A container with a pin-hole contains equal moles of H₂₍𝓰₎ and O₂₍𝓰₎. Find the fraction of oxygen gas escaped at the same time when one-fourth of hydrogen gas escapes

The correct option is 1/16.

To understand why this is correct, we can apply Graham's Law of Effusion. According to Graham's Law, the rate of effusion (or escape) of a gas is inversely proportional to the square root of its molar mass, and directly proportional to its partial pressure (or the number of moles present in the container):

$r\propto \frac{P}{\sqrt{M}}$

Since the container initially contains equal moles of H_2(g) and O_2(g), their initial partial pressures are equal. Thus, the ratio of their rates of effusion depends only on their molar masses:
Molar mass of Hydrogen (H₂), $M_{H_2}=2\text{ g/mol}$
Molar mass of Oxygen (O₂), $M_{O_2}=32\text{ g/mol}$

The ratio of the rate of effusion of oxygen ($r_{O_2}$) to that of hydrogen ($r_{H_2}$) is given by:

$\frac{r_{O_2}}{r_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{O_2}}}$

Substituting the molar masses:

$\frac{r_{O_2}}{r_{H_2}}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$

The rate of effusion can also be expressed as the fraction of gas escaped ($f$) over a given time interval ($t$) since the initial moles are equal:

$\frac{f_{O_2}}{f_{H_2}}=\frac{r_{O_2}}{r_{H_2}}=\frac{1}{4}$

We are given that one-fourth of the hydrogen gas escapes, which means:
$f_{H_2}=\frac{1}{4}$

Now, we substitute this value into our ratio to find the fraction of oxygen gas escaped ($f_{O_2}$):

$f_{O_2}=\frac{1}{4}\times f_{H_2}=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

Therefore, the fraction of oxygen gas escaped at the same time is indeed 1/16.