From the given reaction $2 \mathrm{KMnO}_4 +3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O}_2…
From the given reaction
$2 \mathrm{KMnO}_4 +3 \mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{MnSO}_4 +8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{O}_2$
Find the normality of $\mathrm{H}_2 \mathrm{O}_2$ solution, if $20 \mathrm{~mL}$ of it is required to react completely with 16 $\mathrm{mL}$ of $0.02 \mathrm{M} \mathrm{KMnO}_4$ solution.
$\left(\right.$ Molar mass of $\mathrm{KMnO}_4=158 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$4 \times 10^{-2} \mathrm{~N}$
$2 \times 10^{-2} \mathrm{~N}$
$6 \times 10^{-2} \mathrm{~N}$
$8 \times 10^{-2} \mathrm{~N}$
Solution
Using law of equivalence
Number of gram equivalence of $\mathrm{H}_2 \mathrm{O}_2$
$=\text { number of gram equivalence of } \mathrm{KMnO}_4$
Number of gram equivalence
$=\text { normality } \times \text { volume }(\text { in litres) }$
or normality $=$ molarity $\times$ valence factor for $\mathrm{KMnO}_4$ is 5 in this reaction.
Since, equivalences are equal, considering volume in $\mathrm{mL}$. So,
$\begin{aligned}
N & \times 20=0.02 \times 5 \times 16 \\
N & =\frac{0.02 \times 5 \times 16}{20}=\frac{0.01 \times 5 \times 16}{10} \\
& =0.001 \times 5 \times 16 \\
& =0.001 \times 80=8 \times 10^{-2} \mathrm{~N}
\end{aligned}$