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

Answer

A

Explanation

`A_(x)B_(y)hArrxA^(y+)+yB^(x-)`
`"After dissociation"(1-alpha)" "xalpha" "yalpha`
`i=n(A_(x)B_(y))+n(A^(Y+))+n(B^(x-))`
`=1-alpha+xalpha+yalpha=1+alpha(x+y-1)`
`:.alpha=(i-1)/((x+y-1))`