Question

Which of the following \(0.1 \mathrm{M}\) aqueous solutions will have the lowest freezing point? (a) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{NaCl}\) (c) urea (d) glucose

Short answer

(a) \(\text{K}_2\text{SO}_4\) has the lowest freezing point.

Step-by-step solution

Understanding Freezing Point Depression

Freezing point depression is a colligative property, meaning it depends on the number of solute particles in the solution, not their identity. The formula to calculate the freezing point depression is \[ \Delta T_f = i \cdot K_f \cdot m \]where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_f\) is the freezing point depression constant of the solvent, and \(m\) is the molality of the solution.

Identify Van't Hoff Factor for Each Solute

Identify the van't Hoff factor \(i\) for each compound:- \(\text{K}_2\text{SO}_4\) dissociates into 3 ions: \(2\text{K}^+\) and \(\text{SO}_4^{2-}\), so \(i = 3\).- \(\text{NaCl}\) dissociates into 2 ions: \(\text{Na}^+\) and \(\text{Cl}^-\), so \(i = 2\).- Urea and glucose do not dissociate in solution, so \(i = 1\) for both of them.

Calculate Freezing Point Depression for Each Solution

Assuming the molality \(m = 0.1 \ M\) for simplicity, calculate \(\Delta T_f\) for each:- For \(\text{K}_2\text{SO}_4\), \(\Delta T_f = 3 \cdot K_f \cdot 0.1\).- For \(\text{NaCl}\), \(\Delta T_f = 2 \cdot K_f \cdot 0.1\).- For urea, \(\Delta T_f = 1 \cdot K_f \cdot 0.1\).- For glucose, \(\Delta T_f = 1 \cdot K_f \cdot 0.1\).

Compare Freezing Point Depressions

The greater the van't Hoff factor \(i\), the larger the freezing point depression \(\Delta T_f\), meaning a lower freezing point for the solution. Therefore:- \(\text{K}_2\text{SO}_4\) with \(i = 3\) will have the largest \(\Delta T_f\), leading to the lowest freezing point.- \(\text{NaCl}\) with \(i = 2\), urea, and glucose with \(i = 1\) follow after.

Key concepts

Van't Hoff Factor

The Van't Hoff factor is crucial when examining solutions for freezing point depression. This factor, denoted as \(i\), tells us how many particles a solute splits into when dissolved in solvent. It's essential to understand because the freezing point is affected by the total number of solute particles present.

The Van’t Hoff factor for a non-electrolyte like urea or glucose is \(i = 1\) because these substances do not dissociate into ions. In contrast, electrolytes such as sodium chloride (NaCl) dissociate into ions, resulting in an \(i\) greater than 1. NaCl breaks apart into two separate ions: \(\text{Na}^+\) and \(\text{Cl}^-\), so its \(i\) equals 2.

Potassium sulfate ( \(\text{K}_2\text{SO}_4\)) dissociates into three ions: two \(\text{K}^+\) and one \(\text{SO}_4^{2-}\), making its \(i = 3\). This higher \(i\) indicates more solute particles in solution, influencing the freezing point substantially.

Colligative Properties

Colligative properties are unique because they depend primarily on the number of solute particles rather than their identity. Freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure are examples of colligative properties.

The more solute particles present, the more pronounced the effect on the colligative property. For freezing point depression, the formula \(\Delta T_f = i \cdot K_f \cdot m\) illustrates this: \(\Delta T_f\) increases with the number of particles \((i)\), suggesting a lower freezing point.

Therefore, a solution with a solute that dissociates into more particles, like \(\text{K}_2\text{SO}_4\), has a deeper freezing point depression than solutions of solutes like glucose that remain intact when dissolved.

Ionic Dissociation

Ionic dissociation describes how ionic compounds break into individual ions when dissolved in water. This process tentatively increases the number of particles in solution, affecting its colligative properties.

Some notable aspects of ionic dissociation include:

• Electrolytes vs Non-electrolytes: Electrolytes like NaCl fully dissociate into ions, whereas non-electrolytes like glucose do not dissociate at all.
• Impact on Freezing Point Depression: The more ions formed, the larger the Van't Hoff factor \((i)\), leading to greater freezing point depression.
• Examples in Solutions: Solutions such as \(\text{K}_2\text{SO}_4\) have greater impacts on freezing points due to more ions compared to urea or glucose solutions.

Understanding ionic dissociation helps explain why some solutions significantly lower the freezing point as compared to others.

Molality

Molality (\(m\)) is a measure of concentration often used in colligative properties because it remains unaffected by temperature changes. It's defined as moles of solute per kilogram of solvent.

Unlike molarity, which depends on the total volume of solution that can change with temperature, molality offers a temperature-independent way to express concentration. Therefore, it helps accurately predict the impacts on colligative properties like freezing point depression.

In freezing point calculations, molality plays an important role in the formula \(\Delta T_f = i \cdot K_f \cdot m\). This formula shows how the partnership of molality and the Van’t Hoff factor influences the extent of freezing point reduction in solutions.