One mole of anhydrous ${\mathrm{MgCl}}_2$ dissolves in water and liberates $25 \mathrm{cal}/\mathrm{mol}$ of heat. ${\mathrm{\Delta H}}_{\text{hydration }}$ of ${\mathrm{MgCl}}_2=-30 \mathrm{cal}/\mathrm{mol}$. Heat of dissolution of ${\mathrm{MgCl}}_2.{\mathrm{H}}_2\mathrm{O}$ is :

${\mathrm{MgCl}}_2+{\mathrm{H}}_2\mathrm{O}\stackrel{∆{\mathrm{H}}_{\mathrm{sol}}}{\to }{\mathrm{MgCl}}_2\cdot {\mathrm{H}}_2\mathrm{O}$

${\mathrm{\Delta H}}_{\mathrm{sol}}=-25/\mathrm{mol}$ (given)

$\begin{array}{l}{\mathrm{MgCl}}_2+{\mathrm{H}}_2\mathrm{O}\stackrel{{\mathrm{\Delta H}}_{\text{sol }}}{⟶}{\mathrm{Mg}}_{\left(\mathrm{aq}\right)}^{+2}+2{\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}\\ {\mathrm{\Delta H}}_{\text{hyddration }}=-30 \mathrm{cal}/\mathrm{mol}\\ {\mathrm{\Delta H}}_{\text{dissolution }}={\mathrm{\Delta H}}_{\text{hydration }}-{\mathrm{\Delta H}}_{\text{solution }}\\ =-30-(-25)\\ =-5 \mathrm{cal}/\mathrm{mol}\end{array}$