Which separation should come first? — answer key

1. Q29 · Practice

Options as numbered on the pupil sheet: 1 — Remove water from the separate filtrate to recover L.; 2 — Add enough water and stir to dissolve L.; 3 — Filter to retain M and collect the solution of L..

Model answer. Add enough water and stir to dissolve L. Filter to retain M and collect solution of L in water. Remove water from that separate filtrate to recover L. Filtering first separates the undissolved component while the solubility difference is still useful.

A full answer includes

• Add enough water to dissolve L; M remains insoluble.
• Filter: M residue, L-in-water filtrate.
• Remove water from filtrate to recover L; earlier evaporation would remix L with M.

Also accept

• A complete property-based dissolve/filter/recover plan with each product named.

Watch for

• “Remove water from the unfiltered mixture first.” — That leaves L and M together again. Use filtration while one component is dissolved and the other is not.

2. Q30 · Practice

Model answer. Filter first, collecting O as residue and solution of N as filtrate. Remove water from the separate filtrate by the given approved procedure to obtain N. N is already dissolved, and no magnetic component is specified, so neither an extra dissolution stage nor magnetic collection follows from the facts.

A full answer includes

• Filter the existing mixture: O residue, N-in-water filtrate.
• Remove water from separate filtrate by the approved procedure to obtain N residue.
• I explained N is already dissolved; no magnetic property/component is supplied, so extra stages are not justified.

Also accept

• A minimal justified filter/recover sequence; real washing/drying may be discussed as qualified practical detail, not asserted yield.

Watch for

• “Every mixture must follow all the simulation's stages.” — Methods are chosen from current components and properties. This sample already contains dissolved N and has no stated magnetic component.

3. Q25 · Apply

Model answer. Filter the existing feed to retain Y as residue and collect solution of X in water as filtrate. Heating the whole feed to dryness is unsuitable because the supplied data say Y decomposes; it also doesn’t selectively collect Y. Recovering X is not required when only Y is wanted. Any real washing/drying must respect Y's stated stability constraint.

A full answer includes

• I rejected heating the whole feed: Y would decompose, and solvent removal does not by itself separate the solids.
• I filtered the existing feed under the stated conditions: Y is residue; water with dissolved X is filtrate.
• Recovery of X is unnecessary for the stated goal; qualified teacher-directed washing/drying of Y may be discussed without heating to the prohibited condition.

Also accept

• A property- and goal-based filtration plan that preserves Y and omits unnecessary recovery of X.

Watch for

• “Every separation must finish by heating everything to dryness.” — The wanted product here is the retained insoluble solid, and the supplied heating condition would change it. Select stages from the actual goal and constraints.

4. Q26 · Apply

Model answer. Label it 'solution of E in water — filtrate'. The liquid passed through the filter. D was retained instead of collected below, and E travelled with water, so the collected liquid is not pure water.

A full answer includes

• Receiver contains solution of E in water, not D in pure water.
• It is filtrate because it passed through the filter.
• D was retained and dissolved E passed, contradicting both the D label and purity claim.

Also accept

• The log establishes water plus dissolved E as the filtrate.

Watch for

• “The receiver label overrides the actual observations.” — A label can be wrong. Trace the components from the observations before accepting it.

5. Q27 · Apply

Model answer. Choose B. It selectively dissolves H without reacting, leaving J to be filtered. A changes H, so it fails recovery of unchanged substances. C dissolves both and doesn’t separate them by ordinary filtration.

A full answer includes

• I chose B: selective dissolution without changing H/J.
• A changes H by reaction, violating unchanged-product goal.
• C dissolves both and puts both in filtrate.

Also accept

• Only B satisfies both supplied constraints.

Watch for

• “A and B are equally suitable because their solubility columns match.” — The goal requires unchanged substances; the reaction information rules out A.

6. Q28 · Apply

Model answer. Use B: it retains the fine grains as residue while salt solution becomes filtrate. A would not remove those fine grains. Neither retains the stated dissolved salt, so changing between these papers doesn’t recover salt from solution.

A full answer includes

• I chose B because it retains the fine grains.
• Fine grains are residue; salt solution is filtrate.
• Dissolved salt passes through either stated paper with water.

Also accept

• The choice depends on actual supplied retention, not the word filter alone.

Watch for

• “A finer paper must remove all dissolved salt too.” — The supplied table explicitly says dissolved salt passes both. Do not invent a capability from the grain result.