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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____
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$92 \%$
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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