Question

$H_2{O}_2$ acts as both oxidising and reducing agent. As oxidising agent, its product is $H_{2}O$ but as reducing agent, its product is $O_2$. Volume strength has great significance for chemical reactions. The strength of ${}^{\prime }10V’$ means 1 volume (or litre) of $H_2{O}_2$ on decomposition $\left(H_2{O}_2\to H_{2}O+\frac{1}{2}{O}_2,\right)$ gives 10 volumes (or litre) of oxygen at $NTP$.
$15gBa{\left(Mn{O}_4\right)}_2$ sample containing inert impurity is completely reacting with $100mLof‘11.2V’H_2{O}_2,$ then what will be the $\%$ purity of $Ba{\left(Mn{O}_4\right)}_2$ in the sample: (Atomic mass: $Ba=137,Mn=55$)

• A. $5/$
• B. $/10$
• C. $50/$
• D. none of these

Answer 1

Answer: (C)

milliequivalents of $Ba{\left(Mn{O}_4\right)}_2$ reacted = milliequivalents of $H_2{O}_2$ reacted
$\frac{m}{375}\times 10\times 1000=\frac{11.2}{5.6}\times 100$
$\therefore m=7.5g$
$\therefore wt.ofBa{\left(Mn{O}_4\right)}_2$ in sample $=7.5g$
$\%$ purity of $Ba{\left(Mn{O}_4\right)}_2$ in sample $=\frac{7.5}{15}\times 100=50\%$