For the given electrolyte `A_(x)B_(y)`. The degree of dissociation `'alpha'` can be given as:

Options

A. `alpha=(i-1)/(x+y-1)`
B. `i=(1-alpha)+xalpha+yalpha`
C. `alpha=(1-i)/(1-x-y)`
D. none

Answer

A, B, C

Explanation

`{:(,A_(x)B_(y)hArr,xA^(y+)+,yB^(x+)),("Initial moles",n,0,0),(At eq b.,n(1-alpha),nxalpha,nyalpha):}`
`i=("Total mol at equilibrium")/("Initial mol")=(n[(1-alpha)+xalpha+yalpha])/(n)`
`i=(i-alpha)+xalpha+yalpha`
It can also seen that all other expressions imply the same thing
(A) `alpha=(i-1)/(x+y-1)` (B) `i=(1-alpha)+xalpha+yalpha` (C) `(1-i)/(1-x-y)`