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,
Rate of diffusion of hydrogen ( R) = 28.7 cm³/s
Rate of diffusion of unknown gas (R') = 7.2 cm³/s

Use Graham's law of diffusion ,
R/R' = √{ M'/M}
here M = 2 g/mol

Now, 28.7/7.2 = √{ M'/2}
M' = 2 × (28.7/7.2)²
= 32.09 g/mol
Hence, molecular mass of unknown gas is 32 g/mol , it means gas must be O2 . because molar mass of O2 gas is also 32 g/mol