What is the volume (in $\mathrm{mL}$ ) of $20 \mathrm{vol} \mathrm{H}_2 \mathrm{O}_2$ required to completely…
What is the volume (in $\mathrm{mL}$ ) of $20 \mathrm{vol} \mathrm{H}_2 \mathrm{O}_2$ required to completely react with $500 \mathrm{~mL}$ of $0.02 \mathrm{M}$ acidified $\mathrm{KMnO}_4$ solution?
14.0
7.0
28.0
42.0
Solution
Given,
Volume of acidified $\mathrm{KMnO}_4$ solution $=500 \mathrm{~mL}$
Molarity of acidified $\mathrm{KMnO}_4$ solution $=0.02 \mathrm{M}$
Volume strength of $\mathrm{H}_2 \mathrm{O}_2=20 \mathrm{vol}$
$\because \quad$ Normality $=\frac{\text { Volume strength }}{\text { Equivalent weight }}$
$\therefore$ Normality for $\mathrm{H}_2 \mathrm{O}_2=\frac{20}{5.6}=3.57 \mathrm{~N}$
$\because$ For $\mathrm{KMnO}_4$, reaction is in acidic medium, thus valence factor is 5 .
$\mathrm{Mn}^{7+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}$
Thus, normality for $\mathrm{KMnO}_4=$ Molarity $\times 5$
Now, applying normality equation,
$\begin{aligned}
N_1 V_1\left(\mathrm{H}_2 \mathrm{O}_2\right) & =N_2 V_2\left(\mathrm{KMnO}_4\right) \\
3.57 \times V_1 & =0.02 \times 5 \times 500 \\
V_1 & =\frac{50}{3.57}=14.0 \mathrm{~mL}
\end{aligned}$
Hence, option (a) is the correct answer.