`X` and `Y` are two volatile liquids with molar weights of `10gmol^(-1)` and `40gmol^(-1)` respectively. Two cotton plugs, one soaked in `X` and the other soaked in `Y`, are simultaneously placed at the ends of a tube of length `L=24` cm, as shown in the figure.
The tube is filled with an inert gas at 1 atm pressure and a temperature of `300K`. Vapours of `X` and `Y` react to form a product whichh is first observed at a distance `d`cm from the plug soaked in `X`.
Take `X` and `Y` to have equal molecular diameters and assume ideal behaviour for the inert gas and two vapours.

The value of `d` in cm (shown in figure), as estimated from Graham's law, is

Options

A. 8
B. 12
C. 16
D. 20

Answer

C

Explanation

Let the distance covered by vapours of liquid `X = d` cm,
`:.` The distance covered by vapours of liquid
`Y = (24-d) cm`.
If `r_(X)` and `r_(Y)` be the rate diffusion for X and Y, then
`(r_(X))/(r_(Y)) = sqrt((M_(Y))/(M_(X))) = sqrt((40 g mol^(-1))/(10 g mol^(-1))) = 2`
Distance convered by the gas `prop` Rate of diffusion
`:. (r_(Y))/(r_(X)) = (d)/(24-d) = 2`
or `d+2d=48` or `d = (48)/(3) = 16` cm.