Problem 65 - Chemistry The Molecular Science

Question

Place these aqueous solutions in order of increasing boiling point. (a) \(0.10 \mathrm{~mol} \mathrm{KCl} / \mathrm{kg}\) (b) \(0.10 \mathrm{~mol}\) glucose \(/ \mathrm{kg}\) (c) \(0.080 \mathrm{~mol} \mathrm{MgCl}_{2} / \mathrm{kg}\) (Assume that all of the salts dissociate completely into their ions in solution.)

Step-by-Step Solution

Answer

Order: (b), (a), (c)

Step 1: Identify Boiling Point Elevation Concept

Boiling point elevation occurs when a solute is added to a solvent, raising its boiling point. It is calculated using the formula:\[\Delta T_b = i \cdot K_b \cdot m\]where \(\Delta T_b\) is the boiling point elevation, \(i\) is the van't Hoff factor (number of particles the solute dissociates into), \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution.

Step 2: Determine van't Hoff Factor for Each Solute

- For KCl: dissociates into 2 ions (K\(^+\) and Cl\(^-\)), so \(i = 2\).- For glucose: does not dissociate, so \(i = 1\).- For MgCl\(_2\): dissociates into 3 ions (Mg\(^{2+}\) and 2 Cl\(^-\)), so \(i = 3\).

Step 3: Calculate Boiling Point Elevation for Each

Assuming \(K_b\) is constant for each solution, the boiling point elevation depends on \(i \cdot m\):- (a) KCl: \(0.10 \, \text{mol/kg} \times 2 = 0.20\)- (b) Glucose: \(0.10 \, \text{mol/kg} \times 1 = 0.10\)- (c) MgCl\(_2\): \(0.080 \, \text{mol/kg} \times 3 = 0.24\)

Step 4: Order Solutions by Increasing Boiling Point

Order the solutions based on boiling point elevation calculated:1. Glucose: \(0.10\)2. KCl: \(0.20\)3. MgCl\(_2\): \(0.24\)Thus, order from lowest to highest boiling point increase is (b), (a), (c).

Key Concepts

van't Hoff factor

When studying the effect of solutes on the boiling point of a solution, the van't Hoff factor plays a crucial role. This factor, denoted as \(i\), tells us how many particles a solute breaks into when it dissolves in a solvent. The more particles a solute breaks into, the greater the effect it has on the boiling point elevation.

Let's see how this works in practice:

• KCl dissociates into two ions: \(\text{K}^+\) and \(\text{Cl}^-\), so its van't Hoff factor is \(i = 2\).
• Glucose does not dissociate into ions; it remains as a single particle in solution. Hence, for glucose, \(i = 1\).
• MgCl\(_2\) dissociates into three ions: one \(\text{Mg}^{2+}\) and two \(\text{Cl}^-\) ions, which gives it a van't Hoff factor of \(i = 3\).

The greater the van't Hoff factor, the more the solute influences properties like freezing point depression or boiling point elevation. This is because more particles in solution mean greater interference with the physical properties of the solvent.

solute dissociation

Solute dissociation refers to the process by which a compound breaks into smaller particles, typically ions, when it dissolves in a solvent. Many ionic compounds, like salts, dissociate when placed in water.

This is key for understanding boiling point elevation because:

• Each separated particle in the solution contributes to the overall effect on the physical properties of the liquid, like its boiling point.
• Dissociation increases the number of particles in the solution, influencing the colligative properties.

In the given example:

• KCl fully dissociates into two ions, \(\text{K}^+\) and \(\text{Cl}^-\).
• MgCl\(_2\), with its higher dissociation into three ions \(\text{Mg}^{2+}\) and two \(\text{Cl}^-\), results in a more substantial impact on the boiling point than KCl.
• In contrast, glucose, which does not dissociate, results in just one set of particles, having minimal impact on colligative properties compared to dissociating salts.

Comprehending solute dissociation is vital for predicting and calculating changes in boiling points when substances are dissolved.

molality

Molality is a concentration term used in chemistry to describe how much solute is dissolved in a solvent. It is defined as the number of moles of solute per kilogram of solvent, making it useful for understanding changes in physical properties like boiling point and freezing point. The formula for molality is:
\[m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\]
One major advantage of molality over other concentration measurements like molarity is that it does not change with temperature, since it’s based on mass, not volume. This makes it particularly advantageous for studying thermodynamic properties.

For example:

• The solutions in the exercise are all given in terms of molality \((\text{mol/kg})\), ensuring consistent measurement across different substances.
• The boiling point elevation formula \(\Delta T_b = i \cdot K_b \cdot m\) uses molality directly, combined with the van't Hoff factor \(i\) and the ebullioscopic constant \(K_b\), to calculate changes.

This precise measure allows students to clearly estimate how different solutes and their concentrations affect solutions’ boiling points.