At 300 K and 760 torr pressure, the density | vedclass

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(B) Using the ideal gas equation, $PV = nRT$, for $1 \ mol$ of the mixture, the volume $V$ is given by $V = \frac{RT}{P}$.Given $P = 760 \ torr = 1 \

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$80$

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(B) Using the ideal gas equation, $PV = nRT$, for $1 \ mol$ of the mixture, the volume $V$ is given by $V = \frac{RT}{P}$.Given $P = 760 \ torr = 1 \ atm$, $T = 300 \ K$, and $R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$, we have $V = \frac{0.0821 	imes 300}{1} = 24.63 \ L$.The total mass of $24.63 \ L$ of the mixture is $M = V 	imes d = 24.63 \ L 	imes 0.543 \ g \ L^{-1} = 13.37 \ g$.Let $x$ be the mole fraction of $O_2$. Then the mole fraction of $He$ is $(1 - x)$.The average molar mass of the mixture is $M_{avg} = x(32) + (1 - x)(4) = 13.37$.Solving for $x$: $32x + 4 - 4x = 13.37$ $\Rightarrow 28x = 9.37$ $\Rightarrow x \approx 0.3346$.The mass of $O_2$ in $1 \ mol$ of the mixture is $0.3346 	imes 32 \approx 10.707 \ g$.The mass percent of $O_2$ is $\frac{	ext{mass of } O_2}{	ext{total mass}} 	imes 100 = \frac{10.707}{13.37} 	imes 100 \approx 80.08 \% \approx 80 \%$.

At $300 \ K$ and $760 \ torr$ pressure, the density of a mixture of $He$ and $O_2$ gases is $0.543 \ g \ L^{-1}$. The mass percent of oxygen is approximately $(R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1})$

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