Answer:-1. **The weak electrolytes involve the concept of degree of dissociation (\(\alpha\)) that changes the van’t Hoff factor.** 2. **Consider an electrolyte \(A_xB_y\) that dissociates in aqueous solution as:** \[ \begin{array}{|c|c|} \hline & A_xB_y \, \underset{\text{⇌}}{\text{↔}} \, xA^{y+} + yB^{x-} \\ \hline \text{Initially} & 1 \, \text{mol} \quad 0 \quad 0 \\ \hline \text{At equilibrium} & (1 – \alpha) \, \text{mol} \quad (x\alpha \, \text{mol}) \quad (y\alpha \, \text{mol}) \\ \hline \end{array} \] 3. **If \(\alpha\) is the degree of dissociation of the electrolyte, then the moles of cations are \(x\alpha\), and those of anions are \(y\alpha\) at equilibrium. We have dissolved just 1 mol of electrolyte initially. \(\alpha\) mol of the electrolyte dissociates, and \((1 – \alpha)\) mol remains undissociated at equilibrium.** **Total moles after dissociation:** \[ = (1 – \alpha) + (x\alpha) + (y\alpha) = 1 + \alpha (x + y – 1) = 1 + \alpha (n – 1) \] where \(n = x + y\) is the moles of ions obtained from the dissociation of 1 mole of electrolyte. 4. **The van’t Hoff factor is given as:** \[ i = \frac{\text{actual moles of particles in solution after dissociation}}{\text{moles of formula units dissolved in solution}} = \frac{1 + \alpha (n – 1)}{1} \] Hence, \(i = 1 + \alpha (n – 1)\) or \