$\Delta {\mathrm{H}}_f$ of ${\mathrm{AlCl}}_3\left(\mathrm{ }\mathrm{s}\right),\mathrm{Al}\left(\mathrm{OH}{)}_3\right(s),{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$ and $\mathrm{HCl}\left(\mathrm{aq}\right)$ respectively are $-704,-1276$, $-286$ and $-167\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$. Then $\mathrm{\Delta H}$ for the precipitation of $0.5 \mathrm{mol} \mathrm{Al}(\mathrm{OH}{)}_3$ is (in kJ)

The balanced equation is:

$$\begin{gathered} {\mathrm{AlCl}}_3+3{\mathrm{H}}_2\mathrm{O}\to \mathrm{Al}(\mathrm{OH}{)}_3+3\mathrm{HCl} \\ {\mathrm{\Delta H}}_{\mathrm{f}}\text{ of }{\mathrm{AlCl}}_3(\mathrm{s})=-704 \mathrm{kJ}/\mathrm{mol} \\ {\mathrm{\Delta H}}_{\mathrm{f}}\text{ of }\mathrm{Al}(\mathrm{OH}{)}_3\left(\mathrm{s}\right)=-1276 \mathrm{kJ}/\mathrm{mol} \\ {\mathrm{\Delta H}}_{\mathrm{f}}\text{ of }{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)=-286 \mathrm{kJ}/\mathrm{mol} \\ {\mathrm{\Delta H}}_{\mathrm{f}}\text{ of }\mathrm{HCl}\left(\mathrm{aq}\right)=-167 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$

We know that,

$$\begin{gathered} \mathrm{\Delta H}={\mathrm{\Delta H}}_{\text{product }}-{\mathrm{\Delta H}}_{\text{reactant }} \\ \mathrm{\Delta H}={\mathrm{\Delta H}}_{\mathrm{Al}(\mathrm{OH}{)}_3}+3{\mathrm{\Delta H}}_{\mathrm{HCl}}-{\mathrm{\Delta H}}_{{\mathrm{AlCl}}_3}-3{\mathrm{\Delta H}}_{{\mathrm{H}}_2\mathrm{O}} \\ \mathrm{\Delta H}=-1276+3(-167)-(-704)-3(-286) \\ \Delta H=-215 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$

So, for precipitation of 1 mole of $\mathrm{Al}(\mathrm{OH}{)}_3=-215\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

Thus, for precipitation of $0.5$ mole of $\mathrm{Al}(\mathrm{OH}{)}_3=-215\times 0.5\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

$=-107.5 \mathrm{kJ}/\mathrm{mol}$

Hence, option (B) is correct.