From two identical holes, nitrogen and an unknown gas are leaked into a common vessel of `3 L` capacity for `10 min`, at `27^(@)C`. The resulting pressure is `4.18` bar and the mixture contains `0.4 mol` of nitrogen. What is the molar mass of the unknown gas?

Explanation

Let unknown gas is ‘a’ 
Here, `T = 300K, t_(N_(2)) = t_(a) = 10 "minutes"`, `V = 3 L = 3dm^(3), P_(T) = 4.18bar, n_(N_(2))= 0.4 "mol"`
From Dalton’s law of partial pressures, we known that. 
`P_(T) = R_(N_(2)) + P_(a)`
`P_(T) = n_(N_(2)) (RT)/(V)+n_(2)(RT)/(V)=(n_(N_(2))+n_(2))(RT)/(V)`
`4.18 = (0.4 + n_(2)) (0.083 xx 300)/(3)`
`implies (0.4 + n_(2) ) = (4.18 xx 3)/(0.083 xx 300)= 0.5036 implies n_(2)= 0.5036-0.4=0.1036`
Now, from Graham’s law of diffusion, we now that. 
`(n_(N_(2)))/(n_(2))= (t_(N_(2)))/(t_(a))sqrt((M_(a))/(M_(N_(2))))implies (0.4)/(0.1036)= (10)/(10)sqrt((M_(a))/(28))implies 3.861=sqrt((M_(a))/(28))`
`M_(a)=(3.861)^(2) xx 28= 417.4`
therefore Molar mass of un nown gas `= 417.4 g//"mol"`.