1.

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

Answer»

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

SOLUTION :`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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