Question

A certain volume of $H_2$ effuses from an apparatus in one minute. The same volume of ozonised oxygen ($O_3$ + $O_2$) mixture took 246 sec. 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%

$\frac{r_{H_2}}{r_{O_3+O_2}}=\frac{t_{O_3+O_2}}{t_{H_2}}=\sqrt{\frac{M_{O_3+O_2}}{M_{H_2}}}$

$\frac{r_{246}}{r_{60}}=\sqrt{\frac{M_{O_3+O_2}}{M_2}}$

$M_{O_3+O_2}=33.62g/mol$

$100\times 33.62=48(100-x)+32x$

$x=89.87$%

Hence, the percentage of O2 by a mole in the mixture is 89.87 %

Hence, the correct option is $\text{A}$