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Question

The elevation in boiling point, when 13.44 g of freshly prepared CuCl2 are added to one kilogram of water is:

[Some useful data, Kb = 0.52 kgKmol1 . Molecular weight of CuCl2= 134.4 gm]

A
0.05
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B
0.1
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C
0.16
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D
0.21
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Solution

The correct option is B 0.16
Elevation of boiling point ( ΔTb ) is the difference between the boiling points of the solution and the pure solvent:

ΔTb=TbTb0

Elevation of boiling point can be calculated as

ΔTb=i.m.Kb

Here, m is the concentration of the solute expressed in molality and Kb is the molal boiling point elevation constant of the solvent.

i is van't Hoff factor =3 for CuCl2

No. of moles of CuCl2 is 13.44134.4=0.1

Molality of CuCl2 is m=0.11=0.1

The elevation of boiling point is 3×0.1×0.52=0.1560.16

Hence, the correct option is C

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