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Question
The vapour pressure of a pure solvent at a certain temperature is 0.0227 bar. What is the vapour pressure of a solution containing 6 g of solute (M = 60 g/mol) in 50 g of solvent?
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Solution
Given: Vapour pressure of pure solvent = `"P"_1^0` = 0.0227 bar
Mass of solute = 6 g
Molar mass of solute = 60 g mol–1
Mass of solvent = 50 g
To find: Vapour pressure of the solution
Formula: `("P"_1^0 - "P"_1)/"P"_1^0 = ("W"_2"M"_1)/("M"_2"W"_1)`
Calculation: Molar mass of solvent = 18 g mol–1 (Assuming water as the solvent)
Using formula,
`("P"_1^0 - "P"_1)/"P"_1^0 = ("W"_2"M"_1)/("M"_2"W"_1)`
`(0.0227 "bar" - "P"_1)/(0.0227 "bar") = (6 "g" xx 18 "g mol"^-1)/(50 "g" xx 60 "g mol"^-1)`
∴ `(0.0227 "bar" - "P"_1)/(0.0227 "bar")` = 0.036
∴ 0.0227 bar − P1 = 0.036 × 0.0227 bar
∴ P1 = 0.0227 bar − 8.172 × 10−4 bar = 0.022 bar
Vapour pressure of the given solution is 0.022 bar.
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