STD 12 -1. Solution and colligative properties — JEE STD 11 Science — Question

$\underset{68 g}{2 H_{2}} O_{2} \rightarrow 2 H_{2} O+\underset{22400 m L a t}{O_{2}}$

$68 g$ of $H _{2}$ Ogives $=22400 mL$ of $O _{2}$ at $STP$

$25.5 g H_{2} O_{2}$ gives $=\frac{22400}{68} \times 25.5=8400 mL$ of $O_{2}$ at $S T P$

$25.5 g$ of $H_{2} O_{2}$ is present in $1000 m L$ of $H _{2} O _{2}$ solution $1000 mL$ of $H _{2} O _{2}$ gives $8400 mL$ of $O _{2}$ at $STP$

$1 mL$ of $H _{2} O _{2}$ gives $\frac{8400}{1000} mL$ of $O _{2}$ at $STP$

$=8.4 m L$ of $O_{2}$

Hence, volume strength of $1.5 NH _{2} O _{2}=8.4$ volume. Or mass of $H_{2} O_{2}$ in $1.5 N$ solution $=E W$ of $H_{2} O_{2} \times 1.5 N$

$=17 \times 1.5=25.5 g / L$

Hence, volume strength of $1.5 NH _{2} O _{2}$ solution $=\frac{22.4 \times 25.5}{68}=8.4$