Question

The rate of diffusion of methane at a given temperature is twice that of gas $X$ . The molecular mass of gas $X$ is:
(A) $64$
(B) $32$
(C) $4.0$
(D) $8.0$

Answer 1

Hint : As we know that the rate of change in number of molecules that diffuse per unit time is known as the rate of diffusion. In this question we have to find the molecular mass of the gas $X$ . To solve this we can apply the Graham’s law which states that the rate of diffusion of any gas is inversely proportional to the square root of the molar mass of gas. It can be represented by $\dfrac{{{r_{C{H_4}}}}}{{{r_X}}} = \sqrt {\dfrac{{{M_X}}}{{{M_{CH4}}}}}$ , the chemical formula of methane is $C{H_4}$ .

Complete Step By Step Answer:
We can say that the rate of diffusion of methane at a given temperature is twice that of gas $X$ . So we can write it as ${r_{CH4}} = 2{r_X}$ . And the molar mass of methane i.e. ${M_{CH4}} = 16$ (approx.)
We know that the rate of diffusion in inversely proportional to the square root of its molar mass i.e. $\dfrac{{{r_{C{H_4}}}}}{{{r_X}}} = \sqrt {\dfrac{{{M_X}}}{{{M_{CH4}}}}}$ .
By putting the values we have, $\dfrac{{2{r_X}}}{{{r_X}}} = \sqrt {\dfrac{{{M_x}}}{{16}}}$ , now to remove the square root we will square both the sides, it gives ${(2)^2} = {\left( {\sqrt {\dfrac{{{M_X}}}{{16}}} } \right)^2} \Rightarrow 4 = \dfrac{{{M_X}}}{{16}}$ .
So it gives us the value ${M_X} = 16 \times 4 = 64$ .
Hence the correct option is (A) 64.

Note :
Before solving this kind of question we should be aware of Graham's law and their formulas. We should note that the diffusion is a measure of average speed of the molecule. We should know that the molar mass of methane is $16.04g/mol$ . The diffusion rate depends on several factors such as the concentration gradient, the amount of surface area available etc.