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1 mole heptane (V.P = 92 mm of Hg) is mi...

1 mole heptane `(V.P = 92 mm of Hg)` is mixed with 4 mol. Octane `(V.P = 31` mm of `Hg)`, form an ideal solution. Find out the vapour pressure of solution.

Text Solution

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total mole `=1 +4 = 5`
Mole fraction of heptane `= X_(A) =1//5`
Mole fraction of octane `= X_(B) = 4//5`
`P_(S) = X_(A)P_(A)^(0) +X_(B)P_(B)^(0) = (1)/(5) xx 92 +(4)/(5) xx 31 = 43.2 mm of Hg`.
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