What percentage by mass of methane will be present in a mixture containing $50 %$ helium and $50 %$ methane by volume?

70 %

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40 %

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80 %

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60 %

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Solution

The correct option is C $80 %$
For gases, mole and volume ratios are the same at a fixed temperature and pressure.
Since, the mixture contains $50 %$ helium and $50 %$ methane by volume so, it will have $1 : 1$ mole ratio. Let the mixture contains $1$ mole of $H e$ and $1$ mole of $C H_{4} .$
Mass of $1$ mole of $C H_{4} = 16 g$
Mass of $1$ mole of $H e = 4 g$
Total mass of mixture $= 20 g$
Percentage by mass of $C H_{4}$
$= \frac{mass of C H_{4}}{Total mass} \times 100 = \frac{16}{20} \times 100 = 80$
$∴$ Percentage of $C H_{4}$ present in the mixture by mass $= 80 %$

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What percentage by mass of methane will be present in a mixture containing 50% helium and 50% methane by volume?

What percentage by mass of methane will be present in a mixture containing $50 %$ helium and $50 %$ methane by volume?