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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The solvent $CS_2$ has enthalpy of vapourization of $352\;J\;g^{-1}$ and boiling point equal to $320\;K$. The value of $K_b$ for this solvent is about.

$\begin{array}{1 1} 0.51\;K\;kg\;mol^{-1}\\0.86\;K\;kg\;mol^{-1} \\ 1.23 \;K\;Kg\;mol^{-1} \\ 2.42\;K\;kg\;mol^{-1} \end{array} $

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Answer : $ 2.42\;K\;kg\;mol^{-1}$
$K_b =\large\frac{M_t RT_b^{+2}}{K_b}$
$\quad = \large\frac{(0.076 \;kg\;mol^{-1})(8.314\;J K^{-1} mol^{-1})(320\;K)^2}{(352Jg^{-1})(76\;g\;mol^{-1})}$$=2.42\;K\;kg\;mol^{-1}$
answered Aug 7, 2014 by meena.p
 

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