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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))`

Text Solution

Verified by Experts

The correct Answer is:
a
`AxBy=xA^(Y+)+yB^(x-)`
`(1-alpha)=xalpha, yalpha`
`i=("No.of particles after dissociation")/("No. of particles originally present")`
`=((1-alpha)+xalpha+yalpha)/1=(1+alpha(x+y)-1)/1`
`alpha=-((1-i))/((x+y-1))`
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