A mixture of helium and neon gases has a density of 0.1562 g/L at 40.2°C and 296 torr. What is the mole fraction of neon in this mixture?

Question

akashdeepdeb · Answer

I think you have to use Dalton's law of partial pressures here.
Do you know that?

anonymous · Answer

no i really don't know how to start this

akashdeepdeb · Answer

How are you doing these kinds of questions if you aren't familiar with Dalton's Laws of Partial Pressures?

anonymous · Answer

I know what it is, I just don't know what to do

akashdeepdeb · Answer

Okay then, hold on.

jfraser · Answer

I'd find the densities of pure neon and pure helium by using the IGL. Since:$$PV=nRT$$and $$n = \frac{g}{MM}$$then$$\frac{g}{L} = \frac{MM*R*T}{P}$$ The mole fraction of the mixture will be equal to the weighted average of the two densities

akashdeepdeb · Answer

@JFraser Not by Dalton's law of Partial Pressures?

jfraser · Answer

not without knowing either partial pressure of neon or helium. All that we know is that both gases are present, and the total pressure.

anonymous · Answer

what is MM?

jfraser · Answer

molar mass

anonymous · Answer

I'm not getting it....

jfraser · Answer

if only neon gas were present, it would have a density based on the IGL. if only helium were present, it would have a different density. since it is a mixture, the density will be a blend of the two densities

anonymous · Answer

I'm getting a big number. I think i'm doing it how you told me...

jfraser · Answer

are you converting the pressure into atmospheres and the temperature into kelvin?

anonymous · Answer

oh i didn't convert the pressure

anonymous · Answer

i can;t leave it in torr?

aaronq · Answer

you need to write 2 equations (known laws/relations) using the data given, and substitute one into the other.

anonymous · Answer

pv=nrt?

aaronq · Answer

yeah, i would use but rearrange it as:

$PV=nRT -> PV=\dfrac{mRT}{M} -> (density)\rho=\dfrac{m}{V}=\dfrac{PM}{RT}$

anonymous · Answer

what would the answer come out to be?

aaronq · Answer

the answer to the question?

anonymous · Answer

yes

aaronq · Answer

idk i haven't done the problem

anonymous · Answer

can you please solve it, i want to see if I'm getting the correct answer.

aaronq · Answer

why don't you post what you've done and i'll check if its right?