Solutions Question 82

Question: Calculate the molal depression constant of a solvent which has freezing point $ {{16.6}^{o}}C $ and latent heat of fusion $ 180.75,J{g^{-1}} $ . [Orissa JEE 2005]

Options:

A) 2.68

B) 3.86

C) 4.68

D) 2.86t6

Show Answer

Answer:

Correct Answer: B

Solution:

$ K_{f}=\frac{RT_f^{2}}{1000\times L_{f}} $ , $ R=8.314J{K^{-1}}mo{l^{-1}} $

$ ,T_{f}=273+16.6=289.6K $ ; $ L_{f}=180.75,J{g^{-1}} $

$ K_{f}=\frac{8.314\times 289.6\times 289.6}{1000\times 180.75} = 3.86$



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