UPSEE Solved Model Paper 7

Given: No. of moles of $\mathrm{CH}_4$ = 1 No. of moles of unknown gas = 1 Area of orifice through which $\mathrm{CH}_4$ diffusing = 1 $\mathrm{mm}^2$ Time = 40 sec Area of orifice through which unknown gas effusing = 2 $\mathrm{mm}^2$ Time = 10 sec We have to find the molecular mass of unknown gas Let it be M Molecular mass of $\mathrm{CH}_4$ = 16 g/mol According to Graham's law of diffusion r α Area of orifice (i) where r = Rate of diffusion Also, r α $\frac{1}{\sqrt{M}}$ ... (ii) where. M = Molecular mass Thus, from equation (i) and equation (ii) $\frac{r_1}{r_2}$ = $\frac{\frac{A_1}{\sqrt{M_1}}}{\frac{A_2}{\sqrt{M_2}}}$ or $\frac{r_1}{r_2}$ = $\frac{A_1}{A_2} \cdot \sqrt{\frac{M_2}{M_1}}$ ... (iii) Now, Rat of diffusion r = $\frac{V}{t}$ Thus, equation (iii) can be written as $\frac{\frac{V}{t_1}}{\frac{V}{t_2}}$ = $\frac{A_1}{A_2} \cdot \sqrt{\frac{M_2}{M_1}}$ or $\frac{t_2}{t_1}$ = $\frac{A_1}{A_2} \cdot \sqrt{\frac{M_2}{M_1}}$ On substituting the values in equation (i). we get $\frac{10}{40}$ = $\frac{1}{2} \cdot \sqrt{\frac{M}{16}}$ $\frac{2}{4}$ = $\frac{\sqrt{M}}{4}$ 2 = $\sqrt{M}$ M = 4 Hence, option 'A' is correct.