How can the salt return? — answer key

1. Q15 · Practice

Model answer. Choose water because it dissolves A while B stays undissolved and can be retained. J dissolves both, so filtration would collect both in solution and would not separate A from B.

A full answer includes

• I chose water: A dissolves, B remains insoluble.
• J dissolves both, so both would pass as solutes rather than become separate residue/filtrate components.

Also accept

• Water selectively dissolves A; J lacks the needed contrast.

Watch for

• “Choose J because dissolving more substances always makes a better separation.” — Here the goal requires separating the components, so both dissolving removes the useful difference.

2. Q16 · Practice

Options as numbered on the pupil sheet: 1 — Together in the residue; 2 — Together in the filtrate; 3 — C in residue; D in filtrate; 4 — D in residue; C in filtrate.

Model answer. With T both solids remain together in the residue. With U both dissolve and pass together in the filtrate. Neither gives the selective dissolution needed for the proposed separation.

A full answer includes

• T leaves C and D together as undissolved residue.
• U puts both into solution/filtrate.
• Neither produces the required separate products by ordinary filtration.

Also accept

• No suitable candidate is supplied for this dissolve/filter method.

Watch for

• “A filter always separates two substances regardless of the solvent.” — The given candidates leave both components in the same relevant phase, so the stated ordinary filter can’t split them.

3. Q17 · Practice

Model answer. Choose R because it dissolves E while leaving F undissolved. F becomes the residue and the solution of E in R becomes the filtrate. S would dissolve both and would not give this separation.

A full answer includes

• I chose R.
• E dissolves in R and F does not.
• F is retained; filtrate contains R and dissolved E.

Also accept

• R provides selective dissolution; retained F is separated from solution containing E.

Watch for

• “R cannot be chosen because it is not called water.” — Suitability follows the stated properties and constraints, not a familiar name.

4. Q18 · Practice

Model answer. The note is insufficient: V must leave H undissolved for the proposed dissolve/filter separation. We also need appropriate filter retention and, for a real procedure, compatibility and safety information.

A full answer includes

• Insufficient information to establish suitability.
• Need to know whether H remains undissolved in V, with grains retained by the stated filter; compatibility/safety also need confirmation for a real method.

Also accept

• Knowing only that G dissolves does not establish a useful contrast with H.

Watch for

• “If one solid dissolves, the other must not.” — Some solvents dissolve both components. The missing information can’t be assumed.

5. Q19 · Practice

Model answer. Sodium chloride remains as the residue while water enters the air as vapour. The salt was already present in the solution; removing water makes it visible instead of creating it.

A full answer includes

• Sodium chloride remains as evaporation residue.
• Water leaves as water vapour into the air.
• Salt was already present dissolved in the starting solution.

Also accept

• The existing dissolved salt is recovered; the solvent is removed.

Watch for

• “The hotplate changes water into salt.” — The starting solution already contains salt; there is no source or reaction given to produce it from water.

6. Q20 · Practice

Model answer. The proposed heating is not justified: the supplied data say P decomposes, so it would not be recovered unchanged. A different validated procedure is needed; the salt example doesn’t establish one for P.

A full answer includes

• The stated procedure is unsuitable for recovering unchanged P.
• Decomposition changes P; sodium chloride's model behaviour cannot be generalised.
• A different validated procedure is needed; no unsupported alternative is required.

Also accept

• Reject this procedure on the stated stability constraint.

Watch for

• “Any solid residue proves that the original solute was recovered unchanged.” — A decomposition product can also be solid. Identity matters, and the question explicitly says P changes.

7. Q21 · Practice

Model answer. The basin can retain salt, but the water leaves into the air and is not collected as liquid. The apparatus therefore doesn’t achieve recovery of both products. A suitable solvent-collection method is needed and will be taught later.

A full answer includes

• Stable salt can remain as a collected residue.
• Water disperses into the surrounding air and is not collected as liquid.
• Collecting the solvent requires additional suitable apparatus/method outside this open-basin model.

Also accept

• Evaporation in this open arrangement recovers the stable solute, not a bottle of water.

Watch for

• “Water recorded in the model's air ledger is collected liquid water.” — Accounting for a destination is different from recovering the material in a usable vessel.

8. Q22 · Practice

Model answer. The measured boundary excludes water after it leaves as vapour. Matter has moved from the basin into the air, so mass inside the basin decreases while the total accounting including that water remains conserved. Salt stays in the basin under these conditions.

A full answer includes

• Water leaves the measured basin/contents and enters the surrounding air.
• The basin is an open system; a smaller mass inside that boundary is not destruction of matter.
• Salt remains under the stated model conditions.

Also accept

• Loss from an open measured system is not loss of matter from the complete account.

Watch for

• “Conservation means every container's mass must stay constant.” — Material can cross a container's boundary. Conservation applies to the complete account, not each open vessel separately.