Question

under identical conditions separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. after a certain amount of time it was found that 5.39 mL of O2 had passed through the membrane but only 3.33 mL of the unknown gas had passed through, what is the molar mass of the unknown gas?

Answer 1

Rate of effusion is inversely proportional to square root of molar mass

rate = k*sqrt(1/M)

M is molar mass

If we take ratio, above expression becomes

rate(O2)/rate(unknown) = sqrt (M(unknown)/M(O2))

we have:

M(O2) = 32.0 g/mol

rate(O2)/rate(unknown) = 5.39/3.33

rate(O2)/rate(unknown) = sqrt (M(unknown)/M(O2))

5.39/3.33 = sqrt (M(unknown)/32.0)

sqrt (M(unknown)/32.0) = 1.619

(M(unknown)/32.0) = 2.62

M(unknown) = 83.84 g/mol

Answer: 83.84 g/mol