Standard enthalpy of vapourisation for CCl _ 4 is 30 . 5kJ mol ^ -...

The enthalpy of vapourisation $\left(∆ H\right)$ of {CCl}_{4}= 30 . 5kJ / mol

${q}_{p} = ∆ H = 30 . 5kJ / mol$

No. of moles of ${CCl}_{4} = \dfrac{givenmass}{molarmass}$

                                   $= \dfrac{284}{154} = 1 . 844mol$

Hence, the heat required for the vapourisation of $1 . 844mol of {CCl}_{4}$

$= 1 . 844 \times 30 . 5Heatrequired = 56 . 242kJ / mol \approx 56kJ / mol$