Question

# Calculate the value of ebullioscopic constant of a solvent if enthalpy of vapourisation of that solvent is nearly 3974 cal mol−1. Given : Boiling point of pure solvent is 127∘C Molecular mass of the solvent is 20 g mol−1

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 mol−1Kb=R×T2b×M1000×ΔHvap where, Tb is boiling point of pure solvent =127∘C=400 K M is molar mass of the solvent=20 g mol−1 ΔHvap is the molar enthalpy of vapourisation of solvent =3974 cal mol−1 R is ideal gas constant=1.987 cal mol−1 K−1 Kb=1.987×(400)2×201000×3974Kb=1.6 K kg mol−1

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