Crystallization
Crystallization, explained for NEET
The trap in crystallisation is the solvent choice. A student picks the solvent the compound dissolves in easily — and that is exactly the wrong one. The requirement is the opposite: the compound must be sparingly soluble at room temperature but appreciably soluble at the boiling point of that solvent. Only then does cooling force it out of solution while the impurity stays dissolved.
Crystallisation separates a solid organic compound from its impurities by exploiting this difference in solubility with temperature (NCERT Class 11 Chemistry, Chapter 8, page 279). The solution is concentrated until it is nearly saturated, then cooled. The purified compound crystallises out and is removed by filtration; the impurity, present in smaller amount and still below its saturation level, remains behind in the mother liquor — which is solvent plus impurity plus some dissolved compound, never pure solvent. A second trap: students treat the mother liquor as waste solvent and discard it without recognising it still holds product.
Coloured impurities are handled separately. The hot solution is treated with activated charcoal, which adsorbs the colouring matter. Charcoal decolourises — it does not remove the bulk impurity itself. An option claiming otherwise is a distractor, not chemistry.
When the compound and the impurity have comparable solubilities, one cooling cycle will not do it. Both come down together. The answer is repeated crystallisation — recrystallising the collected solid again, each cycle enriching the compound (NCERT Class 11 Chemistry, Chapter 8, page 279).
For NEET this topic carries about 0.3 questions per year, medium weight, and the skill mix is recall plus interpretation — you are asked to judge a described procedure, not to compute. Negative-marking risk sits in the reversed-solubility option and the charcoal option.
Watch-out: read every solvent statement twice. "Soluble in cold, insoluble in hot" is the reversal, and it reads plausibly if you are skimming.
Can you answer these Crystallization MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Crystallisation as a purification technique for solid organic compounds is based on:
Show answer and why every option is right or wrong
Answer: B. Crystallisation separates a solid from its impurities using the difference in their solubilities in a suitable solvent, as stated in NCERT Class 11 Chemistry, Chapter 8, page 279.
Why A is wrong: A is wrong because a difference in boiling points is the basis of distillation, not crystallisation; crystallisation acts on a solid in a solvent and never separates by volatility.
Why C is wrong: C is wrong because differential sublimation is the basis of the sublimation technique, which applies to solids that pass directly to vapour — a different method in this same chapter.
Why D is wrong: D is wrong because differential adsorption on a solid support is the basis of adsorption chromatography, not crystallisation.
A solid organic compound is to be purified by crystallisation. The correct property for the chosen solvent is that the compound should be:
Show answer and why every option is right or wrong
Answer: D. The solvent must dissolve the compound appreciably when hot and only sparingly when cold, so that cooling the hot saturated solution throws the compound out as crystals (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why A is wrong: A is wrong because it reverses the solubility requirement; a compound freely soluble in the cold would stay in solution on cooling and nothing would crystallise.
Why B is wrong: B is wrong because a compound that never dissolves cannot form a hot saturated solution; direct filtration of an undissolved solid separates nothing from impurities trapped with it.
Why C is wrong: C is wrong because if solubility does not change with temperature, cooling produces no supersaturation and therefore no crystals — the technique needs a solubility gradient.
After the crystals have been separated, the liquid left behind is called the mother liquor. Its composition is best described as:
Show answer and why every option is right or wrong
Answer: D. The impurity remains dissolved in the solvent along with a residual amount of the compound; this solution is the mother liquor (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why A is wrong: A is wrong because it is precisely the impurity staying in solution that makes the technique work; if the liquor were pure solvent, nothing would have been separated.
Why B is wrong: B is wrong because it inverts what is separated — the compound crystallises out and the impurity is what stays dissolved.
Why C is wrong: C is wrong because crystallisation is carried out in solution and involves no melting; the mother liquor is a solution, not a melt.
A coloured impurity is present in an organic compound being purified by crystallisation. The hot solution is treated with activated charcoal. The function of the charcoal is to:
Show answer and why every option is right or wrong
Answer: B. Activated charcoal is added to the hot solution to adsorb the colouring matter, decolourising the solution before it is cooled (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why A is wrong: A is wrong because charcoal acts by physical adsorption on its surface; it is not a chemical reducing agent and the colouring matter is removed intact, not converted.
Why C is wrong: C is wrong because charcoal removes the colour, not the bulk impurity; the impurity separation still depends on the solubility difference exploited by cooling.
Why D is wrong: D is wrong because a small quantity of a solid adsorbent does not meaningfully change the solvent's boiling point, and raising it is not the purpose of the treatment.
An organic compound and its impurity have very nearly the same solubility in the chosen solvent. A single crystallisation gives a solid that is still impure. The correct course of action is:
Show answer and why every option is right or wrong
Answer: A. When solubilities are comparable, both solids come down together in one cycle, so the compound is purified by repeated crystallisation, each cycle enriching it further (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why B is wrong: B is wrong because charcoal adsorbs colouring matter, not a colourless bulk impurity, and it is added to the hot solution rather than the cooled one.
Why C is wrong: C is wrong because a fresh sample faces the identical solubility situation; repeating an unchanged single cycle on new material cannot give a purer product.
Why D is wrong: D is wrong because comparable solubilities make one cycle insufficient, not the compound impossible to purify — repetition is the documented remedy.
In crystallisation, once the impure compound has been dissolved, the solution is concentrated before it is allowed to cool. The reason for concentrating the solution is:
Show answer and why every option is right or wrong
Answer: C. The solution is concentrated to get a nearly saturated solution; cooling that solution is what brings the pure compound out, and the crystals are then removed by filtration (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why A is wrong: A is wrong because crystallisation depends on the impurity STAYING dissolved in the mother liquor. Making it insoluble would bring it down with the product, which is the opposite of what is wanted.
Why B is wrong: B is wrong because concentrating a solution drives off solvent; decomposing the impurity is no part of crystallisation, which separates the two solids while both stay intact.
Why D is wrong: D is wrong because colour is removed by adsorbing it on activated charcoal, a separate step that concentrating the solution does not replace.
A student dissolves an impure solid in a minimum quantity of hot solvent, cools the solution, and obtains crystals. On testing, the crystals are found to be as impure as the starting material, and the mother liquor is almost colourless and nearly free of solute. The most reasonable explanation is:
Show answer and why every option is right or wrong
Answer: C. Crystallisation requires the compound to be appreciably soluble in the hot solvent; if little dissolves, the recovered solid is largely undissolved starting material and the mother liquor holds almost nothing, which matches both observations (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why A is wrong: A is wrong because charcoal addresses colouring matter only, and the mother liquor here is described as almost colourless, so colour was never the problem.
Why B is wrong: B is wrong because slow cooling favours well-formed crystals and does not by itself force impurity out of solution; it also fails to explain why the mother liquor holds almost no solute.
Why D is wrong: D is wrong because discarding the mother liquor happens after the crystals form and cannot change their purity; the liquor is reported as nearly solute-free in any case.
Two students purify the same impure solid. Student I obtains a good yield of crystals, and the mother liquor is deeply coloured. Student II obtains crystals of the same mass, but they are distinctly coloured, and the mother liquor is pale. Assuming both used the same solvent and the same cooling procedure, the difference is best accounted for by:
Show answer and why every option is right or wrong
Answer: A. Treating the hot solution with activated charcoal adsorbs the colouring matter, which is why one student's crystals are clean while the other's carry the colour down with them (NCERT Class 11 Chemistry, Chapter 8, page 279).
Why B is wrong: B is wrong because concentrating the solution changes how much compound crystallises, not whether the colouring matter is adsorbed. The colour would still come down with the crystals.
Why C is wrong: C is wrong because extra solvent dilutes the solution and lowers the crystal yield rather than removing colour, yet both students obtained the same mass of crystals.
Why D is wrong: D is wrong because repeated crystallisation would have improved Student II's crystals, not left them coloured, and would also have reduced the recovered mass below Student I's.
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How do you solve a Crystallization question? A worked example
- 1
Given
A solid organic compound X is contaminated with a small quantity of impurity Y. Measured solubilities in solvent S:
| | at 25 °C | at the boiling point of S |
|---|---|---|
| Compound X | 1.0 g per 1.0 × 10² mL | 2.0 × 10¹ g per 1.0 × 10² mL |
| Impurity Y | 1.8 × 10¹ g per 1.0 × 10² mL | 2.2 × 10¹ g per 1.0 × 10² mL |
Mass of impure sample: 1.0 × 10¹ g, of which 9.0 g is X and 1.0 g is Y. - 2
Required
State whether solvent S is suitable for purifying X by crystallisation, and justify the choice from the solubility data.
- 3
Concept
Crystallisation exploits the change in solubility with temperature. The compound must be sparingly soluble in the cold solvent and appreciably soluble in the hot one, so that cooling a hot saturated solution brings the compound down as crystals while the impurity — present in smaller amount and still below its cold saturation limit — stays dissolved in the mother liquor (NCERT Class 11 Chemistry, Chapter 8, page 279).
- 4
Formula
No algebraic formula applies. The criterion is a comparison:
• Compound: solubility(hot) ≫ solubility(cold) → crystallises on cooling.• Impurity: mass present < solubility(cold) × volume of solvent → stays dissolved. - 5
Substitution
Take 5.0 × 10¹ mL of S, the minimum that dissolves 9.0 g of X when hot.
• X in hot S: 2.0 × 10¹ g per 1.0 × 10² mL → 1.0 × 10¹ g per 5.0 × 10¹ mL. This exceeds the 9.0 g present, so all of X dissolves.• X in cold S: 1.0 g per 1.0 × 10² mL → 0.50 g per 5.0 × 10¹ mL.• Y in cold S: 1.8 × 10¹ g per 1.0 × 10² mL → 9.0 g per 5.0 × 10¹ mL, against only 1.0 g of Y present. - 6
Calculation
Mass of X crystallising on cooling = 9.0 g − 0.50 g = 8.5 g.
Mass of Y crystallising = zero, since 1.0 g of Y is well below the 9.0 g the cold solvent can still hold.
The volume 5.0 × 10¹ mL and the tabulated 1.0 × 10² mL reference volume are problem-defined exact values, and the ratio 5.0 × 10¹ ÷ 1.0 × 10² = 0.500 is an exact scaling factor; none of these contributes to the significant-figure count. The reported masses carry two significant figures, set by the measured solubility and mass data. - 7
Final answer
Solvent S is suitable. Compound X is sparingly soluble cold (0.50 g in the 5.0 × 10¹ mL used) and appreciably soluble hot (capacity 1.0 × 10¹ g), so 8.5 g of the 9.0 g present crystallises out. Impurity Y, at 1.0 g against a cold capacity of 9.0 g, remains entirely in the mother liquor. Recovery of X in one cycle is 8.5 g from 9.0 g.
- 8
Common trap
Two traps sit in this problem. First, a student reading only "X dissolves 2.0 × 10¹ g per 1.0 × 10² mL when hot" may declare S suitable because X dissolves well — the free-solubility error. Suitability rests on the difference between hot and cold, not on the hot figure alone. Second, the 0.50 g of X left behind is often written off as loss; it is dissolved product held in the mother liquor, which is solvent plus impurity plus residual compound — never pure solvent. Had Y's cold solubility been near X's instead of eighteen times higher, both would have crystallised together and repeated crystallisation would be required.
- 9
Similar NEET-style question
A solid compound P is contaminated with impurity Q. In solvent T, P dissolves to the extent of 2.0 g per 1.0 × 10² mL at 25 °C and 2.4 g per 1.0 × 10² mL at the boiling point, while Q dissolves to the extent of 3.0 × 10¹ g per 1.0 × 10² mL at 25 °C. Which statement about purifying P in solvent T is correct?
• A. T is suitable, because Q is far more soluble than P at every temperature.• B. T is unsuitable, because P's solubility barely changes with temperature, so cooling yields very little crystalline P.• C. T is unsuitable, because Q will crystallise out in preference to P.• D. T is suitable, provided activated charcoal is added to adsorb Q.
Answer: B. The whole technique rests on a solubility gradient with temperature; 2.0 g to 2.4 g per 1.0 × 10² mL is too small a change to throw down a useful mass of crystals, whatever Q does.
[provenance-pending] Source provenance not yet catalogued for CHE.U13.CRYSTALLIZATION. NCERT citations, verified PYQs, and exam traps will appear here once the dossier is populated.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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