Question

From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm3 -1. The diffusion of another gas under the same conditions ie measured to have an average rate of 7.2 cm3 -1. Identify the gas. [Hint: Use Graham’* 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

According to graham’* law of diffusion,

(\frac {R_1}{R_2}) = ((\frac {M_1}{M_2})^{1/2})

R1 = rate of diffusion of hydrogen

= 28.7cm3 *-1

R2 = Rate diffusion of unknown gas

= 7.2cm3 *-1

M1 = Molecular mass of hydrogen = 2.02

M2= Molecular mass of unknown gas.

= 32.09

The molecular mass of oxygen gas is 32. Hence, The given unknown gas is oxygen.