${\mathrm{H}}_2{\mathrm{O}}_2$ acts as both oxidising as well as 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 '10 V' means 1 volume (or litre) of ${\mathrm{H}}_2{\mathrm{O}}_2$ on decomposition  $\left({\mathrm{H}}_2{\mathrm{O}}_2⟶{\mathrm{H}}_2\mathrm{O}+\frac{1}{2}{\mathrm{O}}_2\right)$ gives 10 volumes (or litre) of oxygen at STP.

15g $\mathrm{Ba}{\left({\mathrm{MnO}}_4\right)}_2$ sample containing inert impurity is completely reacted with 100 mL of '11.2 V'  ${\mathrm{H}}_2{\mathrm{O}}_2$ then what will be the percentage purity of Ba ${\left({\mathrm{MnO}}_4\right)}_2$ in the sample? (Atomic mass: 137, Mn =55 ) (Assume 1 mole of an ideal gas occupies (Assume 1 mole of an ideal gas occupies $22.4\mathrm{L}\text{ at STP) }$.