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.

One mole of nitrogen gas at `0.8atm` takes `38s` to diffuse through a

1.	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.
Answer» 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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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.

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