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Question
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Solution
Here, Vapour pressure of the solution at normal boiling point (ps) = 1.004 bar
Vapour pressure of pure water at normal boiling point (p0) = 1 atm = 1.013 bar
Let mass of solution = 100 g
Mass of solute, (w2) = 2 g
Mass of solvent (water), (w1) = 100 − 2 = 98 g
Molar mass of solvent (water), (M1) = 18 g mol−1
According to Raoult’s law,
`("p"^0 - "p"_"s")/"p"^0 = "n"_2/("n"_1 + "n"_2) = "n"_2/"n"_1`
`= ("w"_2//"M"_2)/("w"_1//"M"_1) = "w"_2/"M"_2 xx "M"_1/"w"_1`
`=>(1.013-1.004)/1.013=(2xx18)/("M"_2xx98)`
`=>0.009/1.013 = (2xx18)/("M"_2xx98)`
`=>"M"_2 = (1.013xx2xx18)/(0.009xx98)`
= 41.35 g mol−1
Hence, the molar mass of the solute is 41.35 g mol−1.
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