Posted by Chase on .
Problem: A gaseous mixture of O2 and Kr has a density of 1.181 g/L at 485 torr and 300. K. What is the mole percent O2 in the mixture?
Attempt: I have tried everything possible that I know and I can't get anywhere close to the solution. d=m/v=pM/RT... nothing works. Help please?

Chemistry 
DrBob222,
It seems to me you could do the following:
PV=nRT
PV = (grams/molar mass)RT
P= gRT/molar mass*V
P*molar mass = (g/v)RT
molar mass = density*RT/P
I plugged in the numbers and obtained about 45 something. Check my thinking. Check my arithmetic.
So the 45 something molar mass is made up of some percentage (or fraction) of 83.80 Kr (check that number) and some percentage (fraction) of 32.00 oxygen. Let x = fraction of oxygen, then 1x = fraction of Kr.
32.00(x) + 83.80(1x) = 45 something.
Solve for x. You can convert to grams and to mols and to mol fractions and mol percent from there. I kept looking for a way to obtain partial pressure of oxygen and Kr and from there the mol percent but that didn't happen. I think there may be an easier way to do this but I think the above will get it. Let me know.