Match List-I with List-II
| List-I | List-II |
| Solution type | Examples |
| (A) Minimum boiling azeotrope | (I) Chloroform/Acetone |
| (B) Maximum boiling azeotrope | (II) Ethanol/Acetone |
| (C) Negative deviation from Raoult’s law | (III) 95% Ethanol/water |
| (D) Positive deviation from Raoult’s law | (IV) 68% HNO3/H2O |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (I), (D) - (II)
An ideal solution obeys Raoult’s law because the A–B attractions are the same as the A–A and B–B attractions. When they differ, the solution deviates from Raoult’s law:
Positive deviation: A–B attractions are weaker, so molecules escape more easily. The vapour pressure is higher than predicted and the boiling point is lower. A large positive deviation gives a minimum boiling azeotrope.
Negative deviation: A–B attractions are stronger. The vapour pressure is lower than predicted, and a large deviation gives a maximum boiling azeotrope.
(A) Minimum boiling azeotrope — This matches (III) 95% Ethanol/water. Ethanol–water shows a large positive deviation. The mixture with about 95% ethanol boils at a constant 351 K, below both liquids, so ethanol cannot be purified beyond this by fractional distillation.
(B) Maximum boiling azeotrope — This matches (IV) 68% HNO3/H2O. Nitric acid and water attract each other strongly (negative deviation); the 68% mixture boils at 393.5 K, above both components.
(C) Negative deviation from Raoult’s law — This matches (I) Chloroform/Acetone. The three Cl atoms make the H of \( \ce{CHCl3} \) slightly acidic, so it forms a hydrogen bond with the O of acetone. This new A–B attraction is stronger than the original attractions.
(D) Positive deviation from Raoult’s law — This matches (II) Ethanol/Acetone. Acetone molecules get in between ethanol molecules and break some of their hydrogen bonds, so the attractions in the mixture become weaker.
Shortcut: (A)–(III) appears only in Options (2) and (4), and (C)–(I) then leaves only Option (4).