At `27^(@)C`, hydrogen is leaked through a tiny hole into a vessel for `20 min`. Another unknown gas at the same temperature and pressure as that of hydrogen is leaked through the same hole for `20 min`. After the effusion of the gases, the mixture exerts a pressure of `6 atm`. The hydrogen content of the mixture is `0.7 mol`. If the volume of the container is `3 L`, what is the molecular weight of the unknown gas?

Explanation

`PV=nRT`
or `6xx3=nxx0.0821xx300`
or `n=0.7308`
`n_(H_(2))=0.7` (given)
`n_(gas)=0.7308-0.7=0.0308`
Applying Graham's law of diffusion, we get
`(r_(H_(2)))/(r_(gas))=sqrt((M_(gas))/(M_(H_(2))))`
or `(n_(H_(2)))/(n_(gas))=sqrt((M_(gas))/(M_(H_(2))))` (because the time of diffusion is same)
or `(0.7)/(0.0308)=sqrt((M_(gas))/(2))`
or `M_(gas)=1033.05`