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Question

2.5 g of a monobasic acid when dissolved in 100 g of water elevates the boiling point of the solution by 0.256C. If 0.5 g of the acid requires 8 milliequivalents of NaOH for complete neutralisation, then the degree of dissociation of the acid is: [Kb of water =0.512Kkgmol1]

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Solution

To find molar mass of acid we can simply relate gm equivalents of the acid with the base,
0.5M×1 = 8×103
M = 62.5 gm
ΔTb=Kbim0.256=0.512(1+α)m
0.256=0.512(1+α)×2.5×100062.5×100
α=0.25

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