Solutions Question 321

Question: 0.440 g of a substance dissolved in 22.2 g of benzene lowered the freezing point of benzene by $ {{0.567}^{o}}C $ . The molecular mass of the substance $ (K_{f}={{5.12}^{o}}C,mo{l^{-1}}) $ [BHU 2001; CPMT 2001]

Options:

A) 178.9

B) 177.8

C) 176.7

D) 175.6

Show Answer

Answer:

Correct Answer: A

Solution:

$ m=\frac{K_{f}\times w\times 1000}{\Delta T_{f}\times W} $

$ =\frac{5.12\times 0.440\times 1000}{0.567\times 22.2}=178.9 $



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