When acetone is added to ethanol, what type of deviation from Raoult’…

The correct answer is Option (4) → Positive deviation, because acetone breaks some of the hydrogen bonds between ethanol molecules

• 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 then escape more easily and the vapour pressure is higher than Raoult’s law predicts.

• Acetone breaks some of the hydrogen bonds between ethanol molecules, so the attractions in the mixture are weaker than in pure ethanol: positive deviation.

Why the other options are incorrect

Option (1) Negative deviation, because the ethanol–acetone interactions are stronger than those in the pure liquids — Incorrect. The ethanol–acetone interactions are weaker, not stronger, because hydrogen bonds of ethanol are broken. Negative deviation is seen in mixtures such as chloroform and acetone, where a new hydrogen bond forms between the two.

Option (2) No deviation, because the ethanol–acetone interactions are of the same strength as those in the pure liquids — Incorrect. Equal interactions give an ideal solution, such as n-hexane and n-heptane; breaking hydrogen bonds makes the interactions unequal.

Option (3) Positive deviation, because the ethanol–acetone interactions are stronger than those in the pure liquids — Incorrect. The type of deviation is right but the reason is wrong: stronger A–B interactions lower the vapour pressure and give 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 … Mixtures of ethanol and acetone behave in this manner. 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 …”