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Question

The boiling point of pure solvent is is 27C and its heat of vapourisation is 84 cal/g. Determine the boiling point elevation constant (Kb) for the solvent

A
0.23 K kg mol1
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B
5.18 K kg mol1
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C
2.14 K kg mol1
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D
6.89 K kg mol1
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Solution

The correct option is C 2.14 K kg mol1
We know,
Boiling point elevation constant,
Kb=RT2b1000 Lv
where,
Tb is boiling point of pure solvent in K
R in cal/(mol.K)
Lv is latent heat of vaourisation of solvent in cal/g
Thus,
Kb=2×(27+273)21000×84=2.14 K kg mol1

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