T $

where:

$ p = $ pressure in atm (convert 760 torr to 1 atm),

$ M = $ molar mass of the gas mixture,

$ d = $ density ($0.543 \, \text{g/L}$),

$ R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} $,

$ T = $ temperature ($300 \, \text{K}$).

First, calculate the molar mass $ M $:

$ M = \frac{d \times R \times T}{p} = \frac{0.543 \times 0.0821 \times 300}{1} = 13.37 \, \text{g/mol} $

Assume $ y $ is the mole fraction of oxygen ($\text{O}_2$) in the mixture and $ (1-y) $ is the mole fraction of helium ($\text{He}$).

The average molecular weight of the mixture is given by:

$ M = y \times M_{\text{O}_2} + (1-y) \times M_{\text{He}} $

Substitute the known values:

$ 13.37 = y \times 32 + (1-y) \times 4 $

$ 13.37 = 32y + 4 - 4y $

$ 13.37 = 28y + 4 $

$ 9.37 = 28y $

$ y = \frac{9.37}{28} \approx 0.335 $

The mass percent of oxygen is calculated as follows:

$ \text{Mass percent of } \text{O}_2 = \frac{y \times M_{\text{O}_2} \times 100}{(1-y) \times M_{\text{He}} + y \times M_{\text{O}_2}} $

$ = \frac{0.335 \times 32 \times 100}{(1-0.335) \times 4 + 0.335 \times 32} $

Simplifying:

$ = \frac{10.72}{10.72 + 2.66} \times 100 $

$ = \frac{10.72}{13.38} \times 100 \approx 80\% $

Thus, the mass percent of oxygen in the mixture is approximately $80\%$.

States Of Matter

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