Question

$2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$

$100$ mL of $X$ molar $H_2{O}_2$ gives $3$ L of $O_2$ gas under the condition when $1$ mol occupies $24$ L. The value of $X$ is :

• A. 2.5
• B. 1
• C. 0.5
• D. 0.25

Answer 1

The correct option is A 2.5

$2H_2{O}_2\left(l\right)\to 2H_{2}O\left(l\right)+O_2\left(g\right)$

$100$ mL of $X$ molar $H_2{O}_2$ gives $3$ L of $O_2$ gas.
$1$ mol occupies $24$ L.

Hence, 3 L corresponds to $\frac{1\times 3}{24}=\frac{1}{8}$ mol.

This corresponds to the moles of oxygen.

The moles of $H_2{O}_2$ are twice i.e, $2\times \frac{1}{8}=\frac{1}{4}$

The molarity of $H_2{O}_2$ solution is $\frac{1mol}{4\times \frac{100mL}{1000mL/L}}=2.5M$.

Hence, option A is correct.