Question

A solution of a $0.4$ g sample of $H_2{O}_2$ reacted with $0.632$ g of $KMn{O}_4$ in the presence of sulphuric acid. The percentage purity of the sample of $H_2{O}_2$ is:

• A. $83\%$
• B. $84\%$
• C. $82\%$
• D. $85\%$

Answer 1

Answer: (D)

$85\%$

The molar mass of $KMn{O}_4$ is $158$ g/mol.
$0.632$ g of $KMn{O}_4$ corresponds to $\frac{0.632}{158}=0.004$ mol.
In acidic medium, $2Mn{O}_4^{-}+6H^{+}+5H_2{O}_2\to 2M{n}^{2+}+8H_{2}O+5O_2$
$2$ moles of $KMn{O}_4$ will react with $5$ moles of $H_2{O}_2$.
$0.004$ moles of $KMn{O}_4$ will react with $\frac{5}{2}\times 0.004=0.01$ moles of $H_2{O}_2$.
The molar mass of $H_2{O}_2$ is $34$ g/mol.
$0.01$ moles of $H_2{O}_2$ corresponds to $0.01\times 34=0.34$ g.
The percentage purity of the sample is $\frac{0.34}{0.4}\times 100=85$%.