#### Answer

$39.4\,kJ/mol$

#### Work Step by Step

$T_{1}=(50.0+273)K= 323\,K$
$P_{1}= 233\, mmHg$
$T_{2}=(78.3+273)K=351.3\,K$
$P_{2}=1\,atm=760\, mmHg$
$\Delta_{vap}H$ is given by:
$\ln (\frac{P_{2}}{P_{1}})=\frac{-\Delta_{vap}H}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$
That is,
$\ln (\frac{760}{233})=\frac{-\Delta_{vap}H}{8.314\,J\,mol^{-1}K^{-1}}[\frac{1}{351.3\,K}-\frac{1}{323\,K}]$
$\implies 1.18228=-\Delta_{vap}H\times
-2.9998\times10^{-5}\,mol/J$
$\implies \Delta_{vap}H=\frac{1.18228}{2.9998\times10^{-5}\,mol/J}$
$=39400\,J/mol=39.4\,kJ/mol$