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At a certain temperature, the vapour pre...

At a certain temperature, the vapour pressure of pure ether is `640 mm` and that of pure acetone is `280 mm`. Calculate the mole fraction of each component in the vapour state if the mole fraction of ether in the solution is `0.50.`

Text Solution

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In the given solution, both ether and acetone are volatile, so from Raoults's law, we can have vapour pressure of solution` (P_("total"))`.
Note: `chi_(A)_("Liquid phase") = P_(A)/ P_(A)@ , chi_(A_"Vapour phase") = P_(A)/(P_(A) + P_(B))`
Let `A` is ether and `B` is acetone.
`P_("total") = P_(A) + P_(B) = P_(A)@chi_(A) + P_(B)@chi_(B)`
`= 640 xx 0.5 + 280 xx 0.5 = 460.0 mm`
Now mole fraction in vapour state is given as follows:
`chi_(A_"Vapour") = P_(A) / P_("total") = (P_(A)@chi_(A))/ P_("total") = 640 xx 0.5 /460.0 = 0.6956 `
`chi_(B_"Vapour") = P_(B) / P_("total") =( P_(B)@chi_(B))/ P_("total") = 280 xx 0.5 /460.0 = 0.304`
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