In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is _________ $$\times 10^{-2}$$ (Round off to the nearest integer)

Let us begin by assuming that we originally dissolve $$1\;{\text{mol}}$$ of the monoprotic acid $$\mathrm{HA}$$ in the solvent.

According to the statement, the acid undergoes two simultaneous processes: one half of it dimerises while the other half dissociates.

1. Dimerisation
The reaction is $$2\mathrm{HA}\;\longrightarrow\;(\mathrm{HA})_2.$$ If 50 % of the original acid dimerises, the number of moles of $$\mathrm{HA}$$ consumed is $$0.50 \times 1 = 0.50\ \text{mol}.$$ Because the stoichiometry is 2 : 1, every 2 mol of $$\mathrm{HA}$$ give 1 mol of dimer. Therefore the moles of dimer formed are $$\frac{0.50}{2}=0.25\ \text{mol}.$$ No unreacted monomer is left from this portion.

2. Dissociation
The remaining 50 % of the acid, i.e. $$0.50\ \text{mol},$$ now dissociates completely according to $$\mathrm{HA}\;\longrightarrow\;\mathrm{H}^+ + \mathrm{A}^-.$$ So we obtain $$0.50\$$ mol of $$\mathrm{H}^+ \quad$$ and $$\quad 0.50\$$ mol of $$\mathrm{A}^-.$$ No undissociated $$\mathrm{HA}$$ remains from this portion.

3. Counting the total particles present
After both processes are complete, the species and their amounts are: $$\begin{aligned} (\mathrm{HA})_2 &: 0.25\ \text{mol} \;(1\ \text{particle per formula unit}),\\[2pt] \mathrm{H}^+ &: 0.50\ \text{mol},\\[2pt] \mathrm{A}^- &: 0.50\ \text{mol}. \end{aligned}$$ Hence the total number of moles of solute particles is $$0.25 + 0.50 + 0.50 = 1.25\ \text{mol}.$$ The van’t Hoff factor $$i$$ is defined as $$i = \frac{$$ total moles of particles after reaction $$}{$$ initial moles of solute $$}.$$ Substituting the numbers we have $$i = \frac{1.25}{1.00}=1.25.$$

The question asks for the value in the form “$$\times 10^{-2}$$” and then to round to the nearest integer. Writing $$1.25$$ in that form gives $$1.25 = 125 \times 10^{-2}.$$ On rounding, the required integer is $$125.$$

Hence, the correct answer is Option 125.