Question

A gas X diffuses three times faster than another gas Y the ratio of their densities i.e. $dx:dy$ is:
A. $\dfrac{1}{3}$
B. $\dfrac{1}{9}$
C. $\dfrac{1}{6}$
D. $\dfrac{1}{12}$

Answer 1

Use Graham’s law of diffusion. Graham's law states that the rate of diffusion or of effusion of a gas is inversely proportional to the square root of its molecular weight.

Complete step by step answer:
Using Graham’s law of diffusion
Formula for Graham’s law- $\dfrac{{{R}_{1}}}{{{R}_{2}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$
${{R}_{1}}$ is the rate of effusion for the first gas. (Volume or number of moles per unit time).
${{R}_{2}}$is the rate of effusion for the second gas. ( (Volume or number of moles per unit time).
${{M}_{1}}$ is the molar mass of gas 1
${{M}_{2}}$ is the molar mass of gas 2.

So according to the question.
We assume rate of gas Y to be R1 and so rate of gas X will be 3 times rate of gas Y i.e. 3R1

Let the density of X and Y be dx​ and dy ​respectively
In the same condition of temperature and pressure molar mass is proportional to their densities and therefore, rate of diffusion would be inversely proportional to their densities i.e. $\dfrac{{{R}_{1}}}{{{R}_{2}}}=\sqrt{\dfrac{{{d}_{2}}}{{{d}_{1}}}}$

So applying in the formula:-
$\dfrac{{{R}_{1}}}{3{{R}_{1}}}=\sqrt{\dfrac{dx}{dy}}$
${{\left( \dfrac{{{R}_{1}}}{3{{R}_{1}}} \right)}^{2}}=\dfrac{dx}{dy}$
​
$\dfrac{1}{9}=\dfrac{dx}{dy}$
So, the correct answer is “Option B”.

Note: Graham’s law of Diffusion was formulated by Thomas Graham in 1848.
He was a Scottish physical chemist. Diffusion of gas is the movement of one type of gas molecule into either the empty space or into a space of another type of gas molecule present. And hence the Graham’s law is useful in determining calculating molecular weight of unknown gases by either using rates or densities