The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
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List-I Aqueous Solutions
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List-II Depression in freezing point ($ΔT_f$) expression
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(A) $0.2\, mol\, kg^{-1}\, Al_2(SO_4)_3$
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(II) $K_f$
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(B) $0.3\, mol\, kg^{-1}\, Na_2SO_4$
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(III) $0.9K_f$
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(C) $1\, mol\, kg^{-1}\, KCl$
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(I) $2K_f$
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(D) $0.1\, mol\, kg^{-1}\, K_3[Fe(CN)_6]$
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(IV) $0.4K_f$
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Calculate $i \cdot m$ for each solution
(A) 0.2 mol kg⁻¹ Al₂(SO₄)₃
- Dissociation: $Al_2(SO_4)_3 \rightarrow 2Al^{3+} + 3SO_4^{2-}$ → total 5 ions
- $i \cdot m = 0.2 \times 5 = 1$
- ΔTf = 1 Kf
(B) 0.3 mol kg⁻¹ Na₂SO₄
- Dissociation: $Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-}$ → 3 ions
- $i \cdot m = 0.3 \times 3 = 0.9$ → ΔTf ≈ 0.9 Kf
(C) 1 mol kg⁻¹ KCl
- Dissociation: $KCl \rightarrow K^+ + Cl^-$ → 2 ions
- $i \cdot m = 1 \times 2 = 2$ → ΔTf = 2 Kf
(D) 0.1 mol kg⁻¹ K₃[Fe(CN)₆]
- Dissociation: $K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-}$ → 4 ions
- $i \cdot m = 0.1 \times 4 = 0.4$ → ΔTf = 0.4 Kf
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