Question

A gas A having molecular weight 4 diffuses thrice as fast as the gas B at a given T. the molecular weight of gas B is:
A. 36
B. 12
C. 18
D. 24

Answer 1

Hint: The given problem can be solved by using Graham’s Law, which states that the rate of distribution or diffusion of gas is inversely proportional to the square root of the molecular weight of that particular gas.
\[{\rm{r}} = \dfrac{{\rm{1}}}{{\sqrt {\rm{M}} }}\]

Complete step by step answer:
The molecular weight of gas B can be determined by using Graham’s Law of equation is as follows:
\[\dfrac{{{{\rm{r}}_{\rm{A}}}}}{{{{\rm{r}}_{\rm{B}}}}} = \sqrt {\dfrac{{{{\rm{M}}_{\rm{B}}}}}{{{{\rm{M}}_{\rm{A}}}}}} \] ……………………….(1)
Where,
${r_A}$= rate of diffusion of gas A
${r_B}$= rate of diffusion of gas B
${M_A}$ = molecular weight of gas A
${M_B}$ = molecular weight of gas A
From the data given in the problem, let the rate of diffusion of gas A be x and that of gas B be 3x.
Let us substitute the given values in equation (1), we get
$\Rightarrow$ \[\dfrac{{{\rm{3r}}}}{{\rm{r}}} = \sqrt {\dfrac{{{{\rm{M}}_{\rm{B}}}}}{4}} \]
$\Rightarrow$ \[3 = \sqrt {\dfrac{{\rm{M}}}{{\rm{4}}}} \]
Taking square on both side, we get
$\Rightarrow$ \[{\left( 3 \right)^2} = \dfrac{{{{\rm{M}}_{\rm{B}}}}}{{\rm{4}}}\]
$\Rightarrow$ \[9 = \dfrac{{{{\rm{M}}_{\rm{B}}}}}{{\rm{4}}}\]
$\Rightarrow$ \[{{\rm{M}}_{\rm{B}}} = 9 \times 4\]
\[ = 36\]
The molecular weight of gas B will be 36.
So, the correct answer is “Option A”.

Note:
We need to remember about Graham’s Law that this equation holds good for effusion of gases not for diffusion. This equation is an approximation for diffusion of gases. This equation is valid only at constant pressure and temperature.