Match List-I with List-II
|
List-I Solute |
List-II van't Hoff factor, \( i \) |
|
(A) Benzoic acid in benzene |
(I) 3 |
|
(B) \( K_2SO_4 \) (aq) |
(II) 2 |
|
(C) Urea (aq) |
(III) 0.5 |
|
(D) KOH (aq) |
(IV) 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The van’t Hoff factor i is the number of particles one formula unit gives in solution. It is greater than 1 when the solute dissociates, equal to 1 when it does neither, and less than 1 when it associates.
(A) Benzoic acid in benzene — This matches (III) 0.5. In benzene, two molecules join by hydrogen bonds into a dimer: \( \ce{2C6H5COOH <=> (C6H5COOH)2} \). Two molecules give one particle, so i = 1/2 = 0.5.
(B) \( \ce{K2SO4} \) (aq) — This matches (I) 3. \( \ce{K2SO4 -> 2K^+ + SO4^2-} \): 2 + 1 = 3 ions, so i = 3.
(C) Urea (aq) — This matches (IV) 1. Urea is a non-electrolyte; it does not dissociate: 1 particle, i = 1.
(D) KOH (aq) — This matches (II) 2. \( \ce{KOH -> K^+ + OH^-} \): 1 + 1 = 2 ions, so i = 2.
Shortcut: (A)-(III) appears only in Option (4).