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

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(D) Assume the total volume of the mixture is $100 \, mL$ at $STP$.Since the mixture contains $50 \, mL$ of $He$ and $50 \, mL$ of $CH_4$:Mass of $50

Vedclass · Accepted Answer

$80.03$

Vedclass · Answer

(D) Assume the total volume of the mixture is $100 \, mL$ at $STP$.Since the mixture contains $50 \, mL$ of $He$ and $50 \, mL$ of $CH_4$:Mass of $50 \, mL$ of $He = (\frac{4 \, g/mol}{22400 \, mL/mol}) 	imes 50 \, mL = 0.00893 \, g$.Mass of $50 \, mL$ of $CH_4 = (\frac{16 \, g/mol}{22400 \, mL/mol}) 	imes 50 \, mL = 0.03571 \, g$.Total mass of the mixture = $0.00893 \, g + 0.03571 \, g = 0.04464 \, g$.Percentage of methane by weight = $(\frac{	ext{Mass of } CH_4}{	ext{Total mass}}) 	imes 100 = (\frac{0.03571}{0.04464}) 	imes 100 \approx 80.018\% \approx 80.03\%$.

$A$ 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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