# How to solve for mole fraction - ideal gas law?

I am told that there is oxygen in a 0.035 L container at 35 C with a pressure of 800 mm Hg.

Argon is then added to the container and temperature goes up to 50 C with a pressure of 1100 mm Hg.

What is the mole faction of Ar in the container?

### 2 Answers

- Roger the MoleLv 72 months agoFavorite Answer
n = PV / RT = (800 mmHg) x (0.035 L) / ((62.36367 L mmHg/K mol) x (35 + 273) K)) =

0.00146 mol O2 alone

Supposing the container is rigid (not a balloon):

(1100 mm Hg) x (0.035 L) / ((62.36367 L mmHg/K mol) x (50 + 273) K)) =

0.00191 mol total

(0.00191 mol total) - (0.00146 mol O2) = 0.00045 mol Ar

(0.00045 mol Ar) / (0.00191 mol total) = 0.24 (the mole fraction of Ar)

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- Dr WLv 72 months ago
from daltons law of partial pressures

.. Pa = χ a * Ptotal

rearranging

.. χ argon = Pargon / Ptotal

where

.. Pargon = Ptotal - PO2

.. Ptotal = 1100mmHg

and the only thing missing is PO2 at 50°C. From the ideal gas law, we know that at constant volume and moles

.. P1/T1 = P2/T2

.. P2 = P1*(T2/T1)

putting all that together

.. .. ..... ... . 1100mmHg - 800mmHg * (323.15K / 308.15K)

.. χ argon = ------ ----- ---- ----- ---- ---- ---- ---- ----- ---- ---- --- = 0.237

.... .... .... ... .... .... ... ... ... .. 1100mmHg

*******

you get to sort out the sig figs

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