Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

Assume 1 mole of hydrogen and 1 mole of oxygen gases are placed in a container with a pin-hole through which both can escape.

$\frac{n_O t_H}{n_H t_O} = \sqrt{\frac{M_H}{M_O}}$

$\frac{1}{2} = \sqrt{\frac{2}{32}} \approx \frac{1}{4}$

$ \frac{n_O}{n_H} = \frac{t_O}{t_H} \sqrt{\frac{M_H}{M_O}} $

$n_O t_H = n_H t_O \sqrt{\frac{M_H}{M_O}}$

Let $n_O$ be the number of moles of oxygen effused in time $t_O$ and $M_O$ is the molar mass of oxygen. Similarly, $n_H$ represents the number of moles of hydrogen effused in time $t_H$ and $M_H$ is the molar mass of hydrogen.

$\frac{n_O}{n_H} = \sqrt{\frac{M_H}{M_O}} \frac{t_O}{t_H}$

Given that $n_H = 1$ and $n_O = 1$, and half of the hydrogen escapes ($n_H = \frac{1}{2}$) in time $t_H$. We want to find the fraction of oxygen that escapes in the same time $t_H$.

$\frac{n_O}{\frac{1}{2}} = \sqrt{\frac{2}{32}} \frac{t_O}{t_H}$

$2 n_O = \frac{1}{4} \frac{t_O}{t_H}$

If $t_O = t_H$, then

$2 n_O = \frac{1}{4}$

$n_O = \frac{1}{8}$

The fraction of the oxygen that escapes in the time required for one-half of the hydrogen to escape is $\frac{1}{8}$

Note: Here $n_O$ represents the number of moles of oxygen effused in time $t_O$ and $M_O$ is the molar mass of oxygen. Similarly, $n_H$ represents the number of moles of hydrogen effused in time $t_H$ and $M_H$ is the molar mass of hydrogen.

Hence, the correct option is A.