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Topic: Mass of air in 70% humidity  (Read 5450 times)

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Offline vigs77

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Mass of air in 70% humidity
« on: October 12, 2010, 12:17:00 AM »
The question given is: Calculate the weighted average of the mass of a mole of "air molecules." (Assume 24oC room temperature, 70% humidity and one atmosphere pressure) Answer needs to be in g/mole.

Using info from the reading proceeding this:
*"Dry air is composed of N2 (78.1% of the particles), O2 (21.0% of the particles) and Ar (0.9% of the particles)."
*When there is 100% relative humidity, the percentages (and pressures) of all these gases decrease accordingly.
*100% relative humidity provides an increase of 22.38 torr at 24oC.

I came up with:
N2= 2x14.0067 g/mol times 78.1% = 21.878 g/mol
O2= 2x15.9994 g/mol times 21.0% = 6.720 g/mol
Ar= 1x39.948 g/mol times 0.9% = 0.360 g/mol
so that the molar mass of dry air is 28.958 g/mol (from what I can tell, this part is correct)

I then worked out that 70% of 22.38 torr is 15.67 torr and if 1 atm = 760 torr, then the air molecules must make up: 760 torr minus the water vapor contribution of 15.67 torr, which would leave 744.334 torr.
78.1% N2 of 744.334 torr  = 581.325 torr due to N2
21.0% O2 of 744.334 torr = 156.310 torr due to O2
0.9% Ar of 744.334 torr = 6.699 torr due to Ar

But I have no volume to use PV = nRT with and am not sure where to progress from here. Can I even use the torr values I worked out?

As far as I can tell the molar mass of O2 (18.0152 g/mol) is not used in the question/answer.

Thanks for any light you can shed on this.

Offline vigs77

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Re: Mass of air in 70% humidity
« Reply #1 on: October 19, 2010, 11:37:27 AM »
 :(
Unfortunately this forum was extremely unhelpful. Thanks to all who read it and to the none that offered help.

Offline DemonicAcid

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Re: Mass of air in 70% humidity
« Reply #2 on: October 19, 2010, 06:30:26 PM »
You are told that you have one mole of gas. How do you find the volume of a gas when you have the amount?

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