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The freezing point of an aqueous solution of 0.1m ${Hg}_{2} {Cl}_{2}$ will be: (If ${Hg}_{2} {Cl}_{2}$ is 80% ionized in the solution to give ${Hg}_{2}^{2 +}$ and ${Cl}^{-}$ )

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$- 0.26 K_{f}$

$- 2.6 f K_{f}$

$- 4.2 K_{f}$

$0.42 K_{f}$

answer is A.

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Detailed Solution

$α = \frac{i - 1}{n - 1}$

$0.8 = \frac{i - 1}{3 - 1}$

$i = 2.6$

${ΔT}_{f} = {iK}_{f} \cdot m$

$2.6 \times K_{f} \times 0.1$

$Δ T_{f} = 0.26 K_{f}$

$So freezing po int = - 0.26 K_{f}$

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