The degree of dissociation (α) of weak electrolyte AxBy is related to van't Hoff factor (i) by the expression
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a
α=x+y−1i−1
b
α=i−1x+y+1
c
α=i−1(x+y−1)
d
α=x+y+1i−1
answer is C.
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Detailed Solution
van't Hoff factor (i)=[1+(n−1)α]where, n = number of ions from one mole of solute α=degree of ionization/association AxBy⇌xA+yBTotal ions =(x+y)∴ i=1+(x+y−1)α∴ α=i−1(x+y−1)