Problem 53 - Chemistry and Chemical Reactivity

Question

Arrange the following aqueous solutions in order of (i) increasing vapor pressure of water and (ii) increasing boiling point. (a) \(0.35 m\) \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (a nonvolatile solute) (b) \(0.50 m\) sugar (c) \(0.20 m\) \(\mathrm{KBr}\) (a strong electrolyte) (d) \(0.20 m\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (a strong electrolyte)

Step-by-Step Solution

Answer

(i) Vapor Pressure: HOCH₂CH₂OH < KBr < Sugar < Na₂SO₄. (ii) Boiling Point: Na₂SO₄ > Sugar > KBr > HOCH₂CH₂OH.

Step 1: Determine Van't Hoff Factor (i) for Each Solute

The Van't Hoff factor (i) represents the number of particles formed in solution per formula unit. \( \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} \) and sugar are non-electrolytes so \(i = 1\). \( \mathrm{KBr} \) dissociates into \( \mathrm{K^+} \) and \( \mathrm{Br^-} \), so \(i = 2\). \( \mathrm{Na}_{2} \mathrm{SO}_{4} \) dissociates into 2 \( \mathrm{Na^+} \) and 1 \( \mathrm{SO}_{4}^{2-} \), so \(i = 3\).

Step 2: Calculate Effective Molality

Effective molality is given by \(m \times i\). Calculate for each solution: \( \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}: 0.35 \times 1 = 0.35 \)Sugar: \(0.50 \times 1 = 0.50\)\( \mathrm{KBr}: 0.20 \times 2 = 0.40 \)\( \mathrm{Na}_{2} \mathrm{SO}_{4}: 0.20 \times 3 = 0.60\)

Step 3: Arrange Solutions by Increasing Vapor Pressure

Vapor pressure decreases as the number of solute particles increases. Arrange by effective molality: 1. \( \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} (0.35) \)2. \( \mathrm{KBr} (0.40) \)3. Sugar \((0.50)\)4. \( \mathrm{Na}_{2} \mathrm{SO}_{4} (0.60) \)

Step 4: Arrange Solutions by Increasing Boiling Point

Boiling point increases with more solute particles. Follow the same order as vapor pressure but reversed:1. \( \mathrm{Na}_{2} \mathrm{SO}_{4} (0.60) \)2. Sugar \((0.50)\)3. \( \mathrm{KBr} (0.40) \)4. \( \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} (0.35) \)

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. For a solution, the addition of a solute reduces the vapor pressure of the solvent. This is because solute particles occupy space at the surface of the liquid, leaving less room for solvent molecules to escape.

When considering solutions with different solutes, the decrease in vapor pressure is directly related to the number of solute particles in the solution. More particles mean a lower vapor pressure. In the exercise, different aqueous solutions were compared based on their effective molality, which gives us an idea of the concentration of solute particles.
In summary:

• The higher the number of solute particles (greater effective molality), the lower the vapor pressure of the solution.
• This means solution (d) with the highest effective molality of 0.60 has the lowest vapor pressure, whereas solution (a) with a value of 0.35 has the highest vapor pressure.

Boiling Point Elevation

Boiling point elevation occurs because solute particles disrupt the formation of vapor bubbles in a liquid, requiring higher temperatures to reach the boiling point. This is a colligative property which means it depends on the particle number, not the type of solute.
The more solute particles present, the higher the boiling point. This happens because the solute particles lower the solvent's escape rate, meaning more heat energy is needed for the solution to start boiling.
In the given exercise, boiling points of the solutions were compared using effective molalities. Essentially, the solution with the highest effective molality will have the smallest increase in vapor pressure and the greatest elevation in boiling point.
To remember:

• Solution (d) with the highest effective molality (0.60) will have the highest boiling point.
• Solution (a) with the lowest effective molality (0.35) will have the lowest boiling point among the listed solutions.

Van't Hoff Factor

The Van't Hoff factor, represented by the symbol 'i', is a key concept in understanding colligative properties. It indicates the number of particles resulting from one formula unit of solute when dissolved in a solvent.
For non-electrolytes, which do not dissociate in solution, the Van't Hoff factor is 1. This applies to solutions of sugar and \(\text{HOCH}_2 \text{CH}_2 \text{OH}\).
For electrolytes, which do dissociate, the Van't Hoff factor is greater than 1. For example:

• KBr dissociates into K⁺ and Br⁻, giving an 'i' value of 2.
• \(\text{Na}_2 \text{SO}_4\) splits into 2 Na⁺ and 1 SO₄²⁻, resulting in an 'i' value of 3.

Van't Hoff factor allows us to calculate the effective molality of the solutions by multiplying it with the given molality.

Effective Molality

Effective molality is a way to account for how many particles a solute yields in solution, hence combining both the actual concentration (molality) and dissociation (Van't Hoff factor).
To find the effective molality, you simply multiply the solute's molality by its Van't Hoff factor. This gives us a realistic concentration that considers dissociation or association of the solute:

• \(\text{HOCH}_2 \text{CH}_2 \text{OH}\), with a Van't Hoff factor of 1, gives an effective molality of 0.35 × 1 = 0.35.
• Sugar, with a Van't Hoff factor of 1, results in 0.50 × 1 = 0.50.
• KBr, with 'i' of 2, gives 0.20 × 2 = 0.40.
• \(\text{Na}_2 \text{SO}_4\), with 'i' of 3, yields 0.20 × 3 = 0.60.

This calculated effective molality is then used to order the solutions based on vapor pressure and boiling point changes, as higher effective molality indicates more solute particles in the solution.