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

Mole of `H_(2)` diffused `=0.7` in 20 minute
Mole of gas diffused ` = n_(1)` in 20 mintue For gaseous mixture after diffusion
`PV =nRT`
` n = (6 xx3)/(0.0821 xx 300) = 0.731`
`:'` Mixture contains mole of `H_(2)` + mole of gas diffused =n
`:. 0.7 + n_(1) = 0.731`
`:. n_(1) = 0.031`
Now `(r_(H_(2)))/(r_(g))=sqrt((M_((g))/(M_(H_(2))))`
`(n_(H_(2)))/(t) xx t/(n_(g)) =sqrt((M_((g))/(2)))`
` (0.7)/(20) xx (20)/(0.031) =sqrt((M_((g))/(2)))`
`(M_((g)))/(2) = (0.7 xx 0.7)/(0.031 xx 0.031)`
`M_((g)) = 1019.77` .