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 `7.2 cm^(3) s^(-1)`. Identify the gas. [Hint : Use Graham's law of diffusion: `R_(1)//R_(2) = (M_(2) //M_(1) )^(1//2)`, where `R_(1), R_(2)`, are diffusion rates of gases 1 and 2, and `M_(1)` and `M_(2)` their respective molecular masses. The law is a simple consequence of kinetic theory.]

Explanation

Accoriding to Graham.s law of diffusion,
`(r_1)/(r_2) sqrt((M_2)/(M_1))`
Where, `r_(1)` = diffusion rate of hydrogen
`= 28.7 cm^(3)s^(-1)`
`r_(2)` = diffusion rate of unknown gas
`=7.2 cm^(3) s^(-1)`
`M_(1)` = molecular mass of hydrogen = 2u
`M_(2) = ?`
`:. (28.7)/(7.2) = sqrt((M_2)/(2)) or M_(2) = ((28.7)/(7.2))^(2) xx 2`
`= 31.78 ~~ 32`.