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If the freezing point depression constant of a solvent is $20.0\;K\;kg\;mol^{-1}$. If its $\Delta_{fus} H^0_m=2.74\;kJ\;mol^{-1}$ and its molar mass is $84.2\;g\;mol^{-1}$ the normal freezing point of the solvent would be about

$\begin{array}{1 1} 308.2\;K \\ 298.1\;K \\ 249.8\;K \\ 279.8\;K \end{array} $

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Answer : $ 279.8\;K $
$T_f = \bigg[\large\frac{K_f \Delta H_m}{M_f R} ^{1/2}\bigg] $
$ \qquad = \bigg[\large\frac{(20\;K\;kg \;mol^{-1})(2.74 \times 10^3 J mol^{-1})}{(84.2 \times 10^{-3} kg\; mol^{-1}) (8.314 \;J\;K^{-1} mol^{-1})} \bigg]^{1 /2}$
$\qquad= 279.8K$
answered Aug 6, 2014 by meena.p
 

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