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.]

13.12 Rate of diffusion of hydrogen, R1= 28.7 cm3 s–1 Rate of diffusion of another gas, R2 = 7.2 cm3 s–1 According to Graham's law of diffusion, we have: R1R2= M2M1, where M1= molecular mass of hydrogen = 2.02 g and M2 is the molecular mass of the unknown gas M2=M1* (R1R2)2= 2.02 * (28.77.2)2= 32.09 = Molecular mass of Oxygen Hence, the unknown gas is Oxygen.

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

9 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