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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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A solvent of molar mass $84.2 \;g\;mol^{-1}$ and boiling point $81.4^{\circ}C$ has boiling point elevation constant equal to $ 2.79 \;K\;kg\;mol^{-1}$. Its molar enthalpy of vapourization will be about

$\begin{array}{1 1} 22.7 \;22.7\;kJ\;mol^{-1} \\25.7\;kJ\;mol^{-1} \\ 30.7\;kJ\;mol^{-1} \\ 35.7 \;kJ\;mol^{-1} \end{array} $

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Answer : $ 30.7\;kJ\;mol^{-1}$
Since $K_b =(M_t RT_b^{+2}/ \Delta _{vap}H_m)$. we have
$ \Delta _{vap} H_m= \large\frac{M_t RT_b^{+2}}{K_b}$
$\qquad= \large \frac{(0.082 \;Kg\;mol^{-1})(8.314\;J\;K^{-1} \;mol^{-1})(354.55 \;K)^2}{(2.79 \;K\;kg\;mol^{-1})}$
$\qquad =30717\;J\;mol^{-1}$
$\qquad= 30. 72\;kJ \;mol^{-1}$
answered Aug 7, 2014 by meena.p
 

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