The freezing point of an aqueous solution of 0.1m Hg₂ Cl₂ will be: (If Hg₂Cl₂ is 80% ionized in the solution to give ${Hg}_{2}^{2 +}$ and Cl^-)

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0.42 K_f

-0.26 K_f

-4.2 K_f

-2.6 K_f

answer is A.

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Detailed Solution

$\begin{array}{l} α = \frac{i - 1}{n - 1} \\ 0 .8 = \frac{i - 1}{3 - 1} \\ = 2 .6 \\ {ΔT}_{f} = i_{f} \cdot m \\ 2 .6 \times K_{f} \times 0 .1 \end{array}$

$T_{f}^{°}$ - T_f = 0.26K_f

T_f = 0 -0.26K_f = - 0.26K_f

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