The freezing point of an aqueous solution of containing 0.19 mol of KCN in 1 kg of water was found to be –0.704°C. On adding 0.095 mol of Hg(CN)₂ to the above solution, the freezing point of the solution was found to be –0.5300C. If the complex formation takes place according to the following equation: $Hg (CN)_{2} + nKCN ⇌ K_{n} [Hg (CN)_{n + 2}]$ The Van’t Hoffs factor for the complex is ____

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answer is 3.

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Detailed Solution

For KCN, $i = 1 + α, {ΔT}_{f} = K_{f} \times i \times m$
$\begin{array}{l} \Rightarrow 0 .704 = 1 .86 \times i \times 0 .1892 \\ \Rightarrow i = \frac{0 .704}{186 \times 0 .1892} (or) 1 + α = 2 \end{array}$
$∴ α = 1$ indicates 100% ionization of KCN,
Now, ${ΔT}_{f} = 0 {.530}^{\circ} C$
Molality of Hg(CN)₂ = 0.095 mol kg^–1
= 0.095 m
$K_{n} [Hg (CN)_{n + 2}] ⇌ {nK}^{\oplus} + {[Hg (CN)_{n + 2}]}^{- n}$
Here,

$\begin{array}{l} ∴ i = 1 + n; {ΔT}_{f} = K_{f} \times i \times molality \\ \Rightarrow 0 .503 = 1 .86 \times i \times 0 .095 \Rightarrow i = 3 \end{array}$

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