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Question
What will be the vapour pressure of a solution containing 5 moles of sucrose (C12H22O11) in 1 kg of water, if the vapour pressure of pure water is 4.57 mm of Hg? [C = 12, H = 1, O = 16]
Solution
`("P"°_"water" - "P"°_"solution")/("P"°_"water") = "X"_("sucrose")`
∴ `"X"_("sucrose") = n_2/(n_1 + n_2) = 5/(5 + 1000/18) = 5/60.5 = 0.083`
∴ `("P"^circ - "P")/"P"^circ = 0.083`
`"P"^circ - "P" = 0.083 xx "P"^circ = 0.083 xx 4.57 = 0.38`
∴ `"P" = "P"^circ - 0.38 = 4.57 - 0.38 = 4.19` mm of Hg
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