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  4. H2O2 acts as both oxidising and reducing agent. As oxidising agent, its product is H2O but as reducing agent, its product is O2. Volume strength has great significance for chemical reactions. The strength of '10V’ means 1 volume (or litre) of H2O2 on decomposition (H2O2→ H2O+(1/2)O2,) gives 10 volumes (or litre) of oxygen at NTP. 15 g Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of ‘11.2 V’ H2O2, then what will be the % purity of Ba (MnO4)2 in the sample: (Atomic mass: Ba = 137, Mn = 55)

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Q. $H_2O_2$ acts as both oxidising and reducing agent. As oxidising agent, its product is $H_2O$ but as reducing agent, its product is $O_2$. Volume strength has great significance for chemical reactions. The strength of $'10V’$ means 1 volume (or litre) of $H_2O_2$ on decomposition $\left(H_2O_2\to H_2O+\frac{1}{2}O_2,\right)$ gives 10 volumes (or litre) of oxygen at $NTP$.
$15 g Ba\left(MnO_4\right)_2$ sample containing inert impurity is completely reacting with $100 mL of ‘11.2 V’ H_2O_2,$ then what will be the $\%$ purity of $Ba \left(MnO_4\right)_2$ in the sample: (Atomic mass: $Ba = 137, Mn = 55$)

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Solution:

milliequivalents of $Ba\left(MnO_4\right)_2$ reacted = milliequivalents of $H_2O_2$ reacted
$\frac{m}{375}\times10\times1000=\frac{11.2}{5.6}\times100$
$\therefore m = 7.5 \,g$
$\therefore wt. of Ba\left(MnO_{4}\right)_{2}$ in sample $= 7.5 \,g$
$\%$ purity of $Ba\left(MnO_{4}\right)_{2}$ in sample $=\frac{7.5}{15}\times100=50\%$