Question
Two gases $$A$$ and $$B$$ having the same volume diffuse through a porous partition in 20 and
$$10s$$ respectively. The molecular mass of $$A$$ is $$49\,u.$$ Molecular mass of $$B$$ will be
A.
12.25$$\,u$$
B.
6.50$$\,u$$
C.
25.00$$\,u$$
D.
50.00$$\,u$$
Answer :
12.25$$\,u$$
Solution :
According to Graham's law of diffusion,
$$\eqalign{
& {\text{rate of diffusion}} \cr
& r \propto \frac{1}{{\sqrt M }},\,r \propto \frac{V}{t} \cr} $$
where, $$V$$ is the volume of the gas diffused in time $$t.$$
$$\eqalign{
& \frac{{{r_A}}}{{{r_B}}} = \sqrt {\frac{{{M_B}}}{{{M_A}}}} \,\,\,{\text{or}}\,\,\,\frac{{{V_A}}}{{{t_A}}} \times \frac{{{t_B}}}{{{V_B}}} = \sqrt {\frac{{{M_B}}}{{{M_A}}}} \cr
& {\text{Given,}}\,\,{V_A} = {V_B} \cr
& \therefore \,\,\,\,\,\,\,\,\,\,\,\frac{{10}}{{20}} = \sqrt {\frac{{{M_B}}}{{49}}} \Rightarrow \frac{1}{4} = \frac{{{M_B}}}{{49}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,{M_B} = \frac{{49}}{4} = 12.25\,u \cr} $$