Solutions Question 329
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$
BETA
