Hello, for my practice exam there\'s a question that is listed below. At 200 K t

Question

Hello, for my practice exam there's a question that is listed below.

At 200 K the vapor pressure of pure xenon is 3908 torr, and of pure radon is 437 torr. Compute mole fraction of Rn in vapor phase if 2 moles of liquid Xe is mixed with 3 moles of liquid Rn, and the system is at equilibrium with its vapor phase.

IT tells us the answer is x=0.144, but not how to get it

EDIT: I figured it out (thanks all for answering, it did help)

For those who also have this question, I had to do:

Total mol = (Rn) 437 * 3 (moles) + (Xe) 3908 * 2 (moles) = 9127

Then since i'm solving for Rn mole fraction

(437*3)/9127 = .144 (Which was the answer)

Explanation / Answer

mole fraction of Xe=3908/(3908+437)=0.899

mole fraction of Rn=437/(3908+437)=0.101

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