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# What volume of air contains 12.9g of oxygen gas at 273K and 1.00 atm?

mole fraction of oxygen gas in air is 0.21

What volume of air contains 12.9g of oxygen gas at 273K and 1.00 atm

my answer is wrong..

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- skipperLv 71 decade agoFavorite Answer
12.9 g / 32 g/mole = 0.403 mol O2 gas

0.403 mol / 0.21 = 1.92 moles of air gases

1.92 moles x 22.4 liters / mol = 43 liters of air

- 1 decade ago
12.9 g of oxygen gas = 0.403125 mole of oxygen gas. PV = nRT. Volume of O2 gas is 9.03 L. 9.03 divided by 0.21 = 43 L of air. The final answer should be rounded to two significant figures because the mole fraction is only two sig figs.

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