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

Options

A. `alpha= (i-1)/(x+y+1)`
B. `alpha =(x+y-1)/(i-1)`
C. `alpha + (x+y+1)/(i-1)`
D. `alpha = (i-1)/((x+y-1))`

Answer

D

Explanation

The van.t Hoff factor is `i=1- alpah + n alpha =1 + alpha (n-1) rArr alpha = (i-1)/(n-1)`
For the reaction `A_(x)B_(y) rarr xA^(y+) + yB^(x-), n=x + y`, we have
`alpha= (i-1)/(n-1)= (i-1)/(x+y-1)`