Question

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

• A. $0.56$
• B. $5.6$
• C. $0.1$
• D. $10$

Answer 1

The correct option is D $10$

Milliequivalents of $H_2{O}_2=$ Milliequivalents of $Mn{O}_4^{-}$

$10\times N_1\left(H_2{O}_2\right)\equiv 10\times \frac{N}{0.56}Mn{O}_4^{⊝}$
$N_1\left(H_2{O}_2\right)=\frac{1}{0.56}$
$x$ volume $H_2{O}_2$ has normality $=\frac{x}{5.6}$
$\frac{x}{5.6}=\frac{1}{0.56}\Rightarrow x=10$