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

Explanation

Using Graham's law of diffusion
`(R_(1))/(R_(2))=sqrt((M_(2))/(M_(1)))`
Squaring both side, we get.
`((R_(1))/(R_(2)))^(2)=((sqrt(M_(2)))/(M_(1)))^(2),((R_(1))/(R_(2)))^(2)=(M_(2))/(M_(1))`
`M_(2)=M_(1)((R_(1))/(R_(2)))^(2)=((28.7)/(7.2))^(2)=(823.69)/(51.84)=15.88=16`
The gas is identified as oxygen.