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Questions  >>  JEEMAIN and NEET  >>  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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