Question

$10$ mL of $H_2{O}_2$ solution (volume strength $=x$) required $10$ mL of $\frac{N}{0.56}$ $KMn{O}_4$ solution in acidic medium. Hence, $x$ is:

• A. $0.56$
• B. $5.6$
• C. $0.1$
• D. $10.0$

Answer 1

Answer: (D)

$10.0$

The normality of $x$ volume strength $H_2{O}_2=\frac{x}{5.6}$

Hence,

$10\times N_1\left(H_2{O}_2\right)=10\times \frac{1}{0.56}\left({Mn{O}_4}^{-}\right)$

$N_1=\frac{1}{0.56}=\frac{x}{5.6}$ $⟹x=10.0$

Hence, the correct option is $D$