Calculate the value of ebullioscopic constant of a solvent if enthalpy of vapourisation of that solvent is nearly 3974calmol−1.
Given :
Boiling point of pure solvent is 127∘C
Molecular mass of the solvent is 20gmol−1
A
2.4Kkgmol−1
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B
0.5Kkgmol−1
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C
1.6Kkgmol−1
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D
3.2Kkgmol−1
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Solution
The correct option is C1.6Kkgmol−1 Kb=R×T2b×M1000×ΔHvap
where, Tb is boiling point of pure solvent =127∘C=400K M is molar mass of the solvent=20gmol−1 ΔHvap is the molar enthalpy of vapourisation of solvent =3974calmol−1 R is ideal gas constant=1.987calmol−1K−1