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

Answer

B

Explanation

`{:(Ax Bx,hArr,overset(y+)(xA),+,overset(x-)(yB)),((1-alpha),,x alpha,, y alpha):}`
`i=1-alpha+x alpha+y alpha=1+alpha(x+y-1)`
`alpha=(i-1)/((x+y-1))`