Match List-I with List-II
List-I | List-II (van’t Hoff factor) |
(A) \(K_4[Fe(CN)_6]\) | (I) 3 |
(B) \(K_2SO_4\) | (II) 0.5 |
(C) \(CH_3COOH\) in benzene | (III) 5 |
(D) \(KCl\) | (IV) 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-I, C-II, D-IV
The van’t Hoff factor i is the ratio of the number of particles actually present in solution to the number of formula units dissolved. For a strong electrolyte that dissociates completely, i = number of ions per formula unit. For a solute that associates (forms dimers), i is less than 1.
(A) \( \ce{K4[Fe(CN)6]} \) — This matches (III) 5. It gives four \( \ce{K^{+}} \) ions and one complex ion: 4 + 1 = 5 ions, so i = 5.
\( \ce{K4[Fe(CN)6] -> 4K+ + [Fe(CN)6]^4-} \)
(B) \( \ce{K2SO4} \) — This matches (I) 3. Two \( \ce{K^{+}} \) and one \( \ce{SO4^{2-}} \): 2 + 1 = 3 ions, so i = 3.
\( \ce{K2SO4 -> 2K+ + SO4^2-} \)
(C) \( \ce{CH3COOH} \) in benzene — This matches (II) 0.5. In benzene, ethanoic acid molecules pair up through hydrogen bonds to form dimers; two molecules give one particle, so i = 1/2 = 0.5 (for complete dimerisation).
\( \ce{2CH3COOH <=> (CH3COOH)2} \)
(D) KCl — This matches (IV) 2. One \( \ce{K^{+}} \) and one \( \ce{Cl^{-}} \): 1 + 1 = 2 ions, so i = 2.
\( \ce{KCl -> K+ + Cl-} \)
Shortcut: C must be 0.5 (the only value below 1) and A must be 5; only Option (1) has both C-II and A-III.