Problem 16 - Chemistry in 30 days

Question

Which of the following solutions will have the maximum lowering of vapour pressure at \(300 \mathrm{~K} ?\) (a) \(1 \mathrm{M} \mathrm{CaCl}_{2}\) (b) \(1 \mathrm{M} \mathrm{NaCl}\) (c) \(1 \mathrm{M}\) Phenol (d) \(1 \mathrm{M}\) Sucrose

Step-by-Step Solution

Answer

1 M CaCl₂ (option a) will have the maximum lowering of vapor pressure.

Step 1: Understanding the Vapor Pressure Lowering

The lowering of vapor pressure is a colligative property, which means it depends on the number of particles in the solution.

Step 2: Determine the Number of Particles for Each Substance

1. CaCl₂ dissociates into 3 ions: 1 Ca²⁺ and 2 Cl⁻. 2. NaCl dissociates into 2 ions: 1 Na⁺ and 1 Cl⁻. 3. Phenol and sucrose do not dissociate, contributing only 1 particle per molecule.

Step 3: Calculate Effective Concentration

CaCl₂: 1 M × 3 = 3 particles NaCl: 1 M × 2 = 2 particles Phenol: 1 M × 1 = 1 particle Sucrose: 1 M × 1 = 1 particle

Step 4: Identify the Maximum Vapor Pressure Lowering

Since the lowering of vapor pressure is proportional to the number of particles, CaCl₂ with 3 particles per formula unit has the most significant effect.

Key Concepts

Vapor Pressure Lowering

In solutions, vapor pressure lowering is a core concept. It is a colligative property, meaning it depends *not* on what the solute is, but on how many solute particles are present.
When a non-volatile solute is added to a solvent, the vapor pressure of the solvent decreases. This is because solute particles take up space on the surface, making it harder for solvent molecules to escape into the gas phase.
This way, the overall vapor pressure of the solution is lower than that of the pure solvent. A practical result of this is that solutions made with higher concentrations of solute particles demonstrate greater vapor pressure lowering. For instance, in a scenario with solutions of equal molarity but different types of solute, the one that breaks down into more particles upon dissolving will cause greater vapor pressure lowering.

Ionic Dissociation

Ionic dissociation is a fascinating process that occurs when ionic compounds dissolve in water. These compounds split into their respective ions, effectively multiplying the number of particles in the solution.
For example, in our exercise, both \( \text{CaCl}_2 \) and \( \text{NaCl} \) are ionic compounds. When \( \text{CaCl}_2 \) dissolves, it splits into three ions: one \( \text{Ca}^{2+} \) and two \( \text{Cl}^- \). On the other hand, \( \text{NaCl} \) splits into two ions: one \( \text{Na}^{+} \) and one \( \text{Cl}^- \). This dissociation means that for ionic compounds, the observed number of particles in solution is higher than their initial molarity might suggest.As a result, this phenomenon can drastically amplify colligative effects such as vapor pressure lowering, boiling point elevation, and freezing point depression, making ionic dissociates a crucial factor to consider.

Effective Concentration

Effective concentration is all about the number of particles present in a solution.This number is what ultimately influences colligative properties.
For any solution, calculating the effective concentration involves considering how many ions or molecules a solute unit contributes to the solution. In ionic solutions, this calculation is straightforward due to dissociation. Each solute molecule may contribute multiple ions, boosting the effective concentration beyond its molar concentration.For example, in our exercise, \( 1 \text{ M CaCl}_2 \) dissociates to produce three particles, resulting in an effective concentration of \( 3 \text{ M} \). Conversely, \( 1 \text{ M NaCl} \) results in an effective concentration of \( 2 \text{ M} \). As such, solutions with higher effective concentrations exhibit more pronounced colligative effects, such as lowering vapor pressure. Understanding this concept is crucial for accurately predicting and explaining the behavior of different solutions.