The degree of dissociation (alpha) of a...

The degree of dissociation `(alpha)` of a weak electrolyte `A_(x)B_(y)` is related to van't Hoff factor (i) by the expression

A

`alpha = (i-1)/((x+y-1))`

B

`alpha = (i-1)/((x+y+1))`

C

`alpha = (x+y-1)/(i-1)`

D

`alpha = (x+y+1)/(i-1)`

Verified by Experts

The correct Answer is:

A

`A_xB_yrarrxA^(y+)+yB^(x-)`
`1-alpha" "xalpha" "yalpha`
`i=1-alpha+xalpha+yalpha`
`i=1-alpha(x+y-1)`
`alpha=(i-1)/((x+y-1))`

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