WAEC Chemistry 2010 Paper

Concept:Graham’s law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.Explanation:For the same volume of gas, the time of diffusion is directly proportional to the square root of the molar mass.So, $t \propto \sqrt{M}$, where $t$ is time and $M$ is molar mass.Molar mass of oxygen, $\mathrm{O}_2 = 2 \times 16 = 32$Molar mass of carbon(IV) oxide, $\mathrm{CO}_2 = 12 + 2(16) = 44$Given $t_{\mathrm{O}_2} = 8$ seconds, we use the relation:$$\frac{t_{\mathrm{CO}_2}}{t_{\mathrm{O}_2}} = \sqrt{\frac{M_{\mathrm{CO}_2}}{M_{\mathrm{O}_2}}}$$Substitute the values:$$t_{\mathrm{CO}_2} = 8 \times \sqrt{\frac{44}{32}}$$$$t_{\mathrm{CO}_2} = 8 \times \sqrt{1.375} = 8 \times 1.1726 \approx 9.4 \text{ seconds}$$Answer:9.4 seconds, which is option C.