Question

a sample of CO2 gas is observed to effuse through a pourous barrier in 4.05 minutes. Under the same conditions, the same number of moles if an unknown gas requires 4.28 minutes to effuse through the same barrier. The molar mass of the unkown gas is what?

Answer 1

From the definition of rate of effusion we can write , rate of effusion = amount of gas transferred / time

From Graham's Law, we can write

rate of effusion of unknown / rate of effusion of CO ₂ =  

MCO2/Munknown

$\therefore$ (amount of unknown transferred / time) / (amount of CO ₂ transferred / time) =

MCO2/Munknown

We have given , no. of moles of unknown = No. of moles of CO ₂

$\therefore$ (1/ time of unknown gas ) / (1 / time of CO ₂ gas )=

MCO2/Munknown

$\therefore$ time of CO ₂ gas / time of unknown gas =

MCO2/Munknown

Molar mass of CO ₂ = 12.01 + ( 2 $\times$ 16.00 ) = 44.01 g / mol

$\therefore$ 4.05 min / 4.28 min =

44.01g/mol/Munknown

0.946 =

44.01g/mol/Munknown

on squaring both sides , we get

0.895 = ( 44.01 g /mol ) / M unknown

$\therefore$ M unknown gas = 44.01 g / mol / 0.895

$\therefore$ M unknown gas = 49.2 g / mol

ANSWER : Molar mass of unknown gas = 49.2 g /mol