Question

The volume of $O_2$ liberated from $0.96g$ of $H_2{O}_2$ at STP is?

• A. $224.6mL$
• B. $316.2mL$
• C. $390.0mL$
• D. $112.5mL$

Answer 1

The correct option is B $316.2mL$

Solution:- (B) $316.2mL$

As we know that,

$\text{No. of moles}=\frac{\text{mass}}{\text{molecular mass}}$

$2H_2{O}_2⟶2H_{2}O+O_2$

Given mass of $H_2{O}_2=0.96g$

Molecular mass of $H_2{O}_2=34g$

$\therefore \text{No. of moles of}{H}_2{O}_2=\frac{0.96}{34}=0.0282\text{moles}$

From the above reaction-

$2$ mole of $H_2{O}_2$ gives $1$ mole of $O_2$

$\therefore 0.0282$ mole of $H_2{O}_2$ will give $0.0141$ mole of $O_2$

As we know that,

Volume of $1$ mole of a gas at NTP $=22400mL$

$\therefore$ Volume of $0.0141$ mole of $O_2$ at NTP $=22400\times 0.0141=315.84mL\approx 316.2mL$

Hence the correct answer is $316.2mL$.