450 g/mol
Explanation:
To solve the problem the problem, it is necessary to use Graham's Law of effusion
M1/32g(mol) = 120s/32s = 450/mol
A heavier gas will require more time to effuse than a lighter gas
A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 34 s for 1.0 L of O2 gas to effuse.
Part A
Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)
2 answers
oops!
It's 125 sec and 34 sec. I get something like 432 but you should confirm that.
(1/125)/(1/34) = sqrt(32/M)
M = 432
It's 125 sec and 34 sec. I get something like 432 but you should confirm that.
(1/125)/(1/34) = sqrt(32/M)
M = 432
3 answers