Solutions Question 84
Question: The molal elevation constant of water $ ={{0.52}^{o}}C $ . The boiling point of 1.0 molal aqueous $ KCl $ solution (assuming complete dissociation of $ KCl $ ), therefore, should be [BHU 1987]
Options:
A) $ {{100.52}^{o}}C $
B) $ {{101.04}^{o}}C $
C) $ {{99.48}^{o}}C $
D) $ {{98.96}^{o}}C $
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Answer:
Correct Answer: B
Solution:
$ \Delta T_{b}=imk_{b}=0.52\times 1\times 2=1.04 $ . \ $ T_{b}=100+1.04={{101.04}^{o}}C $ .