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(D) Since the mixture contains $50\%$ helium and $50\%$ methane by volume,we can assume a total of $100 	ext{ moles}$ of the gas mixture.This means t

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$80$

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(D) Since the mixture contains $50\%$ helium and $50\%$ methane by volume,we can assume a total of $100 	ext{ moles}$ of the gas mixture.This means there are $50 	ext{ moles}$ of $He$ and $50 	ext{ moles}$ of $CH_4$.The molar mass of $He$ is $4 	ext{ g/mol}$ and the molar mass of $CH_4$ is $16 	ext{ g/mol}$.Mass of $He = 50 	ext{ moles} 	imes 4 	ext{ g/mol} = 200 	ext{ g}$.Mass of $CH_4 = 50 	ext{ moles} 	imes 16 	ext{ g/mol} = 800 	ext{ g}$.Total mass of the mixture $= 200 	ext{ g} + 800 	ext{ g} = 1000 	ext{ g}$.Percent by weight of $CH_4 = \frac{	ext{Mass of } CH_4}{	ext{Total mass}} 	imes 100 = \frac{800}{1000} 	imes 100 = 80\%$.

$A$ 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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