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The molar mass and the freezing point of camphor are $152.2\;g\;mol^{-1} $ and $178.4^{\circ} C $ respectively . The cryoscopic constant of camphor is $ 37.7\;K\;kg\;mol^{-1}$, the molar enthalpy of fusion of camphor is about

$\begin{array}{1 1} 5.24\;kJ\;K^{-1}\;mol^{-1} \\ 6.84\;kJ\;K^{-1} \;mol^{-1} \\ 8.24\;kJ\;K^{-1}\;mol^{-1} \\ 9.8 \;kJ K^{-1} mol^{-1} \end{array} $

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Answer : $ 6.84\;kJ\;K^{-1} \;mol^{-1}$
$\Delta _{fus} H_m =\large\frac{M_1RT_f^{+2}}{K_f} $
$\qquad= \large\frac{(152.2 \times 10^{-3} Kg mol^{-1})(8.314 JK^{-1} mol^{-1} )(451.5 K)^2}{(37.7 K\;kg\;mol^{-1})}$
$\qquad= 6842.2 \;JK^{-1}mol^{-1} =6.84\;kJ \;K^{-1}\;mol^{-1}$
answered Aug 6, 2014 by meena.p

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