Match List-I with List-II
|
List-I Components of binary solution |
List-II Deviation from Raoult's Law |
|
(A) Ethanol and water |
(I) Positive |
|
(B) Chloroform and acetone |
(II) No deviation |
|
(C) n-Hexane and n-heptane |
(III) Negative |
|
(D) Carbon disulphide and acetone |
(IV) Large positive |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
A solution obeys Raoult’s law (ideal solution) when the attraction between unlike molecules (A–B) is about the same as between like molecules (A–A and B–B). If the A–B attractions are weaker, molecules escape more easily, the vapour pressure is higher than Raoult’s law predicts, and the deviation is positive. If the A–B attractions are stronger, the vapour pressure is lower and the deviation is negative.
(A) Ethanol and water — This matches (IV) Large positive. The deviation is so large that the mixture forms a minimum boiling azeotrope (about 95% ethanol by volume), so ethanol cannot be fully separated from water by fractional distillation.
(B) Chloroform and acetone — This matches (III) Negative. The hydrogen of chloroform forms a hydrogen bond with the oxygen of acetone, an A–B attraction stronger than those in either pure liquid; the mixture forms a maximum boiling azeotrope.
(C) n-Hexane and n-heptane — This matches (II) No deviation. Both are non-polar alkanes of similar size held by similar dispersion forces, so the solution is nearly ideal.
(D) Carbon disulphide and acetone — This matches (I) Positive. Non-polar \( \ce{CS2} \) molecules come between the polar acetone molecules and weaken their dipole–dipole attraction, so the A–B attractions are weaker.
Shortcut: (C)-(II) appears only in Option (2).