One mole of nitrogen gas at `0.8atm` takes `38s` to diffuse through a pinhole, while `1 mol` of an unknown fluoride of xenon at `1.6 atm` takes `57 s` to diffuse through the same hole. Calculate the molecular formula of the compound.

Explanation

According to Graham's law of diffusion
`(r_(1))/(r_(2)) =sqrt((M_(2))/M_(1)) xx (P_(1))/(P_(2))`
or `(r_(1))/(r_(2)) xx(t_(2))/(n_(2))sqrt((M_(2))/M_(1)) xx (P_(1))/(P_(2))`
or `or (1)/(38) xx (57)/(1) =sqrt(((M_(g))/(28)))xx (0.8)/(1.6)`
`M_(g) =[(57)/(38) xx (1.6)/(0.8)]xx 28 =252`
Thus copound is `XeF_(6)` becasue it can have only one xenon atom (since for two xenon atoms, 2xxAt wt of `Xe =2 xx 131 =262` greater than 252) .