Jamb Reg Calculate the volume of 02 whose time is 5 sece that passee through plug of of 10 dm 3 of Cl 2 a d time of 8 sec ( molarmass of oxygen is 32 and molarmass nof chlorine is 35.5 )

We first note that for effusion (or diffusion) of gases through a small opening, Graham’s law applies. The rate of diffusion (or effusion) is inversely proportional to the square root of the molar mass. That is, for two gases, \[ \frac{r_{1}}{r_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}. \] In our case, we compare oxygen \(\left(\mathrm{O}_2\right)\) and chlorine \(\left(\mathrm{Cl}_2\right)\). The given data are: - For chlorine: a volume of \(10\,\mathrm{dm}^3\) passes in \(8\,\mathrm{s}\). - For oxygen: an unknown volume \(V_{\mathrm{O}_2}\) passes in \(5\,\mathrm{s}\). The molar masses are given as: - \(M_{\mathrm{O}_2} = 32\,\mathrm{g/mol}\), - \(M_{\mathrm{Cl}}\) is given as \(35.5\,\mathrm{g/mol}\), so \[ M_{\mathrm{Cl}_2} = 2 \times 35.5 = 71\,\mathrm{g/mol}. \] The rates of diffusion are determined by the volume per unit time. Thus the rates are: \[ r_{\mathrm{Cl}_2} = \frac{10}{8}\,\mathrm{dm}^3/\mathrm{s} \quad \text{and} \quad r_{\mathrm{O}_2} = \frac{V_{\mathrm{O}_2}}{5}\,\mathrm{dm}^3/\mathrm{s}. \] By Graham’s law we have \[ \frac{r_{\mathrm{O}_2}}{r_{\mathrm{Cl}_2}} = \sqrt{\frac{M_{\mathrm{Cl}_2}}{M_{\mathrm{O}_2}}}. \] Substituting the expressions for the rates gives \[ \frac{\dfrac{V_{\mathrm{O}_2}}{5}}{\dfrac{10}{8}} = \sqrt{\frac{71}{32}}. \] We now solve the above equation step by step. 1. Write the left‐hand side as \[ \frac{V_{\mathrm{O}_2}}{5} \times \frac{8}{10} = \frac{8\,V_{\mathrm{O}_2}}{50} = \frac{4\,V_{\mathrm{O}_2}}{25}. \] 2. Thus we have \[ \frac{4\,V_{\mathrm{O}_2}}{25} = \sqrt{\frac{71}{32}}. \] 3. Calculate the square root: \[ \sqrt{\frac{71}{32}} \approx \sqrt{2.21875} \approx 1.49. \] 4. Now solve for \(V_{\mathrm{O}_2}\): \[ \frac{4\,V_{\mathrm{O}_2}}{25} \approx 1.49. \] Multiply both sides by \(\frac{25}{4}\): \[ V_{\mathrm{O}_2} \approx 1.49 \times \frac{25}{4}. \] 5. Compute the numerical value: \[ V_{\mathrm{O}_2} \approx 1.49 \times 6.25 \approx 9.31\,\mathrm{dm}^3. \] Thus, the volume of oxygen that passes in \(5\,\mathrm{s}\) is approximately \[ \boxed{9.31\,\mathrm{dm}^3}. \]