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Question

20 g of solvent containing 0.25 g of solute (mol. mass 122 g/mol) freezes at 0.3C below the freezing point of pure solvent. If the solvent exists as dimer in the solvent, find its degree of association. (Kf)solvent=5.12 K kg1 mol1

A
0.86
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B
0.14
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C
0.99
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D
0.01
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Solution

The correct option is A 0.86
Weight of solute(w)=0.25 g Weight of solvent(W)=20 g=201000 kgΔTf=0.3C=0.3 K

Let mobs be observed molecular mass of solute

mobs=Kf×wW×ΔTmobs=5.12×0.25×100020×0.3=213.3 g mol1

Theoretical molecular mass of solute =122 g mol1
So, i=Theoretical molecular massObserved molecular mass=122213.3=0.57

Degree of association (β) :
β=(i1)×n(1n)
Since solute dimerises, so n=2
Putting the values:
β=(0.571)×2(12)β=0.86

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