/L
wrongValueDiagnostics
[
{
"value": 86.125,
"misconception": "you divided grams by volume — the mass must first become moles: 20.67 ÷ 68.9 = 0.300 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.3,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.300 ÷ 0.2400",
"feedback": null
},
{
"value": 13.8889,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 20.67 ÷ 68.9",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 20.67 g\nmolar mass = 68.9 g/mol\nV = 240.0 mL = 0.2400 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 20.67 ÷ 68.9\nn = 0.300 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.300 ÷ 0.2400\nM = 1.25 M"
}
28.0 mL of a 2.0 M K₃PO₄ solution is diluted to a total volume of 320.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.18 mol/L KEY
also accepted: 0.18 mol/L
wrongValueDiagnostics
[
{
"value": 22.8571,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 2.0 M",
"feedback": null
},
{
"value": 2.0,
"misconception": "you carried the starting molarity across — dilution lowers the molarity; run M₂ = M₁V₁ ÷ V₂",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 2.0 M\nV₁ = 28.0 mL\nV₂ = 320.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (2.0 × 28.0) ÷ 320.0\nM₂ = 0.175 M ≈ 0.18 M (2 significant figures — set by the 2.0 M given)\nSanity check: 0.175 M is smaller than the starting 2.0 M, as a dilution demands."
}
A chemist adds 195.0 mL of pure solvent to 65.0 mL of a 3.2 M K₂SO₃ solution. Assuming the volumes simply add, what is the molarity of the diluted solution, in mol/L?
Answer: 0.8 mol/L KEY
also accepted: 0.80 mol/L
wrongValueDiagnostics
[
{
"value": 1.0667,
"misconception": "you divided by the solvent added — V₂ is the TOTAL volume: 65.0 + 195.0 = 260.0 mL",
"feedback": null
},
{
"value": 12.8,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 3.2 M",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 3.2 M\nV₁ = 65.0 mL\nV₂ = 65.0 + 195.0 = 260.0 mL — the TOTAL volume, not the solvent added\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (3.2 × 65.0) ÷ 260.0\nM₂ = 0.80 M\nSanity check: 0.80 M is smaller than the starting 3.2 M, as a dilution demands."
}
SOL-044 · medium · application · seq 17 · authored (stock-volume capstone, parallel slot; Zn(NO₃)₂ reused within form D only — recorded) · numeric
What volume of a 3.0 M Zn(NO₃)₂ stock solution is needed to prepare 760.0 mL of Zn(NO₃)₂ solution at 0.375 M? Give your answer in mL.
Answer: 95 mL KEY
also accepted: 95 mL
wrongValueDiagnostics
[
{
"value": 6080.0,
"misconception": "you swapped the molarities — and the sanity check flags it: the stock is stronger than the target, so V₁ must come out SMALLER than 760.0 mL",
"feedback": null
},
{
"value": 285.0,
"misconception": "you forgot to divide by the stock's molarity M₁ = 3.0 — V₁ = M₂V₂ ÷ M₁",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 3.0 M\nM₂ = 0.375 M\nV₂ = 760.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake V₁ the subject:\nV₁ = M₂V₂ ÷ M₁\nSubstitute in the values, and calculate:\nV₁ = (0.375 × 760.0) ÷ 3.0\nV₁ = 95 mL (2 significant figures — set by the 3.0 M given)\nSanity check: the stock is stronger than the target, so V₁ (95.0 mL) must come out smaller than V₂ (760.0 mL) — and it does."
}
SOL-044 · hard · transfer · seq 18 · authored (finish-the-preparation transfer, parallel slot; fresh values) · mcq4
A student calculates that 117.0 mL of a 5.4 M stock solution is needed to prepare 780.0 mL of a 0.81 M solution. What is the correct way to finish the preparation?
Measure 117.0 mL of stock, then add solvent up to a 780.0 mL total. KEY
Measure 117.0 mL of stock, then add 780.0 mL of solvent to it.
Measure 117.0 mL of stock, then add 117.0 mL of solvent.
Add solvent to the whole stock bottle until its label value falls to 0.81 M.
70.0 mL of a 3.0 M CuCl₂ solution reacts completely with excess KOH: CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq). How many moles of KOH are used up?
Answer: 0.42 mol KEY
also accepted: 0.42 mol
wrongValueDiagnostics
[
{
"value": 0.21,
"misconception": "you ignored the mole ratio — the equation uses 2 mol of KOH for every 1 mol of CuCl₂, so multiply by 2",
"feedback": null
},
{
"value": 0.105,
"misconception": "you applied the ratio upside down — KOH carries the 2 and CuCl₂ the 1, so MULTIPLY the CuCl₂ moles by 2",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 3.0 M\nV = 70.0 mL = 0.0700 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(CuCl₂) = 3.0 × 0.0700\nn(CuCl₂) = 0.210 mol\nUse the mole ratio from the balanced equation:\n1 mol CuCl₂ → 2 mol KOH\nn(KOH) = 0.210 × 2\nn(KOH) = 0.42 mol"
}
SOL-045 · hard · transfer · seq 20 · authored (combined-load moles chain: ratio-2 + mL conversion; parallel slot) · numeric
460.0 mL of a MgCl₂ solution at 1.0 M reacts completely with excess AgNO₃: MgCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s) + Mg(NO₃)₂(aq). How many moles of AgCl form?
Answer: 0.92 mol KEY
also accepted: 0.92 mol
wrongValueDiagnostics
[
{
"value": 0.46,
"misconception": "you ignored the mole ratio — the equation makes 2 mol of AgCl for every 1 mol of MgCl₂, so multiply by 2",
"feedback": null
},
{
"value": 920.0,
"misconception": "you skipped the mL → L conversion — divide the milliliters by 1000 first: 460.0 mL = 0.4600 L",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 1.0 M\nV = 460.0 mL = 0.4600 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(MgCl₂) = 1.0 × 0.4600\nn(MgCl₂) = 0.460 mol\nUse the mole ratio from the balanced equation:\n1 mol MgCl₂ → 2 mol AgCl\nn(AgCl) = 0.460 × 2\nn(AgCl) = 0.92 mol"
}
SOL-040 · medium · application · seq 21 · authored (FRQ1 parallel slot; NaClO₃ reused from seq 2 form B — different form, gram-distinct contexts, recorded) · frq
15.96 g of NaClO₃ (molar mass of NaClO₃ = 106.4 g/mol) is dissolved and the solution is made up to a total volume of 0.600 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
15.96 g of NaClO₃ (molar mass of NaClO₃ = 106.4 g/mol) is dissolved and the solution is made up to a total volume of 0.600 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Part a. 15.96 g of NaClO₃ (molar mass of NaClO₃ = 106.4 g/mol) is dissolved and the solution is made up to a total volume of 0.600 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Model: (a) n = m ÷ molar mass
n = 15.96 ÷ 106.4
n = 0.150 mol
(b) M = n ÷ V
M = 0.150 ÷ 0.600
(c) M = 0.250 M
Full model answer. (a) n = m ÷ molar mass
n = 15.96 ÷ 106.4
n = 0.150 mol
(b) M = n ÷ V
M = 0.150 ÷ 0.600
(c) M = 0.250 M
A technician must prepare 900.0 mL of a 0.22 M solution from a 4.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
A technician must prepare 900.0 mL of a 0.22 M solution from a 4.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Part a. A technician must prepare 900.0 mL of a 0.22 M solution from a 4.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Model: (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.22 x 900.0) / 4.0
(b) V1 = 49.5 mL, reported to the limiting 2 significant figures as 50 mL (smaller than V2 = 900.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 900.0 - 49.5 = 850.5 mL, about 850 mL at 2 significant figures
Full model answer. (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.22 x 900.0) / 4.0
(b) V1 = 49.5 mL, reported to the limiting 2 significant figures as 50 mL (smaller than V2 = 900.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 900.0 - 49.5 = 850.5 mL, about 850 mL at 2 significant figures
265.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
265.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Part a. 265.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Model: (a) M = n / V
Make n the subject:
n = M x V
V = 265.0 mL = 0.2650 L
n(FeCl3) = 2.0 x 0.2650
n(FeCl3) = 0.53 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.53 mol
Full model answer. (a) M = n / V
Make n the subject:
n = M x V
V = 265.0 mL = 0.2650 L
n(FeCl3) = 2.0 x 0.2650
n(FeCl3) = 0.53 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.53 mol
Form E
SOL-037 · easy · recall · seq 1 · authored (definition recall — what n counts; no instance words) · mcq4
In a molarity, which amount of the solute is being counted in each liter of solution?
The number of moles of dissolved solute. KEY
The number of grams of dissolved solute.
The number of milliliters the solute took up before dissolving.
Two identical 380 mL vases are filled with cut-flower feed mixed from the same fully dissolved sachet powder: vase 1 with 11 g, vase 2 with 44 g. Which vase holds the more dilute feed, and why?
Vase 1, because it holds less sachet powder in the same amount of solution. KEY
Vase 2, because more powder spreads the feed thinner.
Neither — equal volumes always mean equal concentration.
Vase 1, because it was filled first, and feed settles weaker in the vase that stands longer.
A flask of KAl(SO₄)₂ solution is already saturated with KAl(SO₄)₂, and a layer of undissolved KAl(SO₄)₂ sits on the bottom. A student stirs in another spoonful of KAl(SO₄)₂, and it too settles undissolved. What happens to the solution's concentration?
It stays the same, because only dissolved solute counts. KEY
It increases, because the container now holds more KAl(SO₄)₂.
It decreases, because the settled layer pulls dissolved solute down with it.
It increases slowly as the settled layer dissolves over the next few days.
A 170 mL flask of CsF solution has more solvent added until the total volume is 680 mL. No CsF is added or removed. What happens to the solution's concentration?
It decreases. KEY
It increases.
It stays the same.
It cannot be predicted without knowing the mass of CsF.
A student pours exactly 750 mL of solvent into a flask, dissolves 0.27 mol of NH₄I in it, and labels the flask 0.36 M. Why is the label not exactly right?
The 750 mL measures the solvent poured in — molarity divides by the volume of the finished solution. KEY
Nothing is wrong — 0.27 mol dissolved in 750 mL of solvent puts exactly 0.36 mol in each liter of solution.
The label should state grams — molarity counts the grams of NH₄I dissolved in each liter, not the moles.
The flask should be labeled with its volume instead, because molarity and volume are the same reading.
SOL-038 · medium · application · seq 8 · authored (M = n ÷ V forward in liters, parallel slot; fresh substance + values) · numeric
A solution contains 0.63 mol of KHCO₃ in 0.28 L of solution. What is its molarity, in mol/L?
Answer: 2.3 mol/L KEY
also accepted: 2.3 mol/L
wrongValueDiagnostics
[
{
"value": 0.4444,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.63 ÷ 0.28",
"feedback": null
},
{
"value": 0.1764,
"misconception": "you multiplied n by V — molarity DIVIDES the moles by the volume: 0.63 ÷ 0.28",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.63 mol\nV = 0.28 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.63 ÷ 0.28\nM = 2.25 M ≈ 2.3 M (2 significant figures — set by the 0.63 mol and 0.28 L givens)"
}
A beaker holds 0.230 mol of Na₃PO₄ dissolved in 920.0 mL of solution. What is the solution's molarity, in mol/L?
Answer: 0.25 mol/L KEY
also accepted: 0.250 mol/L
wrongValueDiagnostics
[
{
"value": 0.00025,
"misconception": "you divided by the milliliters — convert first: 920.0 mL = 0.9200 L, then divide",
"feedback": null
},
{
"value": 4.0,
"misconception": "you divided the volume by the moles — M = n ÷ V puts the moles on top: 0.230 ÷ 0.9200",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nn = 0.230 mol\nV = 920.0 mL = 0.9200 L\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.230 ÷ 0.9200\nM = 0.250 M (3 significant figures — set by the 0.230 mol given)"
}
SOL-039 · medium · application · seq 10 · authored (n = M × V in liters, parallel slot; fresh substance + values) · numeric
A storage bottle holds 5.0 L of a 0.68 M KSCN solution. What amount of KSCN, in moles, does it hold?
Answer: 3.4 mol KEY
also accepted: 3.4 mol
wrongValueDiagnostics
[
{
"value": 0.136,
"misconception": "you divided M by V — make n the subject of M = n ÷ V and MULTIPLY: 0.68 × 5.0",
"feedback": null
},
{
"value": 7.3529,
"misconception": "you divided V by M — make n the subject of M = n ÷ V and MULTIPLY: 0.68 × 5.0",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.68 M\nV = 5.0 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.68 × 5.0\nn = 3.4 mol"
}
SOL-039 · medium · application · seq 11 · authored (n = M × V with mL conversion, parallel slot; fresh substance + values) · numeric
A procedure uses 875.0 mL of a 0.56 M ZnSO₄ solution. How many moles of ZnSO₄ does that volume deliver?
Answer: 0.49 mol KEY
also accepted: 0.49 mol
wrongValueDiagnostics
[
{
"value": 490.0,
"misconception": "you multiplied by the milliliters — convert first: 875.0 mL = 0.8750 L, then multiply",
"feedback": null
},
{
"value": 0.64,
"misconception": "you divided instead of multiplying — make n the subject of M = n ÷ V: n = M × V = 0.56 × 0.8750",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 0.56 M\nV = 875.0 mL = 0.8750 L\nWrite down the equation:\nM = n ÷ V\nMake n the subject:\nn = M × V\nSubstitute in the values, and calculate:\nn = 0.56 × 0.8750\nn = 0.49 mol"
}
15.85 g of SrCl₂ (molar mass of SrCl₂ = 158.5 g/mol) is dissolved and the solution is made up to a total volume of 2.00 L. What is the solution's molarity, in mol/L?
Answer: 0.05 mol/L KEY
also accepted: 0.0500 mol/L
wrongValueDiagnostics
[
{
"value": 7.925,
"misconception": "you divided grams by volume — the mass must first become moles: 15.85 ÷ 158.5 = 0.1000 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.1,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.1000 ÷ 2.00",
"feedback": null
},
{
"value": 5.0,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 15.85 ÷ 158.5",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 15.85 g\nmolar mass = 158.5 g/mol\nV = 2.00 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 15.85 ÷ 158.5\nn = 0.1000 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.1000 ÷ 2.00\nM = 0.0500 M"
}
SOL-040 · medium · application · seq 13 · authored (grams→molarity chain with mL conversion, parallel slot; fresh substance + values) · numeric
45.04 g of ZnBr₂ (molar mass of ZnBr₂ = 225.2 g/mol) is dissolved and the solution is made up to a total volume of 800.0 mL. What is the solution's molarity, in mol/L?
Answer: 0.25 mol/L KEY
also accepted: 0.250 mol/L
wrongValueDiagnostics
[
{
"value": 56.3,
"misconception": "you divided grams by volume — the mass must first become moles: 45.04 ÷ 225.2 = 0.2000 mol, then divide by the liters",
"feedback": null
},
{
"value": 0.2,
"misconception": "you stopped at the moles — finish the chain: M = n ÷ V = 0.2000 ÷ 0.8000",
"feedback": null
},
{
"value": 6.25,
"misconception": "you inverted the molar-mass division — the moles are the mass divided by the molar mass: 45.04 ÷ 225.2",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nm = 45.04 g\nmolar mass = 225.2 g/mol\nV = 800.0 mL = 0.8000 L\nFind the moles of solute first:\nn = m ÷ molar mass\nn = 45.04 ÷ 225.2\nn = 0.2000 mol\nWrite down the equation:\nM = n ÷ V\nSubstitute in the values, and calculate:\nM = 0.2000 ÷ 0.8000\nM = 0.250 M"
}
64.0 mL of a 2.0 M (NH₄)₂SO₄ solution is diluted to a total volume of 512.0 mL. What is the molarity of the diluted solution, in mol/L?
Answer: 0.25 mol/L KEY
also accepted: 0.25 mol/L
wrongValueDiagnostics
[
{
"value": 16.0,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 2.0 M",
"feedback": null
},
{
"value": 2.0,
"misconception": "you carried the starting molarity across — dilution lowers the molarity; run M₂ = M₁V₁ ÷ V₂",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 2.0 M\nV₁ = 64.0 mL\nV₂ = 512.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (2.0 × 64.0) ÷ 512.0\nM₂ = 0.25 M\nSanity check: 0.25 M is smaller than the starting 2.0 M, as a dilution demands."
}
A chemist adds 405.0 mL of pure solvent to 105.0 mL of a 3.4 M NaSCN solution. Assuming the volumes simply add, what is the molarity of the diluted solution, in mol/L?
Answer: 0.7 mol/L KEY
also accepted: 0.70 mol/L
wrongValueDiagnostics
[
{
"value": 0.8815,
"misconception": "you divided by the solvent added — V₂ is the TOTAL volume: 105.0 + 405.0 = 510.0 mL",
"feedback": null
},
{
"value": 16.5143,
"misconception": "you used the volume ratio upside down — and the sanity check flags it: a diluted solution must come out BELOW 3.4 M",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 3.4 M\nV₁ = 105.0 mL\nV₂ = 105.0 + 405.0 = 510.0 mL — the TOTAL volume, not the solvent added\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake M₂ the subject:\nM₂ = M₁V₁ ÷ V₂\nSubstitute in the values, and calculate:\nM₂ = (3.4 × 105.0) ÷ 510.0\nM₂ = 0.70 M\nSanity check: 0.70 M is smaller than the starting 3.4 M, as a dilution demands."
}
SOL-044 · medium · application · seq 17 · authored (stock-volume capstone, parallel slot; NH₄I reused within form E only — recorded) · numeric
What volume of a 2.0 M NH₄I stock solution is needed to prepare 950.0 mL of NH₄I solution at 0.14 M? Give your answer in mL.
Answer: 67 mL KEY
also accepted: 67 mL
wrongValueDiagnostics
[
{
"value": 13571.4,
"misconception": "you swapped the molarities — and the sanity check flags it: the stock is stronger than the target, so V₁ must come out SMALLER than 950.0 mL",
"feedback": null
},
{
"value": 133.0,
"misconception": "you forgot to divide by the stock's molarity M₁ = 2.0 — V₁ = M₂V₂ ÷ M₁",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM₁ = 2.0 M\nM₂ = 0.14 M\nV₂ = 950.0 mL\nWrite down the equation:\nM₁V₁ = M₂V₂\nMake V₁ the subject:\nV₁ = M₂V₂ ÷ M₁\nSubstitute in the values, and calculate:\nV₁ = (0.14 × 950.0) ÷ 2.0\nV₁ = 66.5 mL ≈ 67 mL (2 significant figures — set by the 2.0 M and 0.14 M givens)\nSanity check: the stock is stronger than the target, so V₁ (66.5 mL) must come out smaller than V₂ (950.0 mL) — and it does."
}
SOL-044 · hard · transfer · seq 18 · authored (finish-the-preparation transfer, parallel slot; fresh values) · mcq4
A student calculates that 22.0 mL of a 4.4 M stock solution is needed to prepare 880.0 mL of a 0.11 M solution. What is the correct way to finish the preparation?
Measure 22.0 mL of stock, then add solvent up to an 880.0 mL total. KEY
Measure 22.0 mL of stock, then add 880.0 mL of solvent to it.
Measure 22.0 mL of stock, then add 22.0 mL of solvent.
Add solvent to the whole stock bottle until its label value falls to 0.11 M.
185.0 mL of a 2.0 M CuBr₂ solution reacts completely with excess KOH: CuBr₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KBr(aq). How many moles of KOH are used up?
Answer: 0.74 mol KEY
also accepted: 0.74 mol
wrongValueDiagnostics
[
{
"value": 0.37,
"misconception": "you ignored the mole ratio — the equation uses 2 mol of KOH for every 1 mol of CuBr₂, so multiply by 2",
"feedback": null
},
{
"value": 0.185,
"misconception": "you applied the ratio upside down — KOH carries the 2 and CuBr₂ the 1, so MULTIPLY the CuBr₂ moles by 2",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 2.0 M\nV = 185.0 mL = 0.1850 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(CuBr₂) = 2.0 × 0.1850\nn(CuBr₂) = 0.370 mol\nUse the mole ratio from the balanced equation:\n1 mol CuBr₂ → 2 mol KOH\nn(KOH) = 0.370 × 2\nn(KOH) = 0.74 mol"
}
SOL-045 · hard · transfer · seq 20 · authored (combined-load moles chain: ratio-2 + mL conversion; parallel slot) · numeric
540.0 mL of a MgCl₂ solution at 1.0 M reacts completely with excess AgNO₃: MgCl₂(aq) + 2 AgNO₃(aq) → 2 AgCl(s) + Mg(NO₃)₂(aq). How many moles of AgCl form?
Answer: 1.1 mol KEY
also accepted: 1.1 mol
wrongValueDiagnostics
[
{
"value": 0.54,
"misconception": "you ignored the mole ratio — the equation makes 2 mol of AgCl for every 1 mol of MgCl₂, so multiply by 2",
"feedback": null
},
{
"value": 1080.0,
"misconception": "you skipped the mL → L conversion — divide the milliliters by 1000 first: 540.0 mL = 0.5400 L",
"feedback": null
}
]
feedback
{
"keyExplanation": "Write down the values in the question:\nM = 1.0 M\nV = 540.0 mL = 0.5400 L\nFind the moles of the dissolved reactant first:\nM = n ÷ V\nMake n the subject:\nn = M × V\nn(MgCl₂) = 1.0 × 0.5400\nn(MgCl₂) = 0.540 mol\nUse the mole ratio from the balanced equation:\n1 mol MgCl₂ → 2 mol AgCl\nn(AgCl) = 0.540 × 2\nn(AgCl) = 1.08 mol ≈ 1.1 mol (2 significant figures — set by the 1.0 M given)"
}
SOL-040 · medium · application · seq 21 · authored (FRQ1 parallel slot; NaBr reused from seq 16 form A — different form, gram-distinct contexts, recorded) · frq
20.58 g of NaBr (molar mass of NaBr = 102.9 g/mol) is dissolved and the solution is made up to a total volume of 0.160 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
20.58 g of NaBr (molar mass of NaBr = 102.9 g/mol) is dissolved and the solution is made up to a total volume of 0.160 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Part a. 20.58 g of NaBr (molar mass of NaBr = 102.9 g/mol) is dissolved and the solution is made up to a total volume of 0.160 L.
Calculate the molarity of the solution, showing:
(a) the moles of solute, on its own line.
(b) the molarity equation and the substitution.
(c) the result with its unit.
Model: (a) n = m ÷ molar mass
n = 20.58 ÷ 102.9
n = 0.200 mol
(b) M = n ÷ V
M = 0.200 ÷ 0.160
(c) M = 1.25 M
Full model answer. (a) n = m ÷ molar mass
n = 20.58 ÷ 102.9
n = 0.200 mol
(b) M = n ÷ V
M = 0.200 ÷ 0.160
(c) M = 1.25 M
A technician must prepare 820.0 mL of a 0.39 M solution from a 6.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
A technician must prepare 820.0 mL of a 0.39 M solution from a 6.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Part a. A technician must prepare 820.0 mL of a 0.39 M solution from a 6.0 M stock solution of the same solute. Assume the two volumes add together.
Work out the preparation, showing:
(a) the dilution equation, the rearrangement for the stock volume, and the substitution.
(b) the stock volume with its unit.
(c) the volume of solvent that must be added to reach the total.
Model: (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.39 x 820.0) / 6.0
(b) V1 = 53.3 mL, reported to the limiting 2 significant figures as 53 mL (smaller than V2 = 820.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 820.0 - 53.3 = 766.7 mL, about 770 mL at 2 significant figures
Full model answer. (a) M1V1 = M2V2
Make V1 the subject:
V1 = M2V2 / M1
V1 = (0.39 x 820.0) / 6.0
(b) V1 = 53.3 mL, reported to the limiting 2 significant figures as 53 mL (smaller than V2 = 820.0 mL, as the sanity check demands)
(c) solvent to add = V2 - V1 = 820.0 - 53.3 = 766.7 mL, about 770 mL at 2 significant figures
315.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
315.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Part a. 315.0 mL of a 2.0 M FeCl₃ solution reacts completely with excess LiOH: FeCl₃(aq) + 3 LiOH(aq) → Fe(OH)₃(s) + 3 LiCl(aq).
Calculate the moles of Fe(OH)₃ that form. Show:
(a) the moles of the dissolved reactant, starting from the molarity equation, with the volume converted to liters.
(b) the mole ratio you use, taken from the balanced equation.
(c) the moles of Fe(OH)₃ with the unit.
Model: (a) M = n / V
Make n the subject:
n = M x V
V = 315.0 mL = 0.3150 L
n(FeCl3) = 2.0 x 0.3150
n(FeCl3) = 0.63 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.63 mol
Full model answer. (a) M = n / V
Make n the subject:
n = M x V
V = 315.0 mL = 0.3150 L
n(FeCl3) = 2.0 x 0.3150
n(FeCl3) = 0.63 mol
(b) 1 mol FeCl3 -> 1 mol Fe(OH)3
(c) n(Fe(OH)3) = 0.63 mol