Question

(a) Which solution is expected to have the higher boiling point: \(0.20 \mathrm{m}\) KBr or \(0.30 \mathrm{m}\) sugar? (b) Which aqueous solution has the lower freezing point: \(0.12 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Short answer

(a) KBr has the higher boiling point; (b) Na2CO3 has the lower freezing point.

Step-by-step solution

Identify Relevant Concepts on Boiling Point Elevation

The boiling point elevation depends on the number of solute particles in a solution (colligative property). The formula is given by \( \Delta T_b = i \cdot K_b \cdot m \), where \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution. The solute KBr dissociates into 2 particles (K+ and Br-), so \( i = 2 \). Sugar does not dissociate, so \( i = 1 \).

Calculate the Boiling Point for KBr and Sugar Solutions

For KBr, \( \Delta T_b = 2 \cdot K_b \cdot 0.20 = 0.40 K_b \). For sugar, \( \Delta T_b = 1 \cdot K_b \cdot 0.30 = 0.30 K_b \). The solution with the higher boiling point elevation will have the higher boiling point.

Identify Relevant Concepts on Freezing Point Depression

The freezing point depression is also based on the number of solute particles, given by \( \Delta T_f = i \cdot K_f \cdot m \), where \( K_f \) is the cryoscopic constant. For \( \mathrm{NH}_4\mathrm{NO}_3 \), which dissociates into 2 ions, \( i = 2 \). For \( \mathrm{Na}_2\mathrm{CO}_3 \), which dissociates into 3 ions, \( i = 3 \).

Calculate the Freezing Point for NH4NO3 and Na2CO3 Solutions

For \( \mathrm{NH}_4\mathrm{NO}_3 \), \( \Delta T_f = 2 \cdot K_f \cdot 0.12 = 0.24 K_f \). For \( \mathrm{Na}_2\mathrm{CO}_3 \), \( \Delta T_f = 3 \cdot K_f \cdot 0.10 = 0.30 K_f \). The solution with the higher freezing point depression will have the lower freezing point.

Key concepts

Boiling Point Elevation

Boiling point elevation is a fascinating colligative property that shows how adding a solute to a solvent can change the temperature at which the mixture boils. This happens because the solute particles disrupt the regular evaporating process of the solvent, thus needing more heat to escape into the gas phase. The relationship between the boiling point elevation and the properties of the solution can be described using the formula: \( \Delta T_b = i \cdot K_b \cdot m \).

Here's a quick breakdown of the components in the formula:

• \( \Delta T_b \) is the change in boiling point.
• \( i \) is the van't Hoff factor, which signifies the number of particles a solute breaks into.
• \( K_b \) is the ebullioscopic constant, specific to each solvent.
• \( m \) is the molality of the solution.

For instance, when comparing two solutions, such as solutions of KBr and sugar, it is crucial to recognize that KBr dissociates into more particles compared to sugar, affecting the boiling point noticeably more.

Freezing Point Depression

Freezing point depression is another intriguing concept of colligative properties, which describes how a solute lowers the freezing point of a solvent. This occurs because the solute particles interfere with the orderly crystal formation of the solvent as it transitions to a solid state, thus requiring a lower temperature to achieve this process. The relevant formula for freezing point depression is \( \Delta T_f = i \cdot K_f \cdot m \).

Let's delve into the terms involved:

• \( \Delta T_f \) represents the change in freezing point.
• \( i \) is the van't Hoff factor, indicating the total particles the solute splits into in solution.
• \( K_f \) is the cryoscopic constant, a property of individual solvents.
• \( m \) is the molality of the solution.

In comparing solutions like \( \mathrm{NH}_4\mathrm{NO}_3 \) and \( \mathrm{Na}_2\mathrm{CO}_3 \), the solution with the larger dissociation number (greater value of \( i \)) depresses the freezing point more, leading to a lower temperature required for freezing.

Van't Hoff Factor

The van’t Hoff factor is a crucial concept when discussing colligative properties. It essentially measures the effect of a solute on properties like boiling and freezing points by determining how many particles a solute splits into when dissolved in a solvent.

Let's break it down:

• For non-electrolytes, like sugar, the van’t Hoff factor \( i \) is typically 1 because they do not dissociate into multiple particles.
• For electrolytes, such as KBr or \( \mathrm{Na}_2\mathrm{CO}_3 \), the van’t Hoff factor is greater than 1 because they break down into ions.

Knowing the van’t Hoff factor helps predict the extent of boiling point elevation or freezing point depression, as it indicates the total number of solute particles present in the solution. Consequently, the higher the factor, the more significant the impact on the colligative property.