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

Answer

a

Explanation

Van't half faction (i) is realated degree of dissociation `(alpha)`as
`alpha = (i - 1)/(n- 1)`
Here n are the moles of an elemolyte `(A- B)` dissolved in a solution.
For `A_(x)B_(y), A_(x)B_(y) hArr xA^(+y) + yB^(-x)`
Therefore ` alpha = (i - 1)/((x + y +1))`