A given volume of methane diffuses in 20 s. How long will it take the same volume of sulphur (IV) oxide to diffuse under the same conditions? \([ C =12, H =1, S =32, O =\) 16]
A. 5s B. 20s C. 40s D. \(60 s\)
Correct Answer: C
Explanation
Graham's law of diffusion states that the volume rate of diffusion of gases is inversely proportional to the square roots of their densities (or relative molecular mass). i.e. \(r \alpha \frac{1}{\sqrt{\rho}}\) or \(r \alpha \frac{1}{\sqrt{M}}\) Where \(r\) is the rate of diffusion and \(\rho\) and \(M\) the density and relative molecular mass respectively. \(r=v / t\) where \(v\) is the volume of the gas, and \(t\), the time taken for diffusion thus, \(\frac{r_{1}}{r_{2}}=\sqrt{\frac{\rho_{2}}{\rho_{1}}}\) or \(\frac{r_{1}}{r_{2}}=\sqrt{\frac{M_{1}}{M_{2}}}\) $$ \Rightarrow \frac{v_{1} t_{2}}{v_{2} t_{1}}=\sqrt{\frac{\rho_{2}}{\rho_{1}}}, \frac{v_{1} t_{2}}{v_{2} t_{1}}=\sqrt{\frac{M_{2}}{M_{1}}} $$ for the same volume of the diffusing gas, we have \(V_{1}=V_{2}\) and hence: \(\frac{t_{2}}{t_{1}}=\sqrt{\frac{\rho_{2}}{\rho_{1}}}\) or \(\frac{t_{2}}{t_{1}}=\sqrt{\frac{M_{2}}{M_{1}}}\) \(M_{S O_{4}}=32+(2 \times 16)=64, t_{S O_{2}}=\) ? \(M_{ CH _{4}}=12+(4 \times 1)=16, t_{ CH _{4}}=20 s\) \(t_{50_{1}}=20 \times \sqrt{\frac{64}{16}}=20 \times \sqrt{4}\) \(=20 \times 2=40 s\)