`180ml` of hydrocarbon diffuses through a porous membrane in 15 minutes while `120 ml` of `SO_(2)` under identical conditions diffused in 20 minutes. What is the molecular mass of the hydrocarbon ?

Explanation

`r_(1) =` rate of diffusion of hydrocarbon `= (180)/(15) mL "min"^(-1)`
`r_(2) =` rate of diffusion of `SO_(2) = (120)/(20) mL "min"^(-1)`
`(r_(1))/(r_(2)) = (M_(SO_(2)))/(M)`
Thus, `(180//15)/(120//20) = sqrt((64)/(M))`
`2 = sqrt((64)/(M))`
So, `M = 16`