The ratio of velocities of diffusion of gases A and B is `1 : 4`. If the ratio of their masses present in the mixture is `2 : 3`, calculate the ratio of their mole fractions.

Explanation

From the data : `(W_(A))/(W_(B)) = 2/3` or `W_(A) = 2/3 W_(B)`
According to Graham's Law of diffusion,
`(r_(A))/(r_(B)) = sqrt((M_(B))/(M_(A))) , 1/4 = sqrt((M_(B))/(M_(A)))`
Squaring both sides : `1/16 = (M_(B))/(M_(A))` or `M_(A) = 16 M_(B)`
Ratio of mole fractions : `(x_(A))/(x_(B)) = ((n_(A))/(n_(A) + n_(B)))/((n_(B))/(n_(A) + n_(B)) ) = (n_(A))/(n_(B)) = (W_(A)//M_(A))/(W_(B)//M_(B))`
`= (2//3.W_(B))/(W_(B)) xx (M_(B))/(16M_(B))=2/48 = 1/24`
Thus, the mole friactions of the two gases are in the ratio `1 : 24`