ChemistryICSE Class 9 · Water

Solutions, Solubility and Solubility Curves: Crystallisation on Cooling (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

Solubility is the maximum mass of solute (in grams) that dissolves in 100 g of water at a given temperature to form a saturated solution: S = (mass of solute ÷ mass of water) × 100. A solubility curve plots S against temperature; a steep curve (potassium nitrate) means much solute crystallises when a hot saturated solution cools, a flat curve (sodium chloride) means little does.

Cooling from t₁ to t₂ drops KNO₃ solubility from S₁ to S₂, so S₁ − S₂ grams crystallise per 100 g water; NaCl gives hardly any.

Illustration drawn for QuestionBuddy.

Understand Solutions, Solubility and Solubility Curves: Crystallisation Calculations

Solution, colloid, suspension. A solution is a homogeneous mixture of solute in solvent (water dissolves many substances because its molecules are polar).

Property	True solution	Colloid	Suspension
Particle size	smallest	intermediate	largest
Appearance	clear	translucent	cloudy, opaque
Settles?	no	no	yes
Passes through filter paper?	yes	yes	no
Tyndall effect	no	yes	yes
Example	salt in water	milk, starch solution	chalk in water

Saturated, unsaturated, supersaturated. An *unsaturated* solution can dissolve more solute at that temperature; a *saturated* one cannot; a *supersaturated* one holds more than the saturated amount, is unstable, and crystallises out on shaking or when a seed crystal is added.

Solubility. The maximum mass of solute dissolving in 100 g of water at a given temperature:

$S = \frac{mass of solute}{mass of water} \times 100.$

Note the denominator is the mass of the water, not of the solution. For a saturated solution: mass of solution $=$ mass of solute $+$ mass of water.

Solubility curve. S (y-axis) against temperature (x-axis). Potassium nitrate: steep rise. Sodium chloride: almost flat. Calcium sulphate: rises slightly, then falls at higher temperature. Gases: solubility falls as temperature rises.

Crystallisation on cooling. For a saturated solution at a high temperature $t_{1}$ and a lower temperature $t_{2}$ :
1. Find the masses of solute and water in the hot solution from $S_{1}$ : in $(100 + S_{1})$ g of solution there are $S_{1}$ g of solute.
2. The water stays the same; at $t_{2}$ it can hold only $\frac{S _{2}}{100} \times m_{w a t er}$ of solute.
3. Crystals deposited $=$ solute at $t_{1}$ $-$ solute held at $t_{2}$ .

A salt with a steep curve gives a lot of crystals; one with a flat curve, almost none, so it is obtained instead by evaporating the water.

Gases. Dissolved oxygen falls in warm water, which harms fish.

Solutions, Solubility and Solubility Curves: Crystallisation Calculations: formulas and key facts

Solubility: $S = \frac{m _{so l u t e}}{m _{w a t er}} \times 100$; g per 100 g water at a given temperature.
Solute in a saturated solution: $m_{so l u t e} = m_{so l u t i o n} \times \frac{S}{100 + S}$; From (100 + S) g solution containing S g solute.
Crystals on cooling: $m_{cr y s t a l s} = m_{w a t er} \frac{S _{1} - S _{2}}{100}$; Water unchanged; S₁ at the higher temperature.

How to solve Solutions, Solubility and Solubility Curves: Crystallisation Calculations questions

Keep track of three masses: solute, water and solution.

1Use the solubility to find solute and water in the saturated solution (S g solute with 100 g water).
2Scale to the actual mass: factor = mass given ÷ (100 + S).
3For cooling: the water mass stays fixed; compute solute held at the lower temperature.
4Crystals = original solute − solute still dissolved.
5Check: crystals + remaining solution = original solution.

Solutions, Solubility and Solubility Curves: Crystallisation Calculations in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Making crystals of a salt

Suppose a salt has a solubility of 80 g per 100 g water at 60 °C and 30 g per 100 g water at 20 °C (invented values), and a teacher prepares a saturated solution with 200 g of water at 60 °C.

What mass of crystals appears on cooling to 20 °C?

Try it first, then show the solutionSolution

At 60 °C: $200 \times 0.80 = 160$ g of salt is dissolved.
At 20 °C the 200 g of water holds only $200 \times 0.30 = 60$ g.
Crystals $= 160 - 60 = 100$ g.

Answer: 100 g of crystals.

Memory hook: Water stays; the salt it can hold changes.

Tricks to remember Solutions, Solubility and Solubility Curves: Crystallisation Calculations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Per 100 g of water
Solubility is always per 100 g of water, never per 100 g of solution.
Use it when: Setting up the ratio.
Steep curve, many crystals
A big fall in S on cooling gives many crystals.
Use it when: Curves.
Solution is solute plus solvent
(100 + S) g solution holds S g solute.
Use it when: Scaling.

Solutions, Solubility and Solubility Curves: Crystallisation Calculations: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Definition of solubility · 1 mark · MCQBoard level

The solubility of a solute at a given temperature is the maximum mass of it that dissolves in:
(a) 100 g of solution
(b) 100 g of water
(c) 1 litre of solution
(d) 1 g of water

Try it first, then show the solutionSolution

By definition, grams per 100 g of water forming a saturated solution.

Answer: (b) 100 g of water

Example 2: Which mixture? · 2 marks · Short answerBoard level

Milk, muddy water and sugar in water are mixed samples. Classify each as a solution, colloid or suspension, giving one test for each.

Try it first, then show the solutionSolution

Sugar in water: true solution; passes through filter paper and does not scatter light. (½)
Milk: colloid; passes through filter paper but scatters a beam of light (Tyndall effect) and does not settle. (1)
Muddy water: suspension; settles on standing and is left on the filter paper. (½)

Answer: Solution; colloid; suspension

Example 3: Solubility from evaporation · 3 marks · Short answerHarder board level

A saturated solution of a salt weighing 94.5 g leaves 24.5 g of dry salt on evaporation. Find the solubility of the salt at that temperature.

Try it first, then show the solutionSolution

Mass of water $= 94.5 - 24.5 = 70.0$ g. (1)
$S = \frac{24.5}{70.0} \times 100$ (1)
$= 35$ g per 100 g of water. (1)

Answer: 35 g per 100 g water

Example 4: Crystals on cooling · 3 marks · Short answerHarder board level

A salt has solubilities of 60 g and 20 g per 100 g of water at 70 °C and 30 °C respectively (invented values). 200 g of a saturated solution at 70 °C is cooled to 30 °C. Find the mass of crystals deposited.

Try it first, then show the solutionSolution

At 70 °C: $\frac{200 \times 60}{160} = 75$ g of salt and $125$ g of water. (1)
At 30 °C the 125 g of water holds $\frac{125 \times 20}{100} = 25$ g. (1)
Crystals $= 75 - 25 = 50$ g. (1)

Answer: 50 g

Example 5: Choosing a salt to crystallise · 4 marks · Case-basedHOTS / competency

Invented solubility data (g per 100 g water) at 20, 40, 60, 80 °C: Salt P: 30, 65, 110, 170. Salt Q: 36.0, 36.5, 37.0, 38.0. Salt R falls slowly as temperature rises.
(a) Which salt would you purify by cooling a hot saturated solution?
(b) 270 g of a saturated solution of P at 60 °C is cooled to 20 °C. Find the mass of crystals.
(c) Why is Q recovered by evaporation instead?
(d) Which real salt does the curve of R resemble?

Try it first, then show the solutionSolution

(a) P (steep curve). (½)
(b) At 60 °C: $\frac{270 \times 110}{210} = 141.4$ g salt, $128.6$ g water. At 20 °C the water holds $128.6 \times 0.30 = 38.6$ g. Crystals $= 141.4 - 38.6 \approx 102.9$ g. (2)
(c) Its curve is almost flat, so cooling gives hardly any crystals. (1)
(d) Calcium sulphate (solubility falls at higher temperature). (½)

Answer: (a) P (b) about 102.9 g (c) flat curve (d) calcium sulphate

Where students lose marks in Solutions, Solubility and Solubility Curves: Crystallisation Calculations

The slips examiners see most often, and what to write instead.

Mistake: Mass of solution used instead of water in S
S is per 100 g of water.
Wrong working
S = 24.5/94.5 × 100
Right working
S = 24.5/70 × 100 = 35
Do this: Subtract the solute first.
Usually costs: The answer.
Mistake: Water mass changed after cooling
The water remains; only the dissolved solute changes.
Wrong working
Water at 30 °C = 125 − 50.
Right working
Water stays at 125 g.
Do this: Write "water unchanged".
Usually costs: The answer.
Mistake: Colloid and suspension confused
Colloids do not settle or get filtered out.
Wrong working
Milk settles on standing.
Right working
Milk is a colloid; chalk in water settles.
Do this: Learn the three-column table.
Usually costs: 1 mark.

Exam tips

Define solubility with "100 g of water" and the temperature.
Write the three masses (solute, water, solution) in a small table for every sum.
Say "water unchanged" in cooling questions.
For curves mention steep, flat and falling examples.

Practice questions on Solutions, Solubility and Solubility Curves: Crystallisation Calculations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A solution that cannot dissolve any more solute at the same temperature is called:
(a) unsaturated
(b) saturated
(c) supersaturated
(d) dilute
Hint
Maximum amount dissolved.
Answer
(b) saturated
Q2 · 1 mark · Very short answerBoard level
Name the effect by which a beam of light is scattered by a colloid.
Hint
Named after a physicist.
Answer
Tyndall effect.
Q3 · 2 marks · Short answerHarder board level
The solubility of a salt at 40 °C is 45 g per 100 g of water (invented value). Find the mass of salt in 290 g of its saturated solution at 40 °C.
Hint
100 + 45 = 145 g of solution has 45 g salt.
Answer
$290 \times \frac{45}{145} = 90$ g.
Q4 · 3 marks · Short answerHarder board level
A salt has solubility 60 g at 70 °C and 20 g at 30 °C (per 100 g water, invented). 200 g of the saturated solution at 70 °C is cooled to 30 °C. Find (a) the water in the solution, (b) the mass of crystals, (c) the mass of solution that remains.
Hint
Water stays the same.
Answer
(a) 125 g (solute 75 g). (b) Crystals $= 75 - 25 = 50$ g. (c) Remaining solution $= 200 - 50 = 150$ g (125 g water + 25 g salt).
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): Fish may die in water that has been warmed by hot factory discharge.
Reason (R): Gases are less soluble in water at higher temperatures.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
Dissolved oxygen.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
Using the invented data of the worked example (P: 110 g at 60 °C, 30 g at 20 °C; Q: 37.0 g at 60 °C, 36.0 g at 20 °C), find the mass of crystals obtained when 100 g of each saturated solution at 60 °C is cooled to 20 °C. Compare and comment.
Hint
Do each with the cooling method.
Answer
P: salt $= \frac{100 \times 110}{210} = 52.4$ g, water $47.6$ g; holds $14.3$ g at 20 °C; crystals $\approx 38.1$ g. Q: salt $= \frac{100 \times 37}{137} = 27.0$ g, water $73.0$ g; holds $26.3$ g; crystals $\approx 0.7$ g. P gives about 50 times more crystals because its curve is steep.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Solubility formula?Show answer
(mass of solute ÷ mass of water) × 100.
2.Steep curve example?Show answer
Potassium nitrate.
3.Flat curve example?Show answer
Sodium chloride.
4.A supersaturated solution crystallises on?Show answer
Shaking or adding a seed crystal.
5.Gases in warm water?Show answer
Less soluble.

Solutions, Solubility and Solubility Curves: Crystallisation Calculations: frequently asked questions

What is solubility?

The maximum mass of solute in grams that dissolves in 100 g of water at a given temperature to give a saturated solution.

How do you find crystals deposited on cooling?

Find the solute in the hot saturated solution, the amount the same water holds at the lower temperature, and subtract.

What does a solubility curve show?

How the solubility of a substance changes with temperature; steep curves give many crystals on cooling.