

InterviewSolution
Saved Bookmarks
1. |
The rates of diffusion of two gases `A` and `B` are in the the ratio `1:4` A mixture contains these gases `A` and `B` in the ration `2:3` The ration of mole fraction of the gases `A` and `B` in the mixture is (assume that `P_(A) =P_(B))` .A. `1:24`B. `1:18`C. `1:12`D. `1:6` |
Answer» `(r_(A))/(r_(B)) =sqrt((M_(B))/M_(A)) =(1)/(4)` `(M_(B))/(M_(A)) = (1)/(16)` Now `(n_A)/(n_(B)) = (n_(A))/((n_(A)+n_(B))) xx (n_(A+n_(B)))/(n_(B))` ` =(w_(A))/(m_(A)) xx (M_(B))/(w_(B)) = (2)/(3) xx (1)/(16) =1/(24)` . |
|