A mixture of `50mL` of `H^(2)` and `50mL` of `O^(2)` is allowed to effuse through an effusiometer till the residual gas occupies `90 mL`. What is the composition of (`a`) effused gas, (`b`) the residual gas?

Explanation

According to Graham's law of diffusion,
Rate of diffusion `=sqrt((1)/(Molar weight))`
`:. (Rate of effusion of H_(2))/(Rate of effusion of O_(2))=sqrt((M_(O_(2)))/(M_(H_(2))))=sqrt((32)/(2))=4`
Volume of the mixture effused `=(50+50)-90=10 mL`
Let the volume of `H_(2)` effused be `x mL`.
Then the volume of `O_(2)` effused will be `(10-x) mL`. So
`(x)/(10-x)=4`
or `x=4(10-x)=40-4x`
or `x=8`
Composition of the effused gas
`H_(2)=8 mL` and `O_(2)=2 mL`
Composition of the residual gas
`H_(2)=50-8=42 mL`
and `O_(2)=50-2=48 mL`