Answer
$\Delta H_{vap}=34.5\ kJ/mol$
Work Step by Step
From the Clausius-Clapeyron equation:
$ln (\dfrac{P_1}{P_2})=\dfrac{-\Delta H_{vap}}{R}\times(\dfrac{1}{T_1}-\dfrac{1}{T_2})$
$ln (\dfrac{760\ mmHg}{361.5\ mmHg})=\dfrac{-\Delta H_{vap}}{8.314\ J/K.mol}\times(\dfrac{1}{(273+98.1)K}-\dfrac{1}{(273+75)K})$
$0.743= \dfrac{-\Delta H_{vap}}{8.314\ J/K.mol}\times-1.789\times10^{-4}$
$\Delta H_{vap}=34534\ J/mol=34.5\ kJ/mol$