Problem 52 - Chemistry and Chemical Reactivity

Question

(a) Which aqueous solution is expected to have the higher boiling point: \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\) or \(0.15 \mathrm{m}\) sugar? (b) For which aqueous solution is the vapor pressure of water higher: \(0.30 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(0.15 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4} ?\)

Step-by-Step Solution

Answer

(a) \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}\) has a higher boiling point. (b) \(0.30 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) has a higher vapor pressure of water.

Step 1: Determine Boiling Point Elevation Formula

The formula for boiling point elevation is given by \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the change in boiling point, \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality of the solution.

Step 2: Calculate Van't Hoff Factor for Each Solute in Part (a)

For \( \mathrm{Na}_{2} \mathrm{SO}_{4} \), \( i = 3 \) because it dissociates into three ions (2 \( \mathrm{Na}^+ \) and 1 \( \mathrm{SO}_{4}^{2-} \)). For sugar (nonelectrolyte), \( i = 1 \) because it does not dissociate in solution.

Step 3: Compare Boiling Point Elevations for Part (a)

Calculate boiling point elevation for each solution:- \( 0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}: \Delta T_b = 3 \times K_b \times 0.10 = 0.30 K_b \)- \( 0.15 \mathrm{m} \mathrm{sugar}: \Delta T_b = 1 \times K_b \times 0.15 = 0.15 K_b \)Since \( \Delta T_b \) is greater for \( 0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4} \), this solution has the higher boiling point.

Step 4: Determine Vapor Pressure Lowering Formula

The formula for vapor pressure lowering is given by \( \Delta P = i \cdot X_s \cdot P^0 \), where \( \Delta P \) is the change in vapor pressure, \( X_s \) is the mole fraction of solute, and \( P^0 \) is the vapor pressure of the pure solvent.

Step 5: Calculate Van't Hoff Factor for Each Solute in Part (b)

For \( \mathrm{NH}_{4} \mathrm{NO}_{3} \), \( i = 2 \) as it dissociates into 2 ions (\( \mathrm{NH}_{4}^+ \) and \( \mathrm{NO}_{3}^- \)). For \( \mathrm{Na}_{2} \mathrm{SO}_{4} \), \( i = 3 \) as noted previously.

Step 6: Compare Vapor Pressure Lowering for Part (b)

The vapor pressure lowering impacts inversely the vapor pressure. Calculate for each solution:- \( 0.30 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}: \Delta P = 2 \times X_s \times P^0 \)- \( 0.15 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}: \Delta P = 3 \times X_s \times P^0 \)Due to the greater number of ions and higher molality, \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) causes a larger vapor pressure lowering, so the \( 0.30 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3} \) solution will have a higher vapor pressure.

Key Concepts

Vapor Pressure Lowering

Vapor pressure lowering is an important aspect in studying solutions, especially how solutes affect the properties of solvents. When a solute is added to a pure solvent, the vapor pressure of the solvent above the solution decreases. This phenomenon happens because the solute particles occupy space at the surface of the liquid, reducing the number of solvent molecules that can escape into the vapor phase. The decrease is described by the formula \[ \Delta P = i \cdot X_s \cdot P^0 \]
where \( \Delta P \) is vapor pressure lowering, \( i \) is the Van't Hoff factor, \( X_s \) is the mole fraction of the solute, and \( P^0 \) is the vapor pressure of the pure solvent. In the given exercise, the comparison of vapor pressures between solutions of different solvents and solutes is determined by the amount of solute particles in the solution. For example, \( \mathrm{Na}_2\mathrm{SO}_4 \) dissociates into more ions than \( \mathrm{NH}_4\mathrm{NO}_3 \), causing a greater decrease in vapor pressure. Thus, the solution with \( \mathrm{NH}_4\mathrm{NO}_3 \) retains a higher vapor pressure.

Van't Hoff Factor

The Van't Hoff factor, denoted as \( i \), is a crucial concept when analyzing colligative properties like boiling point elevation and vapor pressure lowering. It indicates how many particles a compound dissociates into in a solution. For substances that do not dissociate, such as sugar, \( i = 1 \). If a substance dissociates, the \( i \) value will be greater than 1.Here are some key points:

• Sugar, as a non-electrolyte, does not dissociate and thus, \( i = 1 \).
• Sodium sulfate \( (\mathrm{Na}_2\mathrm{SO}_4) \) dissociates into 3 ions: 2 sodium ions \( (\mathrm{Na}^+) \) and 1 sulfate ion \( (\mathrm{SO}_4^{2-}) \), so \( i = 3 \).
• Ammonium nitrate \( (\mathrm{NH}_4\mathrm{NO}_3) \) dissociates into 2 ions, making \( i = 2 \).

The Van't Hoff factor directly affects how much a property is altered by the solute. For instance, a higher \( i \) leads to a greater elevation in boiling point and a more significant lowering of vapor pressure.

Colligative Properties

Colligative properties are characteristics of solutions that depend mainly on the number of solute particles, not their identity. These include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure.Key points about colligative properties:

• They are influenced by the concentration of solute particles.
• Van't Hoff factor is crucial as it measures how many particles result from a solute.
• Examples in different contexts, such as boiling point elevation, show higher elevations with more solute particles.

These properties are significant in real-world applications like cooking, where adding salt to water raises its boiling point. In our exercise, understanding colligative properties helped determine why \( 0.10 \mathrm{m} \mathrm{Na}_2\mathrm{SO}_4 \) with a higher Van't Hoff factor exhibited a more substantial impact on the boiling point compared to a sugar solution. This understanding is fundamental to predicting how solutes alter the physical properties of solvents in solutions.