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

according to Grahm'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 unkown 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`
This is the molecular mass of oxygen gas.