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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 $246 \sec$ to effuse from apparatus under identical conditions. Percentage of $O_{2}$ by mole in mixture is:

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$89.87 %$

$10.13 %$

$76.54 %$

$73.12 %$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{60}{246} = \sqrt{\frac{2}{M}} \\ M = 33.62 \end{array}$

Now, $33.62 = \frac{x \times 32 + (100 - x) \times 4}{100}$

$x = 89.87 %$

Hence, the percentage of $O_{2}$ by a mole in the mixture is 89.87 %.

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