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Calculate the molal depression constant of a solvent which has freezing point $16.6^{\circ} \mathrm{C}$ and latent heat of fusion $180.75 \mathrm{Jg}^{-1}$
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3.86
$\begin{gathered}K_f=\frac{R T_f^2}{1000 \times L_f}, R=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \\ T_f=273+16.6=289.6 \mathrm{~K} ; L_f=180.75 \mathrm{Jg}^{-1} \\ K_f=\frac{8.314 \times 289.6 \times 289.6}{1000 \times 180.75} = 3.86\end{gathered}$
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