Standard vaporisation enthalpy of benzene at boiling point is 30.8 kJ ${{mol}^{-1}}$. For how long would 100W electric heater have to operate in order to vaporise a 100 g sample at that temperature? (Power= energy/time and 1W = 1 J${{s}^{-1}}$).
${ ∆ }{ vap }$H°(benzene) = 30.8 kJ${{mol}^{-1}}$
Molar mass of benzene, $C{6}H_{6}$ = (6 x 12+ 6 x 1) g${{mol}^{-1}}$
= 78 g${{mol}^{-1}}$
Energy needed to vaporise benzene = 30.8 kJ${{mol}^{-1}}$ x 100g / 78 g${{mol}^{-1}}$
= 39.49kJ
So Time = energy/power = 39.49 / 100W = 39.49 x ${{10}^{3}}$ J / 100J/s
= 394.9 s = 6.6 min