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Question

At 100C the vapour pressure of a solution of 6.5 g of a solute in 100 g water in 732 mm. If Kb=0.52, the boiling point of this solution will be

A
100C
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B
102C
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C
103C
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D
101C
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Solution

The correct option is D 101C
Elevation in boiling point of a solvent upon addition of solute is given as:
ΔTb=m.Kb
m=molality of the solution=number of moles of solute in 1 kg solvent
Kb=ebullioscopic constant=0.520kg/mol
Vapour pressure of solution Ps=732 mm ofHg
Vapour pressure of pure solvent (at 1000C) P=760 mm of Hg
mass of solute, ms=6.5g, mass of water, mw=100g, molar mass of water, Mw=18g/mol
let molar mass of solute be Ms
thus using PPsPs=nsnw
760732732=ms×MwMs×mw
or 28732=6.5×18Ms×100
or Ms=30.6g/mol
thus ΔTb=0.52×ms×1000Ms×mw
ΔTb=0.52×6.5×100030.6×100=1.1050C
thus boiling point of water is
Tsolvent=100+1.105=101.1050C

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