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S4 Chemistry: Separation and Purification Paper 3 (solubility, recovery and chromatography comparison)

S4 Chemistry: Separation and Purification Paper 3 (solubility, recovery and chromatography comparison)

  • 2026-11-03

S · 4

S4 Chemistry: Separation and Purification Paper 3 (solubility, recovery and chromatography comparison)

This paper links separation methods to data: solubility decides between filtration and evaporation, and Rf decides whether two spots are the same substance.

Quick answer 36 g per 100 cm3; filter then evaporate; Rf 0.40; 70%; 92.0% and 78.3%; 3.0, 6.0 and 9.0 cm; 0.1875 g/cm3, that is 187.5 g per dm3; thermometer at the side arm.

01 Paper details (check answers afterwards)

Item Detail
Questions 10 questions
Suggested time 25 minutes
Difficulty mix Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter Chapter: S4 Chemistry: Separation and Purification
Level S4

02 Questions (10 items)

  1. FoundationAt room temperature 25 cm3 of water dissolves at most 9 g of a solid. What is the solubility per 100 cm3 of water?
  2. FoundationTo separate a mixture of sand and brine and recover both the sand and the salt, what is the correct order?
  3. FoundationAn unknown sample gives one spot 3.2 cm from the start line while the solvent front is 8.0 cm away. What is the Rf and what does it suggest?
  4. FoundationWhere should the thermometer bulb sit in simple distillation, and why?
  5. IntermediateFractional distillation of 250 cm3 of a liquid mixture collects 175 cm3 of distillate. What is the recovery?
  6. IntermediateA 50 g crude product contains 4 g of impurity. What is its purity, and what is the recovery relative to the pure content if recrystallisation gives 36 g, both to one decimal place?
  7. IntermediateOn one chromatogram the solvent front is 12.0 cm from the start and three spots have Rf 0.25, 0.50 and 0.75. How far is each from the start, and how far is the highest from the front?
  8. ChallengeFive hundred cm3 of brine contains 15 g of salt. After 420 cm3 of pure water is distilled off, 80 cm3 of solution remains. What is its concentration in g/cm3 and in g per dm3?
  9. ChallengeTo obtain large copper(II) sulphate crystals, should the solution be heated rapidly to dryness or slowly evaporated and then cooled, and how do crystallisation and evaporation to dryness differ?
  10. Error redoA student writes that the thermometer bulb must be dipped into the liquid to read the boiling point accurately. What is wrong?

03 Answers and one-line notes

# Answer One-line note
1 36 g per 100 cm3 Nine divided by 25 and multiplied by 100 gives 36, since solubility is quoted per 100 cm3 of solvent.
2 Filter first to collect the sand, then evaporate the filtrate to crystallise the salt Sand is insoluble and is trapped by the filter paper, while dissolved salt passes with the filtrate and is recovered by evaporating the water.
3 Rf 0.40, so it may be the same substance as a known sample with Rf 0.40 Three point two over 8.0 gives 0.40, and a matching Rf is only a clue that should be checked in a second solvent.
4 At the side arm of the distillation flask, to read the temperature of the vapour entering the condenser Only at the side arm does the bulb read the boiling point of what actually distils over; inside the liquid it reads the boiling mixture instead.
5 70% One hundred seventy-five over 250 times 100 gives 70%, with the rest left in the flask or on the glassware.
6 Purity 92.0%; recovery 78.3% The pure content is 46 g, and 46 over 50 gives 92.0%; 36 over 46 gives 78.3%, the rest staying in the mother liquor.
7 3.0 cm, 6.0 cm and 9.0 cm, with the highest 3.0 cm below the front Twelve times 0.25, 0.50 and 0.75 gives 3.0, 6.0 and 9.0 cm, and 12.0 minus 9.0 leaves 3.0 cm.
8 0.1875 g/cm3, that is 187.5 g per dm3 The salt mass stays at 15 g, and 15 over 80 gives 0.1875 g/cm3; since 1 dm3 is 1,000 cm3, that is 187.5 g per dm3.
9 Evaporate slowly to near saturation and then cool; evaporating to dryness gives only a fine powder and keeps the impurities Slow cooling lets the lattice grow while the mother liquor carries most impurities away, whereas evaporation to dryness throws everything down together.
10 The bulb belongs at the side arm to read the vapour temperature; dipped in the liquid it reads the boiling mixture instead. What distils over is fixed by the vapour entering the condenser, while the liquid temperature is raised by impurities and superheating.

04 What to do after marking

  • After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
  • Repeat the same item two days later; move to the next paper only when both attempts are correct.
  • Parents only need the answer and note columns; there is no need to explain every item.

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