Question

In acidic medium, 100 ml of 1 M $KMn{O}_4$ solution oxidizes 100 ml of $H_2{O}_2$ solution. The volume of 0.01 M $KMn{O}_4$, required to oxidize the same volume of $H_2{O}_2$ in alkaline medium in ml is :

• A. $\frac{300}{2}$
• B. $\frac{300}{5}$
• C. $\frac{500}{3}$
• D. $\frac{500}{2}$

Answer 1

Answer: (C)

$\frac{500}{3}$

In acidic medium, the equivalent weight of $KMn{O}_4$ is one-fifth of its molecular mass.

$N_1{V}_1=N_2{V}_2$

$100\times 1\times 5=100\times N_2$

Molarity of $H_2{O}_2=5M$

In alkaline medium, the equivalent weight of $KMn{O}_4$ is one-third of its molecular mass.

$N_1{V}_1=N_2{V}_2$

$0.01\times 3\times V_2=5\times 100$

$V_2=\frac{500}{3}$

The volume of 0.01 M $KMn{O}_4$, required to oxidize the same volume of $H_2{O}_2$ in alkaline medium in ml $=\frac{500}{3}mL$.

Hence, option C is correct.