300 mL of nitrogen gas diffuse througha a porous memebrane in `100` seconds. How long will it take for `400 mL` of `CO_(2)` to diffuse through the same membrane under similar conditions of temperature and pressure ?

Answer

`167.14 s`

Explanation

Rate of diffusion of hydrogen `(r_(H_(2))) = (300)/(100) = 3 mL s^(-1)`
Rate of diffusion of `CO_(2)(r_(N_(2))) = (400)/(t) = mL s^(-1)`
Applying Graham's Law of diffusion :
`(r_((N_(2))))/(r_((CO_(2))))=sqrt((M_(CO_(2)))/(M_(N_(2))))` or `(3)/(400//t) = sqrt((44)/(28))` or `t = 400/3 sqrt((44)/(28)) = 167.14s`