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Q.

Calculate the molal depression constant of a solvent which has freezing point 16.6^oC and latent heat of fusion 180.75 Jg^-1.

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a

2.8616

b

2.68

c

4.68

d

3.86

answer is B.

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Detailed Solution

$K_{f} = \frac{{RT}_{f}^{2}}{1000 \times L_{f}}, R = 8 .314 {JK}^{- 1} {mol}^{- 1} \begin{array}{l} T_{f} = 273 + 16 .6 = 289 .6 K; L_{f} = 180 .75 {Jg}^{- 1} \\ K_{f} = \frac{8 .314 \times 289 .6 \times 289 .6}{1000 \times 180 .75} = 3.86 \end{array}$

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Calculate the molal depression constant of a solvent which has freezing point 16.6oC and latent heat of fusion 180.75 Jg-1.