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# The time taken for a certain volume of a gas X to diffuse through a small hole is 2 minutes . It takes 5.65 minutes for oxygen to diffuse under the similiar conditions . The molecular weight of X is

$(a)\;8\qquad(b)\;4\qquad(c)\;16\qquad(d)\;32$

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• 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), $V_1\propto n_1 \;and\; V_2\propto n_2$.
$\large\frac{r_x}{r_{O_2}} =\sqrt{\large\frac{M_{O_2}}{M_X}}$
$\large\frac{V/2}{V/5.65} = \sqrt{\large\frac{32}{M_X}}$
$\large\frac{5.65}{2} = \sqrt{\large\frac{32}{M_X}}$
$\therefore M_X = 4$
Hence answer is (B)
answered Apr 8, 2014
edited Sep 22, 2014 by pady_1