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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 formula of the compound.

Text Solution

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The available data is : `P_(N_(2)) = 0.8` atm, `t_(N_(2)) = 38 s, P_(g) = 1.6` atm , `t_(g) = 57 s`
Rate of diffusion `(r) prop (P)/(sqrt(M))`
`:. (r_(N_(2)))/(r_(g)) = (P_(N_(2)))/(P_(g)) xx sqrt((M_(g))/(M_(N_(2))))`
Since rate of diffusion `(r) prop 1/t`.
`:. (t_(g))/(t_(N_(2))) = (P_(N_(2)))/(P_(g)) xx sqrt((M_(g))/(M_(N_(2))))`
On substituting the value, `57/58 = (0.8)/(1.6) xx sqrt((M_(g))/(28))`
or `sqrt((M_(g))/(28)) = (57)/(38) xx 2 = 3`
Squaring both sides, `(M_(g))/(28) = 9` or `M_(g) = 28 xx 9 = 252`
Let the molecular formula of compound of `Xe` be `Xe F_(n)`.
Molecular mass of `XeF_(n) = 132 + n xx 19`
Comparing (i) and (ii)
`131 + n xx 19 = 252` or `n = (252-131)/(19) = 121/19 = 6.3` or `6`
`:.` Molecular formula `= XeF_(6)`.
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