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

Answer: (C)

92%

Answer 2

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 $C H_{4}=\frac{53.60}{58.08} \times 100=92.28 \%$