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

Total volume of effused gas is 3 litre and its pressure at `27^(@)C` is 6 atm. No. of moles of effuzed mixture may be calculated as
`n=(pV)/(RT)=(6xx3)/(0.0821xx300)`
`=0.73-8`
`"Moles of "H_(2)=0.7`
`therefore" Moles of unknown gas "=0.7308-0.7`
`=0.0308" moles"`
`"Rate of effusion of "H_(2)," "r_(H_(2))=(0.7)/(20)`
Rate of effusion of unknown gas, `r_(g)=(0.0308)/(20)`
According to Graham.s law,
`(r_(H_(2)))/(r_(g))=(0.7//20)/(0.0308//20)=sqrt((M_(g))/(M_(H_(2))))`
`therefore" "(0.7)/(0.0308)=sqrt((M_(g))/(2))`
`therefore" "M_(g)=((0.7)/(0.0308))^(2)xx2=1033.`