KCET · Chemistry · States of Matter
A gas mixture contains \(25 \%\) He and \(75 \% \mathrm{CH}_{4}\) 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 & Solution
Correct Answer
(C) \(92 \%\)
Step-by-step Solution
Detailed explanation
Gas mixture contains \(25 \%\) He and \(75 \% \mathrm{CH}_{4}\) by volumes. We have to calculate the mass percentage of methane in the mixture.
molar mass of \(\mathrm{He}=4\)
molar mass of \(\mathrm{CH}_{4}=16\)
\(\therefore\) Mass percentage of methane in the mixture shall be \(=\frac{75 \times 16}{25 \times 4+75 \times 16} \times 100=92 \%\)
molar mass of \(\mathrm{He}=4\)
molar mass of \(\mathrm{CH}_{4}=16\)
\(\therefore\) Mass percentage of methane in the mixture shall be \(=\frac{75 \times 16}{25 \times 4+75 \times 16} \times 100=92 \%\)
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