a gas mixtuer consists of equal masses of methane(molecular weight 16.0) and argon (atomic weight 40.0) if the partial pressure of argon is 26.6 kPa, what is that of methane in the same units?

Question

10.7
26.7
66.7
34.1

I am going with 26.7
Do you agree and why please
Thanks Andy

Answer 1

You wouldn't think pressure Ar = pressure CH4 because equal masses ( although true) contain different number of moles.
Here is what I would do.
Select a convenient number for grams, say 50 g.
Convert 50 g to moles CH4.
Convert 50 g to moles Ar.

Now find mole fraction Ar.
mole fraction Ar = moles Ar/total moles.


P_Ar = X_Ar*Ptotal
Solve for Ptotal. (P_Ar = 26.6

Then P_Ar + P_CH4 = P_total
Solve for P_CH4

Answer 2

ok so P ch4 = 13.67 kPa
where do i go from there??

Answer 3

No it isn't.
moles CH4 = 50/16 = 3.125
moles Ar = 50/40 = 1.25
total moles = 3.125 + 1.25 = 4.375
XAr = 1.25/4.375 = 0.286

XAr*Ptotal = PAr
0.286*Ptotal = 26.6
Ptotal = 26.6/0.286 = 93
Then 93 = PCH4 + PAr
PCH4 = 93.0 - 26.6 = ??
Answer is in kPa. Where do you go from here. You're here.

Answer 4

so its a trick question as the unwary would say that CH4 and Ar under ideal behavior would be the same and the answer would be 26.7....??
But its obviously worded to catch out the unwary about partial pressure questions?
Finally i can rest and the answer is 66.7 NOT 26.7!!!!!
andy