A sample of co2 gas at 90.7 degree celsius has a volume of 300. ml at 810 mmHg. How many moles of co2 are present R=.08206 L atm/mol K 1 atm=760 mmHg
is this correct?
PV=nRT
(1.07)(0.3)=(44.01)(.08206)(363.7)
0.3424=1313.49
=0.00026068 or 2.61^-4
I thought you were solving for n. You left no unknown. The 1.07 atm is right for P, 0.300 for V, 0.08206 for R, for T I would use (273.2+90.7 = 393.9). 44.01 is the molar mass of CO2. It is the mass of 1 mol CO2 BUT n is what you are to solve for. For what's it worth, personally I don't do these by pieces. For example I start with 810 and 810/760 and I don't write the number down. I leave it in the calculator.
So n = PV/RT = (810/760)*0.300/[(0.08206*(273.2+90.7)] = ??
Then round that number at the end.