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One mole of nitrogen gas at 0.8atm takes...

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.

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From Graham's law of diffusion, we know that for the same volume moles diffused,
Time taken `prop sqrt(M)xx(1)/(P)`
`:. (t_(gas))/(t_("nitrogen"))=(P_("nitrogen"))/(P_(gas))sqrt((M_(gas))/(M_("nitrogen")))`
or `(57)/(38)=(0.8)/(1.6)sqrt((M_(gas))/(28))`
or `M_(gas)=252`
Let the formula of xenon fluoride be `XeF_(n)`.
`:. M=131.29+n(19)=252`
`n~~=6.35`
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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 value of n in the compound if formula is XeF_n .

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