Commercial 11.2 volume ${\mathrm{H}}_2{\mathrm{O}}_2$ solution has a molarity of

10 Vol. ${\mathrm{H}}_2{\mathrm{O}}_2$ = 3.035 g ${\mathrm{H}}_2{\mathrm{O}}_2$

$$\begin{gathered} \therefore 11.2 \mathrm{Vol} {\mathrm{H}}_2{\mathrm{O}}_2 = \frac{3.035 \times 11.2}{10} = 33.99 \mathrm{g}, \\ \mathrm{molarity} = \frac{\mathrm{Strength}}{\mathrm{mol}.\mathrm{wt}} = \frac{33.99}{34}=1 \mathrm{M} \end{gathered}$$