The degree of dissociation `(alpha)` of a weak electrolyte, `A_(x)B_(y)` is related to van't Hoff's 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

` i= 1- alpha + n alpha = 1 + alpha ( n-1)`
`(i-1)/(n-1) = alpha `
`A_(x) B_(y) rarr xA^(+y) + yB^(-x)`
` n = x+y`
So, `alpha = (i-1)/( x+y-1)`