Solubility and solubility curves – IGCSE Chemistry exam questions with mark scheme

1Grade 1–3Triple only3 marks

Sea water contains dissolved sodium chloride.

(a) Give the units used for the solubility of a solid in water. [1]

(b) State what is meant by a saturated solution. [1]

(a) g per 100 g of water (accept g/100 g (of) solvent; reject g/dm³, g/100 cm³, mol/dm³)(1)
(b) a solution in which no more solute will dissolve (at that temperature) (accept contains the maximum mass of solute that can dissolve; ignore 'fully dissolved'; reject 'no more solvent will dissolve')(1)
(c) solubility increases (accept more solid dissolves (per 100 g water))(1)
Answer: (a) g per 100 g of water; (b) no more solute can dissolve at that temperature; (c) it increases
Examiner tip: Writing g/dm³ loses the units mark: solubility in 4CH1 is always quoted per 100 g of solvent.

2Grade 4–5Triple only3 marks

A student found that a maximum of 8.40 g of potassium sulfate dissolves in 75.0 g of water at 20 °C.

(a) Calculate the solubility of potassium sulfate at 20 °C in g per 100 g of water. [2]

(b) Calculate the maximum mass of potassium sulfate that will dissolve in 250 g of water at 20 °C. [1]

(a) 8.40 ÷ 75.0 (× 100) seen (accept 0.112 or 100 ÷ 75.0 × 8.40)(1)
(a) 11.2 (g per 100 g of water) (correct answer with no working scores 2)(1)
(b) 28.0 g (allow ecf: answer to (a) × 2.5)(1)
Answer: (a) 11.2 g per 100 g of water; (b) 28.0 g
Examiner tip: Divide by the mass of WATER (75.0 g), not by the mass of the solution (83.4 g), then scale to 100 g.

3Grade 6–7Triple only6 marks

The graph shows the solubility curves for potassium chloride and sodium chloride in water.

(a) Use the graph to find the solubility of potassium chloride at 60 °C. [1]

(b) Give the temperature at which the two salts have the same solubility. [1]

(d) A student says that crystals of sodium chloride are best obtained from its solution by cooling, but crystals of potassium chloride are best obtained by evaporating the water.

Explain why the student is wrong. Use the graph. [2]

(a) 46 g per 100 g of water (accept 45–47; units not required but penalise wrong units)(1)
(b) 27 °C (accept 25–28 °C)(1)
(c) reads solubility of sodium chloride at 80 °C as 38 (g per 100 g) AND uses 20.0 ÷ 38 (× 100) (allow readings 37.5–38.5)(1)
(c) 53 g (accept 52–53.5 g; allow ecf from their reading)(1)
(d) solubility of sodium chloride changes very little with temperature / curve almost flat, so cooling produces few/hardly any crystals (so evaporation is needed for NaCl)(1)
(d) solubility of potassium chloride decreases a lot as temperature falls / curve is steep, so cooling a hot saturated solution produces many crystals (so cooling suits KCl)(1)
Answer: (a) 46 g per 100 g water; (b) 27 °C; (c) 53 g; (d) NaCl curve nearly flat so little crystallises on cooling; KCl curve steep so lots crystallises on cooling
Examiner tip: In (c) many students multiply 20 by 38 or divide 38 by 20; set it up as 'mass of water = 100 × mass of solid ÷ solubility'.

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4Grade 4–5Triple only6 marks

A student measured the solubility of potassium chlorate at different temperatures.

Temperature in °C	0	20	40	60	80	100
Solubility in g per 100 g of water	3.3	7.3	13.9	23.8	37.5	56.3

(a) On a grid, plot these results and draw a solubility curve. Put temperature on the horizontal axis. [4]

(b) Use your curve to find the solubility of potassium chlorate at 50 °C. [1]

(c) Use your curve to find the temperature at which the solubility of potassium chlorate is 30 g per 100 g of water. [1]

(a) both axes labelled with quantity and unit, and linear scales chosen so the points cover at least half of the grid in each direction(1)
(a) at least 4 points plotted correctly to within ± half a small square(1)
(a) all 6 points plotted correctly to within ± half a small square(1)
(a) single smooth curve passing through or very close to all points (reject straight lines joining points dot-to-dot; reject a straight line of best fit)(1)
(b) 18 g per 100 g of water (accept 17.5–19.5; allow ecf from candidate's own curve)(1)
(c) 70 °C (accept 67–72 °C; allow ecf from candidate's own curve)(1)
Answer: (a) smooth upward curve through all six points; (b) about 18 g per 100 g water; (c) about 70 °C
Examiner tip: Join solubility points with a smooth curve, not ruled straight lines — the curve mark is lost for dot-to-dot.

5Grade 6–7Triple only6 marks

The graph shows the solubility curve for potassium dichromate in water.

(a) Find the lowest temperature at which 30 g of potassium dichromate will completely dissolve in 100 g of water. [1]

(b) A student makes a saturated solution of potassium dichromate using 25.0 g of water at 70 °C. The solution is then cooled to 30 °C.

Calculate the mass of potassium dichromate crystals that form. [3]

(a) 45 °C (accept 44–46 °C)(1)
(b) reads both solubilities from graph: 58 at 70 °C and 18 at 30 °C (accept 57–59 and 17–19)(1)
(b) difference per 100 g of water = 40 g (allow ecf from their readings) OR scales each reading to 25.0 g of water (14.5 and 4.5 g)(1)
(b) 10 g (accept 9.5–10.5 g; correct answer with no working scores 3)(1)
(c) (orange) crystals / solid appear(s) in the solution (ignore 'precipitate forms' unless crystals/solid also stated; ignore colour changes of solution)(1)
(c) solubility decreases as temperature falls, so the water can no longer hold all the dissolved solid / the solution becomes supersaturated and the excess comes out of solution(1)
Answer: (a) 45 °C; (b) 10 g; (c) crystals appear because solubility falls on cooling so the excess solid crystallises
Examiner tip: Students often forget to scale the 40 g per 100 g down to 25.0 g of water — always check the mass of water given.

6Grade 8–9Triple only7 marks

The table shows the solubility of sodium nitrate in water.

Temperature in °C	20	40	60
Solubility in g per 100 g of water	88	104	124

A solution is made by dissolving 140 g of sodium nitrate in 150 g of water at 60 °C.

(a) Show by calculation that this solution is not saturated at 60 °C. [2]

(b) The solution is cooled slowly. Assume that the solubility changes linearly between 20 °C and 40 °C.

Calculate the temperature at which crystals first appear. [3]

(a) maximum mass that dissolves in 150 g at 60 °C = 124 × 1.5 = 186 g (accept 140 g in 150 g is equivalent to 93(.3) g per 100 g)(1)
(a) comparison: 140 g is less than 186 g / 93.3 is less than 124, so more could dissolve (so not saturated)(1)
(b) concentration of solution = 140 ÷ 150 × 100 = 93.3 g per 100 g of water(1)
(b) (93.3 − 88) ÷ (104 − 88) × 20 = 6.7 °C above 20 °C (or equivalent interpolation)(1)
(b) 27 °C (accept 26.7 °C; allow ecf from their concentration)(1)
(c) maximum mass soluble in 150 g at 20 °C = 88 × 1.5 = 132 g(1)
(c) 140 − 132 = 8 g (correct answer with no working scores 2; allow ecf)(1)
Answer: (a) 186 g could dissolve, only 140 g present; (b) about 27 °C; (c) 8 g
Examiner tip: Crystals first appear when the solution becomes saturated — find the solubility equal to the solution's own concentration in g per 100 g, then read the temperature.

7Grade 6–7Triple only6 marks

A student wants to find the solubility of ammonium sulfate in water at 30 °C.

Describe an experiment the student could do to find this solubility. Your answer should include how the student would make sure the result is accurate, and how they would calculate the solubility from their measurements. [6]

put a measured mass / volume of water (e.g. 25 g) in a boiling tube / beaker in a water bath kept at 30 °C (checked with a thermometer)(1)
add ammonium sulfate a little at a time, stirring, until some solid remains undissolved (saturated solution)(1)
filter / decant the solution while it is still at 30 °C(1)
weigh an empty evaporating basin, add (a sample of) the saturated solution and weigh again(1)
heat gently to evaporate the water, then reweigh; repeat heating and weighing until the mass is constant(1)
mass of water = mass of solution − mass of dry solid; solubility = mass of solid ÷ mass of water × 100 (g per 100 g of water)(1)
Answer: Make a saturated solution at 30 °C (water bath, excess solid), filter at 30 °C, weigh a sample of solution, evaporate to constant mass, solubility = mass solid ÷ mass water × 100
Examiner tip: Full-mark answers keep the solution at 30 °C right up to filtering and show the final calculation; many answers stop at 'evaporate the water'.

8Grade 6–7Triple only5 marks

A student measured the solubility of potassium nitrate at 50 °C. She made a saturated solution in a water bath at 50 °C, filtered it, weighed a sample of the solution, then heated it to remove the water and weighed the solid left.

(a) Explain why the solution must be filtered while it is still at 50 °C. [2]

(b) Her value for the solubility was higher than the data-book value. Suggest two mistakes that would make the value too high. [2]

(a) if it cools the solubility decreases / crystals form (in the filter paper or funnel)(1)
(a) so some of the dissolved solid is lost / the mass of solid measured would be too low (so the solubility would be too low) (accept the solution would no longer be saturated at 50 °C)(1)
(b) any one from: solid not completely dry / some water left (not heated to constant mass); undissolved crystals passed into the sample (filter paper torn / not filtered properly); water bath temperature above 50 °C (ignore 'human error', 'misread balance')(1)
(b) a second, different mistake from the list in the previous mark point(1)
(c) to make sure all the water has been removed / evaporated (accept so the mass of solid is accurate)(1)
Answer: (a) on cooling, solubility falls and crystals form, so solid is lost and the value is too low; (b) solid still wet; undissolved solid carried over / temperature above 50 °C; (c) to be sure all the water has gone
Examiner tip: 'Human error' never scores — name the actual mistake and say which mass it makes too big or too small.

9Grade 4–5Triple only4 marks

The table shows the solubility of lithium sulfate in water.

Temperature in °C	0	20	40	60	80
Solubility in g per 100 g of water	36.1	34.8	33.7	32.6	31.4

(a) Describe how the solubility of lithium sulfate changes as the temperature increases. [1]

(b) A saturated solution of lithium sulfate is made using 200 g of water at 20 °C. The solution is heated to 80 °C. No water evaporates.

Calculate the mass of lithium sulfate that comes out of solution. [2]

(c) A student states: 'Heating a solution always makes more solid dissolve.' Use the data to comment on this statement. [1]

(a) solubility decreases (slightly) as temperature increases (ignore 'changes')(1)
(b) 34.8 − 31.4 = 3.4 (g per 100 g of water)(1)
(b) 6.8 g (correct answer with no working scores 2)(1)
(c) statement is not true for all solids: lithium sulfate becomes less soluble when heated, so solid would come out of solution (accept true for most solids but not lithium sulfate)(1)
Answer: (a) decreases; (b) 6.8 g; (c) not always true — lithium sulfate is less soluble when hotter
Examiner tip: Read the table carefully: this solubility goes DOWN as temperature rises, so heating — not cooling — makes solid appear.

10Grade 8–9Triple only4 marks

A mixture contains 80.0 g of potassium nitrate and 10.0 g of sodium chloride. The mixture is dissolved in 50.0 g of water at 90 °C. The solution is then cooled to 20 °C.

Salt	Solubility at 90 °C in g per 100 g of water	Solubility at 20 °C in g per 100 g of water
potassium nitrate	202	31.6
sodium chloride	38.5	35.9

Calculate the mass of potassium nitrate crystals that form, and explain why the crystals do not contain sodium chloride. [4]

mass of potassium nitrate that stays dissolved in 50.0 g of water at 20 °C = 31.6 ÷ 2 = 15.8 g(1)
mass of crystals = 80.0 − 15.8 = 64.2 g (allow ecf from their dissolved mass)(1)
maximum sodium chloride that can dissolve in 50.0 g of water at 20 °C = 35.9 ÷ 2 = 18.0 g (accept 17.95 g)(1)
10.0 g is less than 18.0 g so all the sodium chloride stays in solution (so it does not crystallise)(1)
Answer: 64.2 g of KNO₃; NaCl stays dissolved because only 10.0 g is present and up to 18.0 g can dissolve in 50.0 g water at 20 °C
Examiner tip: The explanation mark needs numbers: compare the 10.0 g of sodium chloride with the 18.0 g that 50 g of water can hold at 20 °C.

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