Problem 108 - Chemistry The Molecular Science

Question

Arrange these aqueous solutions in order of decreasing freezing point. (Assume theoretical values for \(i\).) (a) \(0.20 \mathrm{~mol}\) ethylene \(\mathrm{glycol} / \mathrm{kg}\) (b) \(0.12 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{SO}_{4} / \mathrm{kg}\) (c) \(0.10 \mathrm{~mol} \mathrm{NaBr} / \mathrm{kg}\) (d) \(0.12 \mathrm{~mol} \mathrm{KI} / \mathrm{kg}\)

Step-by-Step Solution

Answer

Order: Ethylene glycol, NaBr, KI, Na2SO4, from highest to lowest freezing point.

Step 1: Understand Freezing Point Depression

The freezing point depression of a solution depends on the concentration of solute particles in the solution, calculated using the formula \( \,\Delta T_f = i imes K_f imes m \,\), where \( i \,\) is the van't Hoff factor, \( K_f \,\) is the freezing point depression constant, and \( m \,\) is the molality of the solution.

Step 2: Determine Van't Hoff Factor (i)

For ethylene glycol, \( i = 1 \) because it is a non-electrolyte. For \( \text{Na}_2\text{SO}_4 \), \( i = 3 \) as it dissociates into 3 ions \( (2\text{Na}^+ + \text{SO}_4^{2-}) \). For \( \text{NaBr} \), \( i = 2 \) as it dissociates into 2 ions \( (\text{Na}^+ + \text{Br}^-) \). For \( \text{KI} \), \( i = 2 \) as it dissociates into 2 ions \( (\text{K}^+ + \text{I}^-) \).

Step 3: Calculate Freezing Point Depression for Each Solution

Use the given molality and calculate \( \Delta T_f \) for each: - Ethylene glycol: \( \Delta T_f = 1 \times m = 0.20 \) - \( \text{Na}_2\text{SO}_4 \): \( \Delta T_f = 3 \times 0.12 = 0.36 \) - \( \text{NaBr} \): \( \Delta T_f = 2 \times 0.10 = 0.20 \) - \( \text{KI} \): \( \Delta T_f = 2 \times 0.12 = 0.24 \)

Step 4: Arrange in Order of Decreasing Freezing Point

The solution with the smallest \( \Delta T_f\) has the highest freezing point. Thus, the order from highest to lowest freezing point is: 1. Ethylene glycol \( (0.20) \)2. NaBr \( (0.20) \)3. KI \( (0.24) \)4. \( \text{Na}_2\text{SO}_4 \) \( (0.36) \).

Key Concepts

Van't Hoff Factor

The Van’t Hoff factor, denoted as \( i \), plays a crucial role in understanding how solutes affect the freezing point of solutions. It describes the number of particles a compound forms in solution. For example, some compounds dissociate into ions when they dissolve in water, impacting the calculation of properties like freezing point depression.

Here's a breakdown:

• Non-electrolyte (e.g., Ethylene glycol): Doesn't dissociate, so \( i = 1 \).
• Electrolytes (e.g., \( \text{Na}_2\text{SO}_4 \)): Dissociates into multiple ions, giving it \( i = 3 \) because it forms 2 \( \text{Na}^+ \) and 1 \( \text{SO}_4^{2-} \) ion.
• Other compounds (e.g., NaBr, KI): Dissociate in solution into 2 ions each, resulting in \( i = 2 \).

Understanding \( i \) is vital because it directly affects the degree of freezing point depression in a solution.

Molality

Molality is a measure of the concentration of a solute in a solution, described in moles of solute per kilogram of solvent. It is denoted by \( m \) and is crucial for calculating freezing point depression.

Unlike molarity, molality is not affected by temperature changes because it depends on mass, not volume. This makes it a reliable measurement for changes in temperature-dependent properties like the freezing point.

In the examples provided:

• Ethylene glycol: \( 0.20 \text{ mol/kg} \)
• \( \text{Na}_2\text{SO}_4 \): \( 0.12 \text{ mol/kg} \)
• NaBr: \( 0.10 \text{ mol/kg} \)
• KI: \( 0.12 \text{ mol/kg} \)

Molality helps us understand the concentration impact on properties such as freezing point.

Solute Particles

Solute particles, resulting from the dissociation of compounds in a solution, are essential in determining properties like freezing point depression. The more particles present, the greater the effect on freezing point.

When a solute dissolves:

• Non-electrolytes: Produce molecules without further dissociation.
• Electrolytes: Break into ions, increasing the number of solute particles.

For instance, \( \text{Na}_2\text{SO}_4 \) dissociates into three particles (ions), impacting the freezing point more significantly than a non-electrolyte like ethylene glycol, which remains as single molecules in solution.

Understanding solute particles is key to predicting how a solution's properties will change.

Aqueous Solutions

Aqueous solutions are those where water acts as the solvent. These solutions are particularly important for studying properties like freezing point depression.

In aqueous solutions:

• Water is the solvent: It provides a medium for solutes to dissolve and dissociate.
• Solutes can be electrolytes or non-electrolytes: Affecting how the solution behaves in terms of temperature-dependent properties.

The structure and interactions of water molecules with dissolved solute particles influence how properties like freezing temperatures are modified. For the problem given, the solutions are aqueous, meaning the dissolving occurs in water, emphasizing the role of water as the solvent in calculating changes in physical properties.