Question

A certain volume of $H_2$ effuses from an apparatus in one minute. The same volume of ozonized oxygen $(O_3+O_2)$ mixture took 246 seconds to effuse from apparatus under identical conditions. Percentage of $O_2$ by mole in mixture is:

• A. 89.87%
• B. 10.13%
• C. 76.54%
• D. 73.12%

Answer 1

The correct option is A 89.87%

$\frac{r_{H_2}}{r_{O_2}}=\sqrt{\frac{M_{av}}{M_{H_2}}}$
$\frac{1}{t_{H_2}}/\frac{1}{t_{mix}}=\sqrt{\frac{M_{av}}{2}}$
$\frac{246}{60}=\sqrt{\frac{M_{av}}{2}}$
${\left(\frac{246}{60}\right)}^2=\frac{M_{mix}}{2}$
$M_{mix}=\frac{246\times 246\times 2}{60\times 60}=33.62$
$100\times 33.62=x\times 32+(100-x)48$
$x=89.87$%