Question

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.]

Answer 1

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:

`R_1/R_2 = sqrt(M_2/M_1)`

Where,

M₁ is the molecular mass of hydrogen = 2.020 g

M₂ is the molecular mass of the unknown gas

`:. M_2 = M_1 (R_1/R_2)^2`

`= 2.02 (28.7/7.2)^2 = 32.09 g`

32 g is the molecular mass of oxygen. Hence, the unknown gas is oxygen.

Answer 2

According to Graham’s law of diffusion of gases, the rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

If R₁ and R₂ be the rates of diffusion of two gases having molecular masses M₁ and M₂ respectively, then

`R_1/R_2 = sqrt(M_2/M_1)`

Now `R_1 = 28.7 cm^3 s^(-1), R_2 = 7.2 cm^3 s^(-1), M_1 = 2, M_2= ?`

`:. 28.7/7.2 = sqrt(M_2/2)`

`or M_2/2 = (28.7xx28.7)/(7.2xx7.2)`

or  `M_2 = (2xx28.7xx28.7)/(7.2xx7.2) = 31.78 ~~ 32`

This molecular mass of oxygen.Therefore the second  gas is oxygen.