The degree of dissociation ($α$) of a weak electrolyte $A_xB_y$ is related to van't Hoff factor (i) by the expression:

$α = (i-1)/(x + y -1)$

The correct answer is Option (1) → $α = (i-1)/(x + y -1)$

For an electrolyte:

$A_x B_y \rightleftharpoons xA^{y+} + yB^{x-}$

If the degree of dissociation = $\alpha$.

From 1 mole:

• Undissociated $= 1 - \alpha$
• Dissociated $= \alpha$
• Ions formed $= x + y$ per mole dissociated

Total particles after dissociation:

$(1 - \alpha) + \alpha(x + y)$

Van’t Hoff factor ($i$)

$i = \text{Total particles after dissociation}$

$i = (1 - \alpha) + \alpha(x + y)$

$i = 1 - \alpha + \alpha x + \alpha y$

$i = 1 + \alpha(x + y - 1)$

Rearranging for $\alpha$

$i - 1 = \alpha(x + y - 1)$

$\alpha = \frac{i - 1}{x + y - 1}$