A 20.0 ml solution containing 0.2 g impure
Question:

A $20.0 \mathrm{~mL}$ solution containing $0.2 \mathrm{~g}$ impure $\mathrm{H}_{2} \mathrm{O}_{2}$ reacts completely with $0.316 \mathrm{~g}$ of $\mathrm{KMnO}_{4}$ in acid solution. The purity of $\mathrm{H}_{2} \mathrm{O}_{2}$ (in \%) is

__________(mol. wt. of $\mathrm{H}_{2} \mathrm{O}_{2}=34 ;$ mol. wt. of$\left.\mathrm{KMnO}_{4}=158\right)$

Solution:

$\mathrm{Eq}$ of $\mathrm{H}_{2} \mathrm{O}_{2}=\mathrm{Eq}$ of $\mathrm{KMnO}_{4}$

$x \times 2=\frac{0.316}{158} \times 5$

$x=5 \times 10^{-3} \mathrm{~mol}$

$\mathrm{m}_{\mathrm{H}_{2} \mathrm{O}_{2}}=5 \times 10^{-3} \times 34=0.17 \mathrm{gm}$

$\% \mathrm{H}_{2} \mathrm{O}_{2}=\frac{0.17}{0.2} \times 100=85$

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