Which one of the following 0.06 M aqueous solutions has lowest freezing point?

<p>To determine which solution has the lowest freezing point, you should consider the van 't Hoff factor $i$, which represents the number of particles the solute dissociates into when dissolved. The freezing point depression is given by:</p> <p>$\Delta T_f = i \cdot K_f \cdot m$</p> <p>where:</p> <p><p>$\Delta T_f$ is the change in freezing point,</p></p> <p><p>$K_f$ is the freezing point depression constant,</p></p> <p><p>$m$ is the molality of the solution,</p></p> <p><p>$i$ is the van 't Hoff factor, which depends on the dissociation of the solute.</p></p> <p>For each solute, the van 't Hoff factor $i$ can be determined as follows:</p> <p><p><strong>Option A: $\text{Al}_2(\text{SO}_4)_3$</strong></p></p> <p><p>Dissociates into 2 Al$^{3+}$ ions and 3 SO$_4^{2-}$ ions.</p></p> <p><p>$i = 2 + 3 = 5$</p></p> <p><p><strong>Option B: $\text{C}_6\text{H}_{12}\text{O}_6$ (glucose)</strong></p></p> <p><p>Does not dissociate in solution (non-electrolyte).</p></p> <p><p>$i = 1$</p></p> <p><p><strong>Option C: $\text{KI}$</strong></p></p> <p><p>Dissociates into K$^+$ and I$^-$ ions.</p></p> <p><p>$i = 1 + 1 = 2$</p></p> <p><p><strong>Option D: $\text{K}_2\text{SO}_4$</strong></p></p> <p><p>Dissociates into 2 K$^+$ ions and 1 SO$_4^{2-}$ ion.</p></p> <p><p>$i = 2 + 1 = 3$</p></p> <p>Comparing the van 't Hoff factors, the solution of $\text{Al}_2(\text{SO}_4)_3$ will have the highest factor ($i = 5$), leading to the greatest freezing point depression. Thus, the $\text{Al}_2(\text{SO}_4)_3$ solution will have the lowest freezing point since the extent of freezing point depression is directly proportional to $i$.</p> <p>Therefore, the solution of $\text{Al}_2(\text{SO}_4)_3$ has the lowest freezing point.</p>