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

For 10 minutes each, at 27^(@)C, from two identical holes nitrogen and

1.	For 10 minutes each, at 27^(@)C, from two identical holes nitrogen and an unknown gas are leaked into a common vessel of 3L capacity. The resulting pressure is 4.18bar and the mixture contains 0.4mol of nitrogen. What is the molar mass of the unknown gas?
Answer» Solution :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"`.