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Question

Calculate the value of ebullioscopic constant of a solvent if enthalpy of vapourisation of that solvent is nearly 3974 cal mol1.
Given :
Boiling point of pure solvent is 127C
Molecular mass of the solvent is 20 g mol1

A
2.4 K kg mol1
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B
0.5 K kg mol1
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C
1.6 K kg mol1
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D
3.2 K kg mol1
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Solution

The correct option is C 1.6 K kg mol1
Kb=R×T2b×M1000×ΔHvap
where,
Tb is boiling point of pure solvent =127C=400 K
M is molar mass of the solvent=20 g mol1
ΔHvap is the molar enthalpy of vapourisation of solvent =3974 cal mol1
R is ideal gas constant=1.987 cal mol1 K1

Kb=1.987×(400)2×201000×3974Kb=1.6 K kg mol1

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