Question

(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?

Short answer

(a) 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 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.

Step-by-step solution

Understand Colligative Properties and Freezing Point Depression Formula

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.

Determine Van't Hoff Factors for \(\mathrm{KCl}\) and Urea

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.

Calculate the Freezing Point Depression for Each Solution

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: \(\Delta T_{f, KCl} > \Delta T_{f, urea}\).

Compare the Freezing Points of the Two Solutions

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.

Explain the Discrepancy between the Experimental Freezing Point and the Calculated Freezing Point for \(\mathrm{KCl}\)

(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.

Key concepts

Freezing Point Depression

Freezing point depression is a key concept in understanding how solutes affect the freezing point of a solvent. It is a colligative property, which means it depends on the number of solute particles rather than the type of particles.
When a solute is added to a solvent, the freezing point of the solution is lowered compared to the pure solvent. This occurs because the solute particles interfere with the formation of the solid structure (crystal lattice) that characterizes the frozen state of the solvent.
The relationship between the freezing point depression (\(\Delta T_f\)) and the concentration of solute particles is given by the formula:

• \(\Delta T_f = K_f \cdot m \cdot i\)

- \(K_f\) is the freezing point depression constant for the specific solvent.
- \(m\) is the molality of the solution.
- \(i\) is the van't Hoff factor, representing the degree to which the solute dissociates.
This principle explains why solutions with a higher number of solute particles will have lower freezing points, which is crucial for various applications, like designing antifreeze solutions.

Van't Hoff Factor

The van't Hoff factor (\(i\)) is a crucial component when predicting the effects of a solute on the freezing point or boiling point of a solution. It essentially tells us how many particles are formed when a compound dissolves in a solvent.
For ionic compounds like potassium chloride (\(\mathrm{KCl}\)), the solute dissociates fully into its ions in ideal conditions, so \(i = 2\) because \(\mathrm{KCl}\) splits into \(\mathrm{K^+}\) and \(\mathrm{Cl^-}\) ions.
Conversely, covalent compounds like urea (\(\mathrm{CO(NH_2)_2}\)) do not dissociate into simpler particles. As a result, the van't Hoff factor for urea is \(i = 1\).

• A greater van’t Hoff factor indicates more particles are created, leading to larger changes in freezing or boiling points.
• Discrepancies can appear when assumptions about the complete dissociation aren’t met, as some solute may remain undissociated in reality.

Understanding the van't Hoff factor helps us predict and explain the magnitude of colligative properties accurately, which is vital in laboratory and industrial settings.

Molality

Molality (\(m\)) is a measure of the concentration of a solute in a solution and is expressed in moles of solute per kilogram of solvent. Unlike molarity, molality is not affected by changes in temperature or pressure due to its mass-based calculation.
This characteristic makes molality particularly useful in colligative properties like freezing point depression as it provides a consistent measure when temperatures vary.

• To calculate molality:

- Determine the number of moles of solute.- Measure the mass of the solvent in kilograms.- Use the formula:\(m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\)By using molality in the freezing point depression formula, we can consistently assess the impact of the solute on the solution regardless of temperature fluctuations, ensuring precise predictions in various scientific calculations.