Consider 1 dm³ of a solution containing m moles of an electrolyte AxBy. The electrolyte on dissociation gives number of A^y+ ions and y number of B^x- ions. Let α be the degree of dissociation.

At equilibrium,

AxBy \(\rightleftharpoons\) xA^y+ + yB^x-

For 1 mole of electrolyte : 1 – α, xα, yα and For ‘m’ moles of an electrolyte : m(1 – α), mxα, myα are the number of particles.

Total number of moles at equilibrium, will be, Total moles = m(1 – α) + mxα + myα

= m[(1 – α) + xα + yα] 

= m[1 – xα + yα – α] 

= m[1 + α(x + y – 1)]

The van’t Hoff factor i will be,

\(i = \frac{\text{Observed colligative property}}{\text{Theoretical colligative property}}\)

\(=\frac{m[1 + \alpha(x + y - 1)]}{m}\)

i = 1 + α (x + y - 1)

If total number of ions from one mole of electrolyte is denoted by n, then (x + y) = n

\(\therefore\) 1 + α (n - 1)

\(\therefore\) α (n - 1) = i - 1

\(\therefore\) α \(=\frac{i - 1}{n - 1}\,....(1)\)

This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.