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

C

Explanation

van't Hoff factor (i)=`[1+(n-1)alpha]`
where,n= number of ions from one mole of solute
`alpha`=degree of ionization //association
`A_(x)B_(y)hArr xA+yB`
Total ions ` =(x+y)`
`:.i=1+(x+y-1)alpha`
`:.alpha =(i-1)/((x+y-1))`