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

Let `p_(H_(2)) " and " p_(un)` be the pressures of hydrogen and unknown gas respectively and `w` be the number of moles of unknown gas.
`p_(H_(2)) = (0.7)/(3) xx 0.0821 xx 300`
`p_(un) = (w)/(3) xx 0.0821 xx 300`
Adding both,
`p_(H_(2)) + p_(un) = 6 = (1//3) xx 0.082 xx 300 (0.7 + w)`
`w = 0.0308` mole
Applying law of diffusion,
`(0.7 //20)/(0.0308 //20) = sqrt((M)/(2)) " or " M = 1033`