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Exercise · Q24

Q.Using Graham's law of diffusion, compare the rate of diffusion of hydrogen gas (M=2 g mol−1M = 2\ \text{g mol}^{-1}) with that of oxygen gas (M=32 g mol−1M = 32\ \text{g mol}^{-1}) under identical conditions.

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By Graham's law, rH2rO2=MO2MH2\dfrac{r_{\text{H}_2}}{r_{\text{O}_2}} = \sqrt{\dfrac{M_{\text{O}_2}}{M_{\text{H}_2}}}. Substituting MH2=2 g mol−1M_{\text{H}_2} = 2\ \text{g mol}^{-1} and MO2=32 g mol−1M_{\text{O}_2} = 32\ \text{g mol}^{-1}: rH2rO2=322=16=4\frac{r_{\text{H}_2}}{r_{\text{O}_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4 Hydrogen, being sixteen times lighter than oxygen by molar mass, diffuses four times faster (the square root of the molar mass ratio) — a direct consequence of both gases sharing the same average kinetic energy at a giv …

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