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Question
The vapour pressure of pure benzene (C6H6) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution
P°C6H6 = 640 mm Hg
W2 = 2.2 g (non volatile solute)
W1 = 40 g (benzene)
`"P"_"solution"` = 600 mm Hg
M2 = ?
`("P"^0 - "P")/"P"^0 = "X"_2`
`(640 - 600)/640 = "n"_2/("n"_1 + "n"_2)` ...[∴ n1 >> n2; n1 + n2 ≈ n1]
`40/640 = "n"_2/"n"_1`
`0.0625 = ("W"_2 xx "M"_1)/("M"_2 xx "W"_1)`
`"M"_2 = (2.2 xx 78)/(0.0625 xx 40)`
= 68.64 g mol-1
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