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?

Answer 1

Step 1: Analyse the formula.

Assume one mole of Hydrogen ($\mathrm{H}$) and one mole of Oxygen ($\mathrm{O}$) gases are placed in a container with a pin-hole through which both can escape.

$\frac{{\mathrm{n}}_{\mathrm{O}}{\mathrm{t}}_{\mathrm{H}}}{{\mathrm{n}}_{\mathrm{H}}{\mathrm{t}}_{\mathrm{O}}}=\sqrt{\frac{{\mathrm{M}}_{\mathrm{H}}}{{\mathrm{M}}_{\mathrm{O}}}}$

Here, ${\mathrm{n}}_{\mathrm{O}}$ represents the number of moles of oxygen effused in time${\mathrm{t}}_{\mathrm{O}}$​ and ${\mathrm{M}}_{\mathrm{O}}$​ is the molar mass of oxygen.

Similarly, ${\mathrm{n}}_{\mathrm{H}}$ represents the number of moles of hydrogen effused in time ${\mathrm{t}}_{\mathrm{H}}$​ and ${\mathrm{M}}_{\mathrm{H}}$​ is the molar mass of hydrogen.

Step 2: Find the fraction of Oxygen escaped.

According to the question, one-half of the Hydrogen has escaped.

The molar mass of Oxygen gas and Hydrogen gas is $32\mathrm{g}{\mathrm{mol}}^{-1}$ and $2\mathrm{g}{\mathrm{mol}}^{-1}$

Also, ${\mathrm{t}}_{\mathrm{O}}$ is equal to ${\mathrm{t}}_{\mathrm{H}}$, according to the question.

So, substituting these values in the formula, we get

:$$\begin{gathered} \frac{{\mathrm{n}}_{\mathrm{O}}{\mathrm{t}}_{\mathrm{H}}}{{\mathrm{n}}_{\mathrm{H}}{\mathrm{t}}_{\mathrm{O}}}=\sqrt{\frac{{\mathrm{M}}_{\mathrm{H}}}{{\mathrm{M}}_{\mathrm{O}}}} \\ \Rightarrow \frac{{\mathrm{n}}_{\mathrm{O}}}{\frac{1}{2}}=\sqrt{\frac{2}{32}} \end{gathered}$$

So, ${\mathrm{n}}_{\mathrm{O}}=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}$

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