(a) Does a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{KCl}\) have a higher freezing point, a lower freezing point, or the same freezing point as a $0.10 \mathrm{~m}$ aqueous solution of urea \(\left(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\right)\), (b) The experimental freezing point of the KCl solution is higher than that calculated assuming that \(\mathrm{KCl}\) is completely dissociated in solution. Why is this the case?

Colligative properties are properties of solutions that depend only on the number of solute particles present, not on the identity of the solute. A common colligative property is the freezing point depression, which states that the freezing point of a solution is lower than the freezing point of the solvent. The formula for freezing point depression is: \(\Delta T_f = K_f \cdot m \cdot i\) Where \(\Delta T_f\) is the freezing point depression, \(K_f\) is the freezing point depression constant, \(m\) is the molality of the solution, and \(i\) is the van't Hoff factor, which represents the number of particles the solute dissociates into when it dissolves in the solvent.

The van't Hoff factor \(i\) represents how many particles a solute dissociates into when it dissolves. For example, a molecule of \(\mathrm{KCl}\) dissociates into one potassium ion \(\mathrm{K^+}\) and one chloride ion \(\mathrm{Cl^-}\), giving a van't Hoff factor of 2. Urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\), does not dissociate when dissolved, so its van't Hoff factor is 1.

We are given that both solutions have the same molality (\(0.10 \mathrm{~m}\)). The freezing point depression formula gives us: (For \(\mathrm{KCl}\)) \(\Delta T_{f, KCl} = K_f \cdot 0.10 \mathrm{~m} \cdot 2\) (For urea) \(\Delta T_{f, urea} = K_f \cdot 0.10 \mathrm{~m} \cdot 1\) Since the freezing point depression constant \(K_f\) is the same for both solutions, the freezing points can be compared directly: \(\Delta T_{f, KCl} > \Delta T_{f, urea}\).

Since the freezing point depression of the \(\mathrm{KCl}\) solution is greater than that of the urea solution, its freezing point will be lower. The aqueous \(\mathrm{KCl}\) solution has a lower freezing point than the aqueous urea solution. (a) Answer: The \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{KCl}\) has a lower freezing point than the \(0.10 \mathrm{~m}\) aqueous solution of urea.

(b) The experimental freezing point of the \(\mathrm{KCl}\) solution is higher than that calculated, assuming \(\mathrm{KCl}\) is completely dissociated. The discrepancy occurs because some \(\mathrm{KCl}\) does not fully dissociate in the solution, leading to fewer ions than expected for a fully dissociated solute. As a result, the calculated van't Hoff factor is higher than the actual one, and the experimental freezing point is higher than the calculated freezing point, assuming complete dissociation. Answer: The experimental freezing point for the \(\mathrm{KCl}\) solution is higher than the calculated freezing point because not all \(\mathrm{KCl}\) dissociates into ions in the solution, resulting in a lower actual van't Hoff factor and hence a higher freezing point than predicted.