Question

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 ?

• A. 44565
• B. 44628
• C. 44563
• D. 44569

Answer 1

The Correct Option is D

Diffusion of Gases Calculation 

The ratio of the rates of diffusion of oxygen (\(O_2\)) and hydrogen (\(H_2\)) is given by Graham's law of diffusion:

\(\frac{r_{O_2}}{r_{H_2}} = \sqrt{\frac{M_{H_2}}{M_{O_2}}}\)

Where: - \( r_{O_2} \) and \( r_{H_2} \) are the rates of diffusion of oxygen and hydrogen, respectively, - \( M_{H_2} \) and \( M_{O_2} \) are the molar masses of hydrogen and oxygen, respectively.

Substitute the values for the molar masses of hydrogen and oxygen:

\(\frac{\frac{n_{O_2}}{t}}{\frac{n_{H_2}}{t}} = \sqrt{\frac{2}{32}} = \sqrt{\frac{1}{16}} = \frac{1}{4}\)

This simplifies to:

\(\frac{n_{O_2}}{n_{H_2}} = \frac{1}{4}\)

Calculation of Moles of \( O_2 \) Diffused

If \( \frac{1}{2} \) moles of \( H_2 \) are diffused in a given time, we can find the moles of \( O_2 \) diffused in the same time using the ratio:

\(\frac{n_{O_2}}{1/2} = \frac{1}{4} \Rightarrow n_{O_2} = \frac{1}{8}\)

Conclusion:

The number of moles of oxygen (\( O_2 \)) diffused in the same time is \( \frac{1}{8} \) moles.