Match List-I with List-II
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List-I Aqueous Solutions |
List-II Depression in freezing point (\( ΔT_f \)) expression |
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(A) \( 0.2\, mol\, kg^{-1}\, Al_2(SO_4)_3 \) |
(I) \( 2K_f \) |
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(B) \( 0.3\, mol\, kg^{-1}\, Na_2SO_4 \) |
(II) \( K_f \) |
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(C) \( 1\, mol\, kg^{-1}\, KCl \) |
(III) \( 0.9K_f \) |
|
(D) \( 0.1\, mol\, kg^{-1}\, K_3[Fe(CN)_6] \) |
(IV) \( 0.4K_f \) |
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Depression in freezing point is a colligative property. For an electrolyte it depends on the molality m and on the van’t Hoff factor i, the number of ions each formula unit gives:
\( \displaystyle \Delta T_f = i\,K_f\,m \)
So for each solution the product i × m gives the multiple of Kf.
(A) \(0.2\, mol\, kg^{-1}\, Al_2(SO_4)_3\) — This matches (II) \(K_f\). \( \ce{Al2(SO4)3 -> 2Al^3+ + 3SO4^2-} \): 2 + 3 = 5 ions, i = 5; \( \Delta T_f = 5 \times 0.2\,K_f = 1.0\,K_f \).
(B) \(0.3\, mol\, kg^{-1}\, Na_2SO_4\) — This matches (III) \(0.9K_f\). \( \ce{Na2SO4 -> 2Na^+ + SO4^2-} \): 2 + 1 = 3 ions, i = 3; \( \Delta T_f = 3 \times 0.3\,K_f = 0.9\,K_f \).
(C) \(1\, mol\, kg^{-1}\, KCl\) — This matches (I) \(2K_f\). \( \ce{KCl -> K^+ + Cl^-} \): 2 ions, i = 2; \( \Delta T_f = 2 \times 1\,K_f = 2\,K_f \).
(D) \(0.1\, mol\, kg^{-1}\, K_3[Fe(CN)_6]\) — This matches (IV) \(0.4K_f\). \( \ce{K3[Fe(CN)6] -> 3K^+ + [Fe(CN)6]^3-} \): 3 + 1 = 4 ions (the complex ion does not break up), i = 4; \( \Delta T_f = 4 \times 0.1\,K_f = 0.4\,K_f \).
Shortcut: (C)-(I) appears in Options (2) and (4); (D)-(IV) then leaves only Option (2).