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 conditions is measured to have an average rate of `7.2cm^(3)s^(-1)`. Identify the gas (NCERT).

Explanation

From Graham's law of diffusion.
`(M_(2))/(M_(1))=((R_(1))/(R_(2)))^(2)"where" R_(1),R_(2)` are diffusion rates of gases.
hence `(M_(2))/(M_(2))=((28.7)/(7.2))^(2)=15.89`
i.e.`M_(2) = 16M_(1)`
Hence the other gas is oxygen.