Which of the following aqueous solutions having same molality exhibits maximum boiling point elevation? (Assume complete dissociation)

Boiling point elevation is a colligative property given by the formula $\Delta T_b = i \cdot K_b \cdot m$, where $K_b$ and $m$ are constant for all aqueous solutions. The elevation is therefore proportional to the van't Hoff factor $i$.

For each solute:

KCl dissociates as $\mathrm{KCl} \to \mathrm{K}^+ + \mathrm{Cl}^-$, giving $i = 2$.

NaCl dissociates as $\mathrm{NaCl} \to \mathrm{Na}^+ + \mathrm{Cl}^-$, giving $i = 2$.

AlCl₃ dissociates as $\mathrm{AlCl_3} \to \mathrm{Al}^{3+} + 3\mathrm{Cl}^-$, giving $i = 4$.

BaCl₂ dissociates as $\mathrm{BaCl_2} \to \mathrm{Ba}^{2+} + 2\mathrm{Cl}^-$, giving $i = 3$.

The highest van't Hoff factor is $i = 4$ for AlCl₃, which therefore exhibits the maximum boiling point elevation.