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=(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_(y),hArr,xA^(y+),+,yB^(x-)),("Moles at equilibrium",1-alpha,,x alpha,,yalpha):}`
Total no. of moles `=1-alpha+x alpha+y alpha`
`i=(1-alpha+x alpha+y alpha)/(1)`
`:.(i-1)=alpha(x+y-1)`
`:.alpha=(i-1)/((x+y-1))`