Given data:

Enthalpy of solution for ${\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O}=8.8\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

Enthalpy of solution for ${\mathrm{BaCl}}_2=-20.6\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

From this, we have to calculate the heat of hydration of ${\mathrm{BaCl}}_2$.

Formula/concept used: Concept of thermochemistry.

Detailed explanation:

From the given data,

(i) ${\mathrm{BaCl}}_2\left(s\right)+aq\to {\mathrm{BaCl}}_2\left(aq\right) \Delta {\mathrm{H}}_1=-20.6\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

(ii) ${\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)+aq\to {\mathrm{BaCl}}_2\left(aq\right)+2{\mathrm{H}}_2\mathrm{O} \Delta {\mathrm{H}}_2=8.8\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

(iii) ${\mathrm{BaCl}}_2\left(aq\right)+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)+aq \Delta {\mathrm{H}}_2=-8.8\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{mol}$

On adding equation (i) and (iii), we get,

$$\begin{gathered} {\mathrm{BaCl}}_2\left(s\right)+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{BaCl}}_2·2{\mathrm{H}}_2\mathrm{O}\left(s\right) \\ \Delta H=\Delta H_1+\Delta H_2 \end{gathered}$$

Thus, $\Delta H=-20.6-8.8$

$\Rightarrow \Delta H=-29.4 \mathrm{kJ}/\mathrm{mol}$

Hence, option (A) is correct.