chem question

[masterMood]

76. from examkrackers lecture 4.

The air we breahte is approximately 21% O2 and 79% N2. If the partial pressure of nitrogen in air is 600 torr, then all of the following are true:

B. The mass of nitrogen in a 22.4 sample of air is 22.1 grams at 0 degrees Celsius.

The book says its true, and the explanatin is that

"Standard Molar Volume is the volume occupied by one mole of any gas at STP"

so if that's true then wouldn't there be one mole of N2 (g) or 28 grams instead of 22.1 grams?

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[nfg05]

What you're forgetting about is that air is only 79% N2, as stated in the question. So in one mol of air (which equals 22.4 L of air at STP), there are actually only 0.79 mol of N2. (0.79 mol)(28 g/mol) = 22.1 g/mol.

 

[Quantum Chem]

Air is 21% O2 by mole and 79% N2 by mole. If you have 22.4L of an air sample, you would have one mole of air. You would then multiply .79 moles by 28 g/mole to get 22.1g of N2 and you could do the same to get the mass of O2.

 

[masterMood]

thanks guys!

 

[Yolisept]

The explanation also says that the percentages given are by particle and NOT by mass, so we can't actually use that logic to eliminate B. B is dependent on the percentages being given by mass not by particle.