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WHAT IS THE VALUE OF VANT HOFF FACTOR FOR K4 [FE(CN)6} IF IT IS IONIZED 50% IN AUEOUS SOLUTION

WHAT IS THE VALUE OF VANT HOFF FACTOR FOR K4
[FE(CN)6} IF IT IS IONIZED 50% IN AUEOUS SOLUTION
 

Grade:12

1 Answers

Anany
19 Points
4 years ago
Use the concept :- Vant Hoof Factor = [1+( y - 1) x] where y = no.of particles produced by one molecule and x = degree of dissociation. Here, in this question molecule K4[Fe(CN)6] dissociates into 4K+ and 1 [Fe(CN)6] raised to power -4. Therefore no. of particles (y) = 5 and degree of dissociation (x) = 50%. Hence, Vant Hoff Factor = [ 1+(5-1)50/100 ] = 3

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