13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.

[Hint : Use Graham’s law of diffusion: R1/R2 = ( M₂ /M₁ )^1/2, where R₁, R₂ are diffusion rates of gases 1 and 2, and M₁ and M₂ their respective molecular masses. The law is a simple consequence of kinetic theory.]

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Payal Gupta | Contributor-Level 10

5 months ago

13.12 Rate of diffusion of hydrogen, $R_{1}$ = 28.7 cm³ s^–1

Rate of diffusion of another gas, $R_{2}$ = 7.2 cm³ s^–1

According to Graham's law of diffusion, we have:

$\frac{R_{1}}{R_{2}}$ = $\sqrt{\frac{M_{2}}{M_{1}}}$ , where $M_{1}$ = molecular mass of hydrogen = 2.02 g and $M_{2}$ is the molecular mass of the unknown gas

$M_{2} = M_{1} * ({\frac{R_{1}}{R_{2}})}^{2}$ = 2.02 $* ({\frac{28.7}{7.2})}^{2}$ = 32.09 = Molecular mass of Oxygen

Hence, the unknown gas is Oxygen.

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