The enthalpy change for the reaction ${\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+(1/2){\mathrm{O}}_2\left(\mathrm{g}\right)$ is $-98.3 \mathrm{kJ} {\mathrm{mol}}^{-1}$ If the enthalpy of formation of ${\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)$ is $-187.4 \mathrm{kJ} {\mathrm{mol}}^{-1}$the enthalpy of formation of ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ would be

The reaction to be considered is ${\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Given data are 

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+(1/2){\mathrm{O}}_2\left(\mathrm{g}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta }H=-98.3\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)⟶{\mathrm{H}}_2{\mathrm{O}}_2\left(\mathrm{l}\right)\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta }H=-187.4\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

Adding these two reactions gives the required reaction. Hence

$\mathrm{\Delta }H=-(98.3+187.4) \mathrm{kJ} {\mathrm{mol}}^{-1}=-285.7 \mathrm{kJ} {\mathrm{mol}}^{-1}$