S3 Science | Solubility and Crystallisation: Curves and Cooling Crystals

S3 Science | Solubility and Crystallisation: Curves and Cooling Crystals — Tutor Prep Guide

2026-11-15

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Students can quote a solubility figure but freeze when a curve appears. This lesson makes saturation a moving boundary and turns cooling crystallisation into one subtraction.

Objectives — Students can: (1) define a saturated solution as one that cannot dissolve more solute at that temperature and read any value from a solubility curve; (2) calculate the mass of crystals formed on cooling as the difference between the two solubilities; (3) choose filtration, evaporation, cooling crystallisation or chromatography for a mixture and justify the choice with the property difference.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Solute and solvent	Salt dissolves in water: which is solute and which is solvent?	Salt is the solute, water the solvent
Saturation	What do we call a solution that will not dissolve more salt?	A saturated solution, with excess solid at the bottom
Reading graphs	What do the two axes of a graph show?	Read the label and unit of each axis first, then the points

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-5 min	Start with a phenomenon	Add salt spoon by spoon to 100 g of water and ask when it stops dissolving	Name the saturated state and note the excess solid
5-17 min	Reading the curve	Plot 32 g at 20 C and 110 g at 60 C, then ask what the slope means	Read an intermediate value and explain a steep slope
17-30 min	Cooling calculation	100 g of water holds 110 g at 60 C but only 32 g at 20 C	Subtract to get 78 g and write 110 minus 32
30-37 min	Choosing a method	Compare salt water, a sand-salt mixture and ink	Pick evaporation, filtration, chromatography with a reason each
37-42 min	Real applications	Ask about sea salt, sugar crystals and drug purification	Explain evaporation or cooling crystallisation and why filtration fails
42-45 min	Exit check	How much salt can 200 g of water hold at 20 C?	Answer 64 g by doubling

03 Teaching script (read it out)

Scene 1: Saturation is a boundary, not an adjective

This section removes the vague idea that a saturated solution is simply a strong one. Run the demonstration: add salt to 100 g of water until solid stays at the bottom, and name that state. Then move the boundary: warm the same beaker to 60 C and more salt dissolves, cool it back and solid reappears. Students should state both conclusions themselves: saturation gives the maximum at that temperature, and the maximum moves with temperature. The common misconception is equating saturated with concentrated, yet a cold sugar solution can be saturated while still dilute. Nailing this early stops students confusing saturation with strength in every later curve question.

Ask for the definition of a saturated solution and whether saturated means concentrated

Scene 2: The curve: the slope is half the answer

Give data before giving a printed curve. Write 32 g at 20 C and 110 g at 60 C, have students plot and join the points, then ask three questions: roughly what dissolves at 40 C (about 60 g by interpolation), which section is steepest, and what a flat curve implies. Students must see that the slope tells you how much more dissolves per degree, so a steep curve suits cooling crystallisation. Compare potassium nitrate (steep) with sodium chloride (almost flat): salt cannot be crystallised by cooling and must be obtained by evaporation. This comparison drives every separation question later, so spend the time here.

Ask students to point at the steepest part and explain why salt is not obtained by cooling

Scene 3: Cooling: one subtraction, not a formula to memorise

The only relationship needed is that crystals formed equal the high-temperature solubility minus the low-temperature solubility, provided the mass of water is unchanged. From 60 C to 20 C: 110 minus 32 gives 78 g. Ask why not all 110 g appears: because 32 g still dissolves at 20 C. Then scale the solvent: 200 g of water doubles to 156 g, 50 g halves to 39 g. Follow with purity: crystals carry mother liquor, so they are filtered and rinsed with a little cold water. Finally compute the recovery 78 out of 110, about 71 per cent, and warn that a question asking for the mass crystallised differs from one asking for recovery, so read the stem carefully.

Ask students to redo the calculation for 50 g of water and explain why the crystals are rinsed

04 Question types and worked steps

Type	Example	Steps	Answer
Reading a value	A curve shows 60 g per 100 g of water at 40 C. How much dissolves in 150 g of water at 40 C?	1) Note the solvent is 150 g. 2) The figure is per 100 g. 3) Multiply 60 by 1.5	90 g
Cooling crystallisation	100 g of water holds 110 g at 60 C and 32 g at 20 C. What mass crystallises on cooling?	1) Write both solubility figures. 2) Solvent unchanged, so subtract. 3) 110 minus 32	78 g
Mass fraction	A saturated solution at 20 C is 132 g in total (32 g solute, 100 g water). What is the solute mass fraction?	1) The denominator is total solution mass. 2) Divide 32 by 132. 3) Convert to a percentage	About 24.2 per cent
Choosing a method	To obtain dry salt from salt water, do you filter, evaporate or cool?	1) Salt solubility barely changes with temperature. 2) Filtration cannot catch dissolved salt. 3) Evaporate the water	Evaporation, because the salt curve is almost flat

Sanity check (teach this): Three checks: the basis is always per 100 g of water, so scale for other solvent masses; the denominator of a mass fraction is the whole solution, not just the solvent; and cooling always gives crystals while warming dissolves more, so any answer that says warming crystallises should stop you. Two common-sense checks: the mass crystallised cannot exceed the solute originally present, and the mother liquor left behind is still saturated.

05 Independent practice (3 questions)

Question	Answer	What to watch
How much salt dissolves in 200 g of water at 20 C (32 g per 100 g)?	64 g	Answering 32 g means the solvent was not scaled
110 g dissolves at 60 C, 32 g at 20 C. What crystallises?	78 g	Answering 110 g means the low-temperature solubility was ignored
Why is salt not purified by cooling crystallisation?	Its solubility barely changes with temperature	Answering that salt is expensive misses the slope idea

06 Common mistakes and what to say

Mistake	Student writes	Say this
Saturated means concentrated	A saturated solution must be a strong solution.	Saturated only means no more dissolves at that temperature. Check concentration with the mass fraction.
Denominator is the solvent	32 g of salt in 100 g of water gives 32 per cent.	The denominator is 132 g of solution, so about 24.2 per cent.
All solute crystallises	Cooling from 60 C must give all 110 g.	32 g still dissolves at 20 C, so only 110 minus 32, that is 78 g, crystallises.
Choosing methods by habit	Salt water can be filtered to collect the salt.	Dissolved salt passes through the paper. Evaporate, or cool if the curve is steep.

07 Homework (paper, section and question numbers)

Pressure, upthrust and floating, question 2 — Both need the conditions stated before any calculation
Reflection and refraction, question 1 — Both read data off a diagram before computing
The rock cycle and Earth resources, question 3 — Matching processes to conditions uses the same answer pattern

08 Exit check (last 5 minutes)

Question	Answer	Decision
Define a saturated solution	One that cannot dissolve more solute at that temperature	Saying a strong solution means saturation is still confused with concentration
110 g at 60 C, 32 g at 20 C: what crystallises?	78 g	Saying 110 g means the low-temperature figure was ignored
Why evaporate salt water instead of cooling it?	Salt solubility changes very little with temperature	Saying salt dissolves easily misses the slope idea

09 Tutor FAQ

Should students memorise solubility values?

No. They need the per-100 g basis and the subtraction, and the figures come with the question.

How do I demonstrate saturation without a lab?

A clear cup, salt and warm water show both effects in two minutes: no more dissolves, then more appears on cooling.

How should curve and mass fraction questions be sequenced?

Pure curve reading, then pure mass fraction, then mixed. Most mixed errors come from using the solvent as the denominator, so require students to write the total solution mass first.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.