Solutions Question 335
Question: KBr is 80% ionized in solution. The freezing point of 0.4 molal solution of KBr is: $ K_{f}(H_2O)=1.86Kkg/mole $
Options:
A) 274.339K
B) $ -1.339K $
C) 257.3 K
D) $ -1.339{}^\circ C $
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Answer:
Correct Answer: D
Solution:
$ \Delta T_{f}=iK_{f}m $
$ i=1+\alpha $ Here $ \alpha =0.8 $ (80% ionization) =1.8 $ \Delta T_{f}=( 1.8 )( 1.86 )( 0.4 )=1.339 $
$ T_{f}=-1.339{}^\circ C $
