Match List-I with List-II
| List-I | List-II |
| Type of solution | Example |
| (A) Ideal solution | (I) 95% by volume of ethanol |
| (B) Positive deviation from Raoult’s law | (II) Bromoethane and chloroethane |
| (C) Negative deviation from Raoult’s law | (III) Ethanol and acetone |
| (D) Azeotrope | (IV) Phenol and aniline |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (III), (C) - (IV), (D) - (I)
A solution’s behaviour depends on how the A–B attractions compare with the A–A and B–B attractions in the pure liquids. Equal attractions give an ideal solution; weaker ones give positive deviation from Raoult’s law; stronger ones give negative deviation. Large deviations lead to azeotropes.
(A) Ideal solution — This matches (II) Bromoethane and chloroethane. The two molecules are very similar, so the A–B attractions are nearly the same as A–A and B–B. The mixture obeys Raoult’s law, with ΔmixH = 0 and ΔmixV = 0.
(B) Positive deviation from Raoult’s law — This matches (III) Ethanol and acetone. Acetone molecules come between ethanol molecules and break some of their hydrogen bonds. The weaker attractions let molecules escape more easily, so the vapour pressure is higher than Raoult’s law predicts.
(C) Negative deviation from Raoult’s law — This matches (IV) Phenol and aniline. The phenolic H forms a hydrogen bond with the lone pair on the nitrogen of aniline. This attraction is stronger than in the pure liquids, so the vapour pressure is lower than predicted.
(D) Azeotrope — This matches (I) 95% by volume of ethanol. Ethanol and water form a minimum boiling azeotrope at about 95% ethanol by volume. At this composition the liquid and vapour have the same composition, so fractional distillation cannot concentrate the ethanol further.
Shortcut: The pair (A)–(II) appears in Options (1) and (2), and (B)–(III) then leaves only Option (2).