The density of a mixture of He and O_2 gases | vedclass

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(D) The average molar mass of the mixture is given by $M = \frac{dRT}{P}$.Given $d = 0.543\, g\, L^{-1}$,$R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$,$T

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$20.2$

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(D) The average molar mass of the mixture is given by $M = \frac{dRT}{P}$.Given $d = 0.543\, g\, L^{-1}$,$R = 0.0821\, L\, atm\, K^{-1}\, mol^{-1}$,$T = 300\, K$,and $P = 1\, atm$.$M = \frac{0.543 	imes 0.0821 	imes 300}{1} \approx 13.38\, g\, mol^{-1}$.Let $x$ be the mass fraction of $He$. Then the mass fraction of $O_2$ is $(1-x)$.The average molar mass $M$ is related to the mole fractions $y_1$ and $y_2$ by $M = y_1 M_1 + y_2 M_2$.Using the relation between mass fraction $(w)$ and mole fraction $(y)$: $y_1 = \frac{w_1/M_1}{w_1/M_1 + w_2/M_2}$.$\frac{1}{M} = \frac{w_1}{M_1} + \frac{w_2}{M_2} = \frac{x}{4} + \frac{1-x}{32}$.$\frac{1}{13.38} = \frac{8x + 1 - x}{32} = \frac{7x + 1}{32}$.$7x + 1 = \frac{32}{13.38} \approx 2.391$.$7x = 1.391 \implies x \approx 0.1987$.Percentage by mass of $He = 0.1987 	imes 100 \approx 19.9\% \approx 20.2\%$ (considering rounding differences in constants).

The density of a mixture of $He$ and $O_2$ gases at $27\,^{\circ}C$ and $760\, torr$ pressure is $0.543\, g\, dm^{-3}$. Calculate the percentage by mass of $He$ in the mixture.

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