Question

$34g$ hydrogen peroxide is present in $1120mL$ of the solution. The volume strength of $H_2{O}_2$ is :

• A. 10 V
• B. 20 V
• C. 30 V
• D. 32 V

Answer 1

The correct option is A 10 V

$H_2{O}_2\left(aq\right)\to H_{2}O\left(l\right)+\frac{1}{2}{O}_2\left(g\right)$
Molecular mass of $H_2{O}_2=34gmo{l}^{-1}$
Moles of $H_2{O}_2=\frac{34}{34}=1mol$
Weight of $H_2{O}_2$ in $1mL=\frac{34}{1120}g$
In $1120mL$ of solution there is $1mol$ of $H_2{O}_2$
So, in $1000mL$ of solution, moles of $H_2{O}_2=\frac{1\times 1000}{1120}=\frac{100}{112}mol$
$1mol$ of $H_2{O}_2$ gives $11.2L$ of $O_2$ at STP
So,
$\frac{100}{112}mol$ of $H_2{O}_2$ will give $11.2\times \frac{100}{112}=10L$ of $O_2$ at STP

By definition , volume strength of $H_2{O}_2$ is $10V$