Question

Some moles of $O_2$ diffuse through a small opening in $18$ second. Same number of moles of an unknown gas diffuse through the same opening in $45$ second. Molecular mass of the unknown gas is:

• A. $32\times \frac{(45{)}^2}{(18{)}^2}$
• B. $32\times \frac{(18{)}^2}{(45{)}^2}$
• C. $(32{)}^2\times \frac{45}{18}$
• D. $(32{)}^2\times \frac{18}{45}$

Answer 1

The correct option is A $32\times \frac{(45{)}^2}{(18{)}^2}$

The rate of diffusion is the ratio of number of moles diffuesd to the time taken for diffusion
The rate of diffusion is inversely proportional to the square root of its molar mass.
$\frac{r_{O2}}{r_X}=\frac{n_{O2}}{t_{O2}}\times \frac{t_X}{n_X}=\sqrt{\frac{M_X}{M_{O2}}}$
$n_{O2}=n_X$
$\frac{n_X}{18}\times \frac{45}{n_X}=\sqrt{\frac{M_X}{32}}$
$\frac{45}{18}=\sqrt{\frac{M_X}{32}}$
$\frac{(45{)}^2}{(18{)}^2}=\frac{M_X}{32}$
The molecular mass of gas X is $M_X=32\times \frac{(45{)}^2}{(18{)}^2}$ g/mol.