🛠️ JEE🧪 Chemistry

Solutions

3 solved JEE Chemistry previous year questions on Solutions, each with the correct answer and a full explanation.

For a dilute solution, Raoult's law states that

athe lowering of vapour pressure is equal to the mole fraction of solute

bthe relative lowering of vapour pressure is equal to the mole fraction of solute

cthe relative lowering of vapour pressure is proportional to the amount of solute in solution

dthe vapour pressure of the solution is equal to the mole fraction of solvent

✓ Correct answer: b) the relative lowering of vapour pressure is equal to the mole fraction of solute

ExplanationFor a dilute solution, Raoult's law states that the relative lowering of vapour pressure is equal to the mole fraction of solute.

During depression of freezing point in a solution the following are in equilibrium

aliquid solvent, solid solvent

bliquid solvent, solid solute

cliquid solute, solid solute

dliquid solute, solid solvent

✓ Correct answer: a) liquid solvent, solid solvent

ExplanationDuring depression of freezing point, the equilibrium is established between: liquid solvent ⇌ solid solvent Hence, the correct answer is: Option A — liquid solvent, solid solvent Explanation: Freezing point is the temperature at which a solvent changes from liquid to solid.When a non-volatile solute is added, the equilibrium between liquid solvent and solid solvent shifts, causing a lowering of freezing point.

Identify the mixture that shows positive deviations from Raoult's Law

aCHCl3 + (CH3)2CO

b(CH3)2CO + C6H5NH2

c(CH3)2CO + CS2

dCHCl3 + C6H6

✓ Correct answer: c) (CH3)2CO + CS2

ExplanationMixing polar acetone with non-polar CS₂ weakens intermolecular attractions (A–B < A– A/B–B) → positive deviation from Raoult’s law.The other pairs (acetone–chloroform, acetone– aniline, chloroform–benzene) form specific associations/H-bonding → negative deviation.

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