A gaseous mixture contains 50% Helium and 50% | vedclass

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(D) For a gas mixture,the weight $W$ is proportional to the product of volume $V$ and molar mass $M$ $(W \propto V 	imes M)$.Let the total volume be

Vedclass · Accepted Answer

$80.03$

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(D) For a gas mixture,the weight $W$ is proportional to the product of volume $V$ and molar mass $M$ $(W \propto V 	imes M)$.Let the total volume be $100 \ mL$. Then,$V_{He} = 50 \ mL$ and $V_{CH_4} = 50 \ mL$.Molar mass of Helium $(M_1)$ = $4 \ g/mol$.Molar mass of Methane $(M_2)$ = $16 \ g/mol$.Weight of Helium $(W_1)$ = $50 	imes 4 = 200$.Weight of Methane $(W_2)$ = $50 	imes 16 = 800$.Percentage of Methane by weight = $\frac{W_2}{W_1 + W_2} 	imes 100 = \frac{800}{200 + 800} 	imes 100 = \frac{800}{1000} 	imes 100 = 80\%$.

$A$ gaseous mixture contains $50\%$ Helium and $50\%$ Methane by volume. What is the percentage of Methane by weight in the mixture?

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