$H_2{O}_2$ acts as both oxidising and reducing agent. As oxidising agent, its product is ${\mathrm{H}}_2\mathrm{O}$

but as reducing agent, its product is ${\mathrm{O}}_2.$ Volume strength has great significance for chemical reactions. The strength of ‘10V’ means 1 volume (or liter) of ${\mathrm{H}}_2{\mathrm{O}}_2$ on decomposition (${\mathrm{H}}_2{\mathrm{O}}_2$$\stackrel{ }{\to }{\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2)$ gives 10 volumes (or liter) of oxygen at NTP.

15 g $\mathrm{Ba}{\left({\mathrm{MnO}}_4\right)}_2$ sample containing inert impurity is completely reacting with 100 mL of ‘11.2 V’ ${\mathrm{H}}_2{\mathrm{O}}_2$, then what will be the % purity of $\mathrm{Ba}{\left({\mathrm{MnO}}_4\right)}_2$ in the sample : (Atomic mass : Ba = 137, Mn = 55)