Question

A sample of N2O effuses from a container in 49 seconds.

How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

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(I2)/rate(N2O) = sqrt (M(N2O)/M(I2))

we have:

M(I2) = 253.8 g/mol

M(N2O) = 44.02 g/mol

rate(I2)/rate(N2O) = sqrt (44.02/253.8)

rate(I2)/rate(N2O) = sqrt (0.1734)

rate(I2)/rate(N2O) = 0.4165

time and rate and inversely related.

So,

time(N2O)/time(I2) = rate(I2)/rate(N2O) = 0.4165

time(N2O)/time(I2) = 0.4165

49 s / time(I2) = 0.4165

time(I2) = 118 s

Answer: 118 s