If same moles of $O_2$ diffuse in 18 sec and same mole of other gas diffuse in 45 sec then what is the molecular weight of the unknown gas?

Question

$\begin {array}{1,1}(a)\;\large\frac{45^2}{18^2}\times32\\(b)\;\large\frac{18^2}{45^2}\times32\\(c)\;\large\frac{18^2}{45^2\times32}\\(d)\;\large\frac{45^2}{18^2\times32}\end {array}$

Answer 1

Toolbox:

• Graham's law states that "at constant pressure and temperature the rate of diffusion or effusion of a gas is inversly proportional to the square root of its density" Rate of diffusion $\propto \large\frac{1}{\sqrt{\alpha}}$ If $r_1 \; and \; r_2$ represent the rates of diffusion of two gases and $d_1 \;and\;d_2$ are their respective densities then $\large\frac{r_1}{r_2} = \sqrt{\large\frac{d_2}{d_1}}$ (at constant P and T) , $\large\frac{r_1}{r_2} = \sqrt{\large\frac{M_2}{M_1}}\times \large\frac{P_1}{P_2}$ (at constant T) . $\large\frac{V_1\times t_2}{V_2\times t_1} = \sqrt{\large\frac{d_2}{d_1}} = \sqrt{\large\frac{M_2}{M_1}} \quad V\propto n$ (where n is no. of moles), $V_1\propto n_1 \;and\; V_2\propto n_2$.

$\large\frac{r_{O_2}}{r_x} = \sqrt{\large\frac{M_x}{M_{O_2}}}$

Volume are in the same ratio , the no. of moles:

$=\large\frac{n/18}{n/45} = \sqrt{\large\frac{M_x}{32}}$

$\large\frac{45}{18} = \sqrt{\large\frac{M_x}{32}}$

$\therefore M_x = \large\frac{45^2}{18^2} \times 32$

= 200

Hence answer is (A)