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

Rate of diffusion `prop(P)/(sqrt(M))`
`therefore(r_(N_(2)))/(r_(X))=(P_(N_(2)))/(P_(X))xxsqrt((M_(X))/(M_(N_(2))))`
But `r_(N_(2))=(1)/(38)mols^(-1)`
and `r_(X)=(1)/(57)mols^(-1)`
`therefore(r_(N_(2)))/(r_(X))=(1//38)/(1//57)=(57)/(38)=(3)/(2)`
`(3)/(2)=(0.8)/(1.6)sqrt((M_(X))/(28))` ltbr or `M_(X)=9xx28=252g`
Let the formula of the compound of Xe and F be `XeF_(n)`
`therefore131xx19x=252` or `xapprox6`
Hence the formula is `XeF_(2)`