A gas mixture contains 25 % He and 75 % CH_4 | vedclass

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(C) Assume the total volume of the gas mixture is $100 \ mL$. Since the volume percentage is equal to the mole percentage for ideal gases,we have $25

Vedclass · Accepted Answer

$92$

Vedclass · Answer

(C) Assume the total volume of the gas mixture is $100 \ mL$. Since the volume percentage is equal to the mole percentage for ideal gases,we have $25 \ mol$ of He and $75 \ mol$ of $CH_4$. Molar mass of He $= 4 \ g/mol$. Molar mass of $CH_4 = 16 \ g/mol$. Mass of He $= 25 	imes 4 = 100 \ g$. Mass of $CH_4 = 75 	imes 16 = 1200 \ g$. Total mass of the mixture $= 100 + 1200 = 1300 \ g$. Mass percentage of $CH_4 = \frac{	ext{Mass of } CH_4}{	ext{Total mass}} 	imes 100 = \frac{1200}{1300} 	imes 100 \approx 92.3 \%$. Thus,the percentage by mass of methane is approximately $92 \%$.

$A$ gas mixture contains $25 \%$ He and $75 \% CH_4$ by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately . . . . . . (in $\%$)

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