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October 20, 2014

October 20, 2014

Posted by **Thao** on Monday, April 25, 2011 at 2:49pm.

A 4.0 liter container has two gases inside, neon and argon. It is known that at 18 °C, the total pressure of the combined gases is 0.850 atm. If it is known that there are 0.100 moles of neon in the container, how many moles of argon are in the container?

n= RT/PV

n= (0.0821 LxATM/KxMOL x 291K) / (0.850 ATM x 4.0 L

= 7.026 mol

I am stuck from here. I don't know how to isolate the 0.100 mols of Neon from the Argon. Initially I subtracted 0.100 mols from 7.026 mol but that is not correct.

- Chemistry -
**DrBob222**, Monday, April 25, 2011 at 4:48pmNote that you inverted the formula (again?).

The formula is PV = nRT and if you solve for n, that will be

n = PV/RT. You can work this two ways.

n(total) = (0.850)(4.0)/(0.08206)(291) = 0.142 total moles. You know there is 0.100 mol neon; therefore, there must be 0.142-0.100 = 0.042 mol Ar.

Second way.

PV = nRT. Solve for P

P = nRT/V and calculate P for the 0.100 mol Ne.

P = (0.1 x 0.08206 x 291)/4.0 = 0.597 atm

Total P = 0.850 atm so PAr must be the difference. That will be 0.850-0.597 = 0.253 atm. Now plug this 0.253 atm for PAr into n = PV/RT = (0.253)(4.0)/(0.08206)(291) = 0.042 mol.

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