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Author: Subject: Easy P-chem HW question
DeAdFX
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[*] posted on 25-10-2008 at 21:11
Easy P-chem HW question


Howdy..

I am having trouble with this question.

A certain mixture of He and Ne in a 356-cm^3 bulb weighs .1480 grams and is at 20.0C and 748 Torr. Find the mass and mole fraction of He present.

The equation that should be used is this PV = mRT/M

where little m = mass and big M = Molar mass.

From here

m = PVM / RT

m = (.9842 atm * .356 L * 4 g/mol) / (.0821 * 293K)
m = .0583 grams

The book says that the answer is suppose to be .0361 grams of He and .691 for mol fraction. Where am I messing up? I triple checked if I did the conversions right and I did. I am using the right gas constant too. (I like using L*atm). Thanks
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Magpie
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[*] posted on 25-10-2008 at 21:34


m = your mass in grams, which is given.

M = the molecular weight, which is your unknown.

Solve for M, which will be a mole proportioned average for the two gases.

Using M, determine the mole ratio of the two gases algebraicaly.

[Edited on 25-10-2008 by Magpie]
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[*] posted on 26-10-2008 at 14:35


Here's my solution in case you are having problems:

M=mRT/(PV) = [(0.1480g)(0.0821 L-atm/(K-mole)(293K)]/[(0.9842 atm)(0.356 L)]

= 10.16 g/mole

For the mole fractions:

Let X = mole fraction He

then 4.003X + 20.18(1-X) = 10.16
-16.177X + 20.18 = 10.16; 16.177X = 10.02; X = 0.619

For the weights:

mole ratio of He/Ne = 0.619/0.3806 = 1.626

Let W = weight of He

then W/4.003 = 1.626[(0.1480-W)/20.18]

W = 0.0477 - 0.3226W; 1.3226W = 0.0477; W = 0.0361g

Did you perhaps transpose the last 2 digits of the mole fraction for the He?
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[*] posted on 27-10-2008 at 11:25


ahh thanks I kinda botched up on the second unknown equation.

I thought it was x + y = .1480g/10.16(g/l).

Right process wrong value :spaz:
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