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

Given that the diffusion rate of hydrogen is 28.7  cm 3 /s and the diffusion rate of unknown gas is 7.2  cm 3 /s .

Let R h be the rate of diffusion for the hydrogen gas and R un be the rate of diffusion for unknown gas, then by Graham’s law of diffusion,

R h R un = M un M h M un M h = ( R h R un ) 2 M un = ( R h R un ) 2 × M h

Here, M un is the molecular mass of unknown gas and M h is the molecular mass of hydrogen which is 2.02 g.

Substitute the values in the above expression.

M un = ( 28.7 7.2 ) 2 ×2.02 =32.09 g

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