Chemistry - Thermochemistry Question with Solution | TestHub

The enthalpy of vapourisation $\left(∆\mathrm{H}\right)$ of ${\mathrm{CCl}}_4$$=30.5 \mathrm{kJ}/\mathrm{mol}$

${\mathrm{q}}_{\mathrm{p}}=∆\mathrm{H}=30.5 \mathrm{kJ}/\mathrm{mol}$

No. of moles of ${\mathrm{CCl}}_4=\frac{\mathrm{given} \mathrm{mass}}{\mathrm{molar} \mathrm{mass}}$

                                   $=\frac{284}{154}=1.844 \mathrm{mol}$

Hence, the heat required for the vapourisation of $1.844 \mathrm{mol}$ of ${\mathrm{CCl}}_4$

$$\begin{gathered} =1.844\times 30.5 \\ \mathrm{Heat} \mathrm{required}=56.242 \mathrm{kJ}/\mathrm{mol}\approx 56 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$