For 10 minutes each at `27^(@)C` from two identical holes nitrogen and an unknow gas are leaked into a common vessel of 3 litre capacity The resulting pressure is 4.18 bar and the mixture contains 0.4 mole of nitrogen What is the molar mass of the unknown gas ? .

Explanation

For the diffusion of both gases for 10 minutes under identical conditions
`(r_(g))/(r_(N_(2))) =sqrt(M_(N_(2))/(M_(g)))`
`:. (n_(g))/(10) xx (10)/(n_(N_(2))) = sqrt(28/(M_(g))) ( :' r =n//t)..(1)`
For mixture in vessel
`PV = nRT`
`:. 4.18 xx 3 =n xx 0.083 xx 300`
`:. n_(mix) = 0.50` mole
`n_(g) = n_(mix) - n_(N) = 0.50 -0.40 =0.10 .. (2)`
By eqs (2) and (1) `(0.10)/(0.4) = sqrt((28)/(M_(g)))`
`:. M_(g) = (28 xx 0.4 xx 0.4)/(0.10 xx 0.10) = 448g mo1^(-1)` .