To solve this problem, we need to apply Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.$\newline$Given: Equal moles of hydrogen ($H_2$) and oxygen ($O_2$) gases are placed in a container with a pin-hole.$\newline$We need to find the fraction of oxygen that escapes when half of the hydrogen escapes.$\newline$Step-by-step solution:$\newline$1. According to Graham's law, the rate of effusion $(r)$ is given by:$\newline r \propto \frac{1}{\sqrt{M}} \newline$where $M$ is the molar mass of the gas.$\newline$2. The molar mass of $H_2$ is 2 g/mol, and the molar mass of $O_2$ is 32 g/mol.$\newline$3. The ratio of the rates of effusion of $H_2$ and $O_2$ is:$\newline \frac{r_{H_2}}{r_{O_2}} = \frac{\sqrt{M_{O_2}}}{\sqrt{M_{H_2}}} = \frac{\sqrt{32}}{\sqrt{2}} = \frac{4}{1} \newline$This means that $H_2$ effuses 4 times faster than $O_2. \newline$4. If half of the $H_2$ escapes, then the fraction of $H_2$ remaining is $\frac{1}{2}. \newline$5. Since $H_2$ effuses 4 times faster, the time taken for half of the $H_2$ to escape will allow $O_2$ to effuse by a fraction of:$\newline \frac{1}{4} \times \frac{1}{2} = \frac{1}{8} \newline$6. Therefore, the fraction of $O_2$ that escapes is $\frac{1}{8}. \newline$7. However, the question asks for the fraction of $O_2$ that escapes, not the fraction remaining.$\newline$8. The fraction of $O_2$ that escapes is $1 - \frac{7}{8} = \frac{1}{8}. \newline$Therefore, the correct option is Option 2: $\frac{1}{4}.$