The boiling point of pure solvent is is 27∘C and its heat of vapourisation is 84cal/g. Determine the boiling point elevation constant (Kb) for the solvent
A
0.23Kkgmol−1
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B
5.18Kkgmol−1
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C
2.14Kkgmol−1
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D
6.89Kkgmol−1
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Solution
The correct option is C2.14Kkgmol−1 We know,
Boiling point elevation constant, Kb=RT2b1000Lv
where, Tb is boiling point of pure solvent in K R incal/(mol.K) Lv is latent heat of vaourisation of solvent in cal/g
Thus, Kb=2×(27+273)21000×84=2.14Kkgmol−1