Question

A gas mixture contains $25\% \,He$ and $75\%\, CH_4$ by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately____

• A. 75%
• B. 25%
• C. 92%
• D. 8%

Answer 1

The Correct Option is C

Since temperature and pressure are constant, we can assume it to be in STP condition.

Now, let the total volume be $100\, L$

Moles of $H e$ present $=\frac{25}{22.4}=1.12\, mol$

Mass of $H e$ present $=1.12 \times 4=4.48 \,g$

Moles of $C H_{4}$ present $=\frac{75}{22.4}=3.35\, mol$

Mass of $C H_{4}$ present $=3.35 \times 16=53.60 \,g$

Hence, the mass percentage of $C H_{4}=\frac{53.60}{58.08} \times 100=92.28 \%$

Answer 2

To find the percentage by mass of methane (CH$_4$) in the mixture, we will first calculate the molar masses of the components:
Molar mass of He = 4 g/mol,
Molar mass of CH$_4$ = 16 g/mol. 
Assuming 100 L of the gas mixture (since percentages are by volume, it simplifies calculations): 
- Volume of He = 25 L,
- Volume of CH$_4$ = 75 L.
Now, using the ideal gas law, the number of moles of each gas will be proportional to the volume, assuming the same temperature and pressure:
- Moles of He = \(\frac{25}{22.4} = 1.116 \, mol\),
- Moles of CH$_4$ = \(\frac{75}{22.4} = 3.348 \, mol\).
Next, we calculate the mass of each gas:
- Mass of He = \(1.116 \times 4 = 4.464 \, g\),
- Mass of CH$_4$ = \(3.348 \times 16 = 53.568 \, g\).
Total mass of the mixture = \(4.464 + 53.568 = 58.032 \, g\).
Mass percentage of CH$_4$ = \(\frac{53.568}{58.032} \times 100 = 92\%\).

Thus, the percentage by mass of methane in the mixture is approximately 92%.