`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

PLAN: This problem can be solved by using the concept of Graham’s law of diffusion according to which rate of diffusion of non-reactive gases under similar conditions of temperature and pressure are inversely proportional to square root of their density.
Rate of diffision `prop (1)/(sqrt("molar weight of gas"))`
Let distance covered by X is d, then distance covered by Y is 24-d.
If `r_(X) and r_(y)` are the rate of diffusion of gases X and Y,
`(r_(x))/(r_(y)) = (d)/(24-d) = sqrt((40)/(10)) =2` [ `:'` Rate of diffusion distance travelled]
`d= 48 - 2d rArr 3d - 48 rArr d = 16 cm`