Question

10 mL of a solution of $H_2{O}_2$ labelled '10 volume' just decolorises 100 mL of potassium permanganate solution acidified with dilute $H_{2}S{O}_4$. Calculate the amount of potassium permanganate in the given solution.

• A. 0.1563 g
• B. 0.563 g
• C. 5.63 g
• D. 0.256 g

Answer 1

The correct option is B 0.563 g

Given reaction is:
$2H_2{O}_2\left(aq\right)\stackrel{\Delta }{\to }2H_{2}O\left(l\right)+O_2\left(g\right)$
10 mL (0.01 L) volume $H_2{O}_2$ aqueous solution means on heating 1 mL of the $H_2{O}_2$ aqueous solution will produce 10 mL of $O_2$ gas at STP.
Volume strength of $H_2{O}_2$ aqueous solution can be related to the molarity of $H_2{O}_2$ solution as
Volume strength of $H_2{O}_2=$ Molarity of $H_2{O}_2\times 11.210=\text{Molarity}\times 11.2$
Molarity of $H_2{O}_2$ solution = $\frac{0.01}{11.2}M$
$\underset{n=2}{H_2{O}_2\left(aq\right)}+\underset{n=5}{KMn{O}_4}\left(aq\right)\stackrel{H^{+}}{\to }M{n}^{2+}\left(aq\right)+O_2$
n - factor of $H_2{O}_2\text{in the above reaction is}2$
n - factor of $KMn{O}_4$ in the above reaction is 5
$\text{Equivalents of}{H}_2{O}_2=\text{equivalents of}KMn{O}_4$
$n-factor\times Molarity\times Volume=\text{no. of moles of}KMn{O}_4\times n-factor$
$2\times \frac{0.01}{11.2}\times 10=\frac{x}{158}\times 5x=0.564g$