From a certain apparatus, the diffusion rate of hydrogen has an average value of `28.7 cm^(3) s^(-1)`. The diffusion of another gas under the same condition is measured to have an average rate of `7.2 cm^(3) s^(-1)`. Identify the gas.

Explanation

Rate of diffusion of hydrogen, `R_1 = 28.7 cm^3 s^–1`

Rate of diffusion of another gas,` R_2 = 7.2 cm^3 s^–1`

According to Graham’s Law of diffusion, we have:
`(R_1)/R_(2)=sqrt(M_2)/(M_1)`
`M_1` is the molecular mass of hydrogen = 2.020 g
` M_2` is the molecular mass of the unknown gas
` therefore M_2=M_1((R1)/(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.