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

Answer

C

Explanation

`A_(x)B_(x) hArr xA^(y+) + yB^(x-)`
Moles at `1 - alpha" "xalpha" "yalpha`
equi :
Total
Total no . Of moles = `1 - alpha + x alpha + y alpha`
` I = (1-alpha + xalpha + yalpha)/(1)`
`therefore " "(i-1) = a(x+y-1)`
`therefore " "alpha = (i-1)/((x+y-1))`