350.0 mg of an unknown gas effuses through a small hole into a 2.00 L flask. It effuses at a rate 1.27 times faster than sulfur dioxide at the same temperature, 20.0 degree C.

Question

350.0 mg of an unknown gas effuses through a small hole into a 2.00 L flask. It effuses at a rate 1.27 times faster than sulfur dioxide at the same temperature, 20.0 degree C.

What pressure would it exert in the 2.00 L flask?

Answer 1

I would make up a number for rate of SO2. Make it convenient one like 5 for SO2.

Then the unknown gas will have a rate of 1.27*5 = 6.35.
5/6.35 = sqrt(M2/molarmass SO2)
Solve for M2.
Then 0.350/M2 = moles = n of the unknown gas and plug that into PV = nRT and solve for P.