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A gas mixture contains 50% helium and 50% methane by volume. What is the percent by weight of methane in the mixture

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80.03%

20.05%

50%

75%

answer is A.

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Detailed Solution

$Solution contain He + {CH}_{4} Their mol . wt . = 4 + 16 = 20 % wt . of {CH}_{4} = \frac{wt of {CH}_{4}}{Total wt .} \times 100 = \frac{16}{20} \times 10 = 80.0 %$

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