Answer
9.83 kJ
Work Step by Step
Number of moles of $C_{6}H_{6}=\frac{25.0\,g}{78.11\,g/mol}$
$=0.320\,mol$
30.72 kJ is needed to vaporize 1 mol of $C_{6}H_{6}$ at 1 bar pressure and at its boiling point.
$\implies$ heat transfer required=
$0.320\,mol\times 30.72\,kJ/mol=9.83\,kJ$
