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Question

The boiling point of 0.1 molal K4[Fe(CN)6] solution will be (Kb for water = 0.52 K.Kg mole1)
  1. 273C
  2. 100.26C
  3. 273K
  4. 500K


Solution

The correct option is B 100.26C
ΔTb=Kb.m.i=0.52×0.1×5=0.26C
Boiling point = 100.26C
 

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