Question

A $20.0mL$ solution containing $0.2g$ impure $H_{2}O_{2}$ reacts completely with $0.316g$ of $KMnO_{4}$ in acid solution. The purity of $H_{2}O_{2}\left(\right.in\%\left.\right)$ is (mol. wt. of $H_{2}O_{2}=34^{'}$ mole wt. of $KMnO_{4}=158$ )

Answer 1

$H_{2}O_{2}+KMnO_{4} \rightarrow Mn^{+ 2}+O_{2}$
[moles of $H_{2}O_{2}$ ] $\times 2=\frac{0 . 316}{158}\times 5$
moles of $H_{2}O_{2}=5\times 10^{- 3}$
mass of $H_{2}O_{2}=170\times 10^{- 3}g$
$\%$ purity $=\frac{170 \times 10^{- 3}}{0 . 2}\times 100=85\%$