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 formation of the compound.

Explanation

We know that
`r prop (P)/(sqrt(M))`
Here, `r_(N_(2))=(1)/(38)mol s^(-1)`
`r_(X)=(1)/(57)mols^(-1)`
(rate of diffusion of unknown fluoride of xenon)
For `N_(2)`
`(1)/(38)=(Kxx0.08)/(sqrt(28))` (`i`)
For unknown gas,
`(1)/(38)=(Kxx0.08)/(sqrt(M_(x)))` (`ii`)
Dividing (`i`) by (`ii`), we get
`(1)/(38)xx(1)/(57)=(0.8)/(1.6)xx((M_(x))/(28))^(1//2)`
` :. (3)/(2)=(1)/(2)xx((M_(x))/(28))^(1//2)`
or `M_(x)=252 g mol^(-1)`
Molar weight of xenon fluoride `(XeF_(x))` is `252`.
`131+19 x=252 (Xe=131, F=19)`
or `19x=252-131=121`
`:. x=(121)/(19)=6.36`
Hence, the xenon fluoride is `XeF_(6)`.