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The degree of dissociation ( α ) of a weak electrolyte, AxBy is related to van't Hoff factor (i) by the expression: 1. α=x+y+1i-1 2.α=x+y-1i-1 3. α=i-1x+y+1 4. α=i-1x+y-1

Correct Answer: 4 is correct

Explanation:

Van't Hoff factor (i) =  O b s e r v e d   C o l l i g a t i v e   P r o p e r t y N o r m a l   C o l l i g a t i v e   P r o p e r t y


 


A x B y 1 - α x A + y x α + y B - x y α   


 


Total Moles =  1 - α + x α + y α  =  1 - α x + y - 1


 


i = 1 + α x + y - 1 1


 


α = i - 1 x + y - 1


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