DeAdFX
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Easy Pchem HW question
Howdy..
I am having trouble with this question.
A certain mixture of He and Ne in a 356cm^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


Magpie
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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 25102008 by Magpie]


Magpie
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Here's my solution in case you are having problems:
M=mRT/(PV) = [(0.1480g)(0.0821 Latm/(Kmole)(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(1X) = 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[(0.1480W)/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?


DeAdFX
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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:

