Question

$X$ $ml$ of $H_2$ effuse out through a hole in container in $5$ seconds, the time taken for the effusion of same volume of gas specified below under identical conditions is

• A. $10$ sec; $H_2$
• B. $20$ sec; $O_2$
• C. $25$ sec; $CO$
• D. $55$ sec; ${CO}_2$

Answer 1

Answer: (B)

$20$ sec; $O_2$

$\frac{t_{H_2}}{t_a}=\sqrt{\frac{M_{H_2}}{M_a}}$ where $a=$ the other gas
$\frac{5}{t_a}=\sqrt{\frac{2}{M_a}}$
$\frac{25}{t_a^2}=\frac{2}{M_a}$
$\frac{M_a}{t_a^2}=\frac{2}{25}$
For (b) $\frac{M_{O_2}}{{\left(20\right)}^2}=\frac{32}{400}=\frac{2}{25}$