Problem 64 - Chemistry The Molecular Science

Question

List these aqueous solutions in order of decreasing freezing point. (a) \(0.10 \mathrm{~mol}\) methanol/kg (b) \(0.10 \mathrm{~mol} \mathrm{KCl} / \mathrm{kg}\) (c) \(0.080 \mathrm{~mol} \mathrm{BaCl}_{2} / \mathrm{kg}\) (d) \(0.040 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{SO}_{4} / \mathrm{kg}\) (Assume that all of the salts dissociate completely into their ions in solution.)

Step-by-Step Solution

Answer

Order: (a) methanol, (d) Na₂SO₄, (b) KCl, (c) BaCl₂.

Step 1: Understand Freezing Point Depression

The freezing point depression of a solution is given by the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the decrease in freezing temperature, \( K_f \) is the freezing point depression constant, \( m \) is the molality, and \( i \) is the van 't Hoff factor representing the number of particles the solute breaks into in solution.

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

(a) Methanol is a nonelectrolyte, so \( i = 1 \).(b) KCl dissociates into 2 ions, so \( i = 2 \).(c) BaCl₂ dissociates into 3 ions (Ba²⁺ and 2Cl⁻), so \( i = 3 \).(d) Na₂SO₄ dissociates into 3 ions (2Na⁺ and SO₄²⁻), so \( i = 3 \).

Step 3: Calculate Freezing Point Depression for Each Solution

Using the formula \( \Delta T_f = i \cdot K_f \cdot m \), and assuming \( K_f \) is the same for all solutions, we compare \( i \cdot m \) for each:(a) \( i = 1, m = 0.10 \) -> \( i \cdot m = 0.10 \)(b) \( i = 2, m = 0.10 \) -> \( i \cdot m = 0.20 \)(c) \( i = 3, m = 0.080 \) -> \( i \cdot m = 0.24 \)(d) \( i = 3, m = 0.040 \) -> \( i \cdot m = 0.12 \)

Step 4: List Solutions in Decreasing Order of Freezing Point

Higher \( i \cdot m \) values mean a lower freezing point. Thus, the order of decreasing freezing point is:1. Solution (a) Methanol: \( 0.10 \)2. Solution (d) Na₂SO₄: \( 0.12 \)3. Solution (b) KCl: \( 0.20 \)4. Solution (c) BaCl₂: \( 0.24 \)

Key Concepts

van 't Hoff factor

The van 't Hoff factor, represented by the symbol \( i \), is an important concept in understanding colligative properties such as freezing point depression. It expresses the number of particles a compound dissociates into when dissolved in a solution. For example, in the case of electrolytes like salts:

• Methanol, being a nonelectrolyte, does not dissociate and thus has an \( i \) value of 1.
• Potassium chloride (KCl) dissociates into two ions (K⁺ and Cl^-), resulting in an \( i \) of 2.
• Barium chloride (BaCl₂) separates into three ions (one Ba²⁺ and two Cl^-), giving an \( i \) of 3.
• Sodium sulfate (Na₂SO₄) also dissociates into three ions (two Na⁺ and one SO₄^2-), leading to an \( i \) of 3.

Understanding the van 't Hoff factor is crucial for predicting how a solute will affect the colligative properties of a solution, like decreasing the freezing point.

molality

Molality, indicated by \( m \), is a measure of concentration defined as the number of moles of solute per kilogram of solvent. It differs from molarity, which measures concentration per liter of solution. This distinction is significant especially in colligative properties because:

• Molality is unaffected by changes in temperature and pressure, making it more reliable for calculations involving changes in physical states.
• The freezing point depression, a colligative property, is calculated using molality in the formula \( \Delta T_f = i \cdot K_f \cdot m \).
• In our exercise, each solution's molality was already provided, such as \(0.10\, \text{mol} / \text{kg}\) for KCl.

Considering molality ensures more accurate insights into how solute particles influence a solution under varying environmental conditions.

aqueous solutions

Aqueous solutions play a significant role in chemistry, referring simply to any solution where water acts as the solvent. Understanding their character is crucial when dealing with colligative properties like freezing point depression:

• Water, as a universal solvent, can dissolve a variety of solutes including salts and other electrolytes.
• Aqueous solutions offer a medium where ionic compounds dissociate into ions, impacting the solution's properties.
• In the context of freezing point depression, it’s important to remember that the nature of water allows for significant interactions and changes in properties such as boiling points and freezing points.

Through complete dissociation of solutes in aqueous solutions, students can observe the real-world application of the concepts like the van 't Hoff factor and molality, providing a concrete learning experience in chemical studies.