Which of the following pairs of liquids shows a positive deviation from Raoult’s law?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Benzene–methanol
A solution shows positive deviation from Raoult’s law when the A–B attractions are weaker than the A–A and B–B attractions; the molecules escape more easily and the vapour pressure is higher than Raoult’s law predicts.
In pure methanol the molecules are hydrogen bonded. Benzene cannot form hydrogen bonds with methanol; its molecules get in between the methanol molecules and break some of the hydrogen bonds. The attractions in the mixture are weaker, so the vapour pressure rises: positive deviation.
Why the other options are incorrect
Option (1) Water–nitric acid — Incorrect. This mixture shows a large negative deviation and forms a maximum boiling azeotrope (about 68% nitric acid, 393.5 K).
Option (2) n-Hexane–n-heptane — Incorrect. The A–A, B–B and A–B forces are nearly equal, so this solution is nearly ideal and shows no marked deviation.
Option (4) Acetone–chloroform — Incorrect. Chloroform forms a hydrogen bond with acetone, so the A–B attractions are stronger and the solution shows negative deviation.
NCERT (2026-27 edition):
1.5.2 Non-ideal Solutions – “… In case of positive deviation from Raoult’s law, A-B interactions are weaker than those between A-A or B-B, i.e., in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules.”
“… In pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them. Due to weakening of interactions, the solution shows positive deviation from Raoult’s law …”