Problem 18 - Chemistry in 30 days

Question

Which of the following aqueous solutions will have the lowest freezing point? (a) \(0.10 \mathrm{M} \mathrm{KCl}\) (b) \(0.10 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (c) \(0.10 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (d) \(0.10 \mathrm{M} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{1}\)

Step-by-Step Solution

Answer

The solution with \(0.10 \mathrm{M} \mathrm{Al}_2(\mathrm{SO}_4)_3\) will have the lowest freezing point.

Step 1: Understanding Freezing Point Depression

Freezing point depression depends on the number of solute particles in a solution. It can be calculated using the formula \[\Delta T_f = i \cdot K_f \cdot m\] where \( \Delta T_f \) is the freezing point depression, \( i \) is the van 't Hoff factor (number of particles the solute dissociates into), \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution.

Step 2: Calculate the van 't Hoff Factors

For each compound, determine \(i\): - \(\mathrm{KCl}\) dissociates into 2 ions (\(\mathrm{K^+} + \mathrm{Cl^-}\)), so \(i = 2\).- \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) dissociates into 5 ions (2 \(\mathrm{Al}^{3+}\) and 3 \(\mathrm{SO}_4^{2-}\)), so \(i = 5\).- \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\) (glucose) does not dissociate, so \(i = 1\).- \(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\) (sucrose) does not dissociate, so \(i = 1\).

Step 3: Determine the Effect on Freezing Point

The greater the value of \( i \), the more the freezing point will be depressed because more particles disrupt the freezing process. \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) has the highest \(i\) value, therefore, its solution will have the lowest freezing point.

Key Concepts

Van 't Hoff Factor

To understand the concept of freezing point depression, we first need to discuss the van 't Hoff factor, denoted as \( i \). This factor represents the number of particles a solute dissociates into when it is dissolved in a solvent. For example:

• When \( \mathrm{KCl} \) is dissolved in water, it dissociates into two ions: \( \mathrm{K^+} \) and \( \mathrm{Cl^-} \). Therefore, the van 't Hoff factor, \( i \), for \( \mathrm{KCl} \) is \( 2 \).
• \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \) dissociates into five ions: two \( \mathrm{Al^{3+}} \) ions and three \( \mathrm{SO}_4^{2-} \) ions, resulting in an \( i \) value of \( 5 \).
• Non-ionic compounds like glucose and sucrose do not dissociate in water, so their \( i \) value is \( 1 \).

The larger \( i \) is, the more the freezing point of the solution will be depressed because more particles interfere with the formation of a solid structure in the solution. Thus, the van 't Hoff factor is crucial in predicting how a solute will affect the temperature at which a solution freezes.

Aqueous Solutions

In the context of chemistry, an aqueous solution is one where water acts as the solvent. It is a common medium for chemical reactions and processes since many substances dissolve well in water. When a substance dissolves in water, it breaks up into its constituent ions or stays intact, depending on its molecular structure.
Here's how it works:

• Ionic compounds, like \( \mathrm{KCl} \) and \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \), dissociate into ions. This increases the number of particles in the solution and affects properties like boiling and freezing points.
• In contrast, molecular compounds such as glucose (\( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \)) and sucrose (\( \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \)) do not dissociate in water.

Being an aqueous solution means a substance is interacting with water molecules, which affects its behavior during physical processes such as freezing and boiling. The more solute particles present, the more disruption there is to these processes.

Molecular Dissociation

Molecular dissociation refers to the breaking apart of a compound into individual ions or molecules when dissolved in a solvent like water. This is especially important in determining how a solute alters the freezing point of a solution. For ionic compounds:

• Compounds like \( \mathrm{KCl} \) readily dissociate into their constituent ions when they dissolve in water, \( \mathrm{K^+} \) and \( \mathrm{Cl^-} \).
• \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \) dissociates into five ions: two \( \mathrm{Al^{3+}} \) ions and three \( \mathrm{SO}_4^{2-} \) ions, showing a greater degree of molecular dissociation compared to \( \mathrm{KCl} \).

Non-dissociating molecular compounds:

• Unlike ionic compounds, organic compounds like glucose and sucrose do not separate into ions when they dissolve. Instead, they dissolve as whole molecules and have less effect on freezing point depression.

Understanding how molecular dissociation works is key to predicting the changes in physical properties of solutions, especially when considering the freezing or boiling points in these systems.