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For water , $ \Delta _{fus} H^3_m=2.74 \;kJ\;mol^{-1}$ and its molar mass is $ 84.2\;g \;mol^{-1} $ the normal freezing point depression constant at 1 atm pressure would be about

$\begin{array}{1 1} 4.71\;K\;kg\;mol^{-1} \\ 1.2\;K\;kg\;mol^{-1} \\ 1.83\;K\;kg\;mol^{-1} \\ 2.5\;K\;kg\;mol^{-1} \end{array} $

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Answer :$ 1.83\;K\;kg\;mol^{-1}$
$K_f = \large\frac{M_tRT_t ^{+2}}{\Delta_{fus}H_m }$$=\large\frac{(0.018)kg \;mol^{-1})(8.314\;JK^{-1}mol^{-1})(273.15 K)^2}{6.09 \times 10^{3} J mol^{-1} )}$
$\qquad = 1.83 K \;kg\;mol^{-1}$
answered Aug 6, 2014 by meena.p
 

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