Answer
$33.85\,kJ/mol$
Work Step by Step
The Clapeyron-Clausius equation can be used to find $\Delta H^{\circ}_{vap}$.
$\ln (\frac{P_{2}}{P_{1}})=\frac{\Delta H^{\circ}_{vap}}{R}(\frac{T_{2}-T_{1}}{T_{1}T_{2}})$
$\implies \ln (\frac{48.1\,Torr}{133.0\,Torr})=-1.017$
$=\frac{\Delta H^{\circ}_{vap}}{8.3145\,Jmol^{-1}K^{-1}}(\frac{273.15\,K-293.15\,K}{293.15\,K\times273.15\,K})$
That is,
$-1.017=\frac{\Delta H^{\circ}_{vap}}{8.3145\,Jmol^{-1}}\times-0.00024977$
Or $\Delta H^{\circ}_{vap}=\frac{-1.017\times8.3145\,J/mol}{-0.00024977}$
$=33.85\times10^{3}\,J/mol= 33.85\,kJ/mol$
