Question

A certain volume of $H_2$ effuses from an apparatus in one minute. The same volume of ozonized oxygen $(O_3+O_2\text{mix})$ took $246\text{s}$ to effuse from the apparatus under identical conditions. Find the % of oxygen in the ozonized oxygen.

• A. 89.87
• B. 99.87
• C. 79.87
• D. 69.87

Answer 1

The correct option is A 89.87

Let the molecular weight of the mixture $(O_3+O_2)$ be $M_2$.

From
$\frac{r_{H_2}}{r_g}=\sqrt{\frac{M_2}{M_1}}$
$\frac{V/t_1}{V/t_2}=\sqrt{\frac{M_2}{M_1}}$
$\Rightarrow \frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$
$\Rightarrow \frac{246}{60}=\sqrt{\frac{M_2}{2}}\Rightarrow M_2=33.62$ (average mol. wt. of $O_3+O_2$)
Let the mol % of $O_3$ be $a$ and the mol % of $O_2=100-a$
The average molecular weight $=\frac{a\times 48+(100-a)32}{100}$
$\Rightarrow 33.62=\frac{a\times 48+(100-a)32}{100}$
$\Rightarrow a=10.125$
$\Rightarrow O_3=10.125\therefore O_2=89.875$