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

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(D) Let the total volume of the mixture be $100 \ mL$.Since the mixture contains $50\%$ helium and $50\%$ methane by volume,the volume of $He$ is $50

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(D) Let the total volume of the mixture be $100 \ mL$.Since the mixture contains $50\%$ helium and $50\%$ methane by volume,the volume of $He$ is $50 \ mL$ and the volume of $CH_4$ is $50 \ mL$.Using the ideal gas law at constant temperature and pressure,the number of moles is proportional to the volume $(n \propto V)$.Let the number of moles of $He$ be $50 \ mol$ and $CH_4$ be $50 \ mol$.Mass of $He = 50 \ mol 	imes 4 \ g/mol = 200 \ g$.Mass of $CH_4 = 50 \ mol 	imes 16 \ g/mol = 800 \ g$.Total mass of the mixture $= 200 \ g + 800 \ g = 1000 \ g$.Percentage by weight of $CH_4 = \frac{	ext{Mass of } CH_4}{	ext{Total mass}} 	imes 100$.Percentage by weight of $CH_4 = \frac{800}{1000} 	imes 100 = 80.0\%$.

$A$ gaseous mixture contains $50\%$ helium and $50\%$ methane by volume. Find the percentage by weight of methane in this gaseous mixture.

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