Question

Arrange the following aqueous solutions in order of decreasing freezing point. (The last three are all assumed to dissociate completely into ions in water.) (a) \(0.20 \mathrm{m}\) ethylene glycol (nonvolatile, nonelectrolyte) (b) \(0.12 \mathrm{m} \mathrm{K}_{2} \mathrm{SO}_{4}\) (c) \(0.10 \mathrm{mgCl}_{2}\) (d) \(0.12 \mathrm{m} \mathrm{KBr}\)

Short answer

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

Step-by-step solution

Understanding Freezing Point Depression

The freezing point of a solution is lower than that of the pure solvent, and this decrease (freezing point depression) is dependent on the molality and the number of particles (ions or molecules) in solution. The formula is \[\Delta T_f = i \cdot K_f \cdot m\]where \(\Delta T_f\) is the freezing point depression, \(i\) is the van't Hoff factor (number of particles per formula unit), \(K_f\) is the freezing point depression constant, and \(m\) is the molality of the solution.

Calculating the Van't Hoff Factor

For each solution:- Ethylene glycol: Non-electrolyte, \(i = 1\).- \(\mathrm{K}_2\mathrm{SO}_4\): Dissociates into 3 ions (2K\(^+\) and 1 \(\mathrm{SO}_4^{2-}\)), \(i = 3\).- \(\mathrm{MgCl}_2\): Dissociates into 3 ions (1Mg\(^{2+}\) and 2Cl\(^-\)), \(i = 3\).- \(\mathrm{KBr}\): Dissociates into 2 ions (K\(^+\) and Br\(^-\)), \(i = 2\).

Determining Freezing Point Depression for Each

To compare, calculate the product \(i \cdot m\) for each solution:- Ethylene glycol: \(1 \cdot 0.20 = 0.20\).- \(\mathrm{K}_2\mathrm{SO}_4\): \(3 \cdot 0.12 = 0.36\).- \(\mathrm{MgCl}_2\): \(3 \cdot 0.10 = 0.30\).- \(\mathrm{KBr}\): \(2 \cdot 0.12 = 0.24\).

Arranging Solutions by Decreasing Freezing Point

The higher the value of \(i \cdot m\), the greater the freezing point depression. Therefore, the solutions in order of decreasing freezing point (less depression to more) are:1. Ethylene glycol (\(0.20\))2. \(\mathrm{KBr}\) (\(0.24\))3. \(\mathrm{MgCl}_2\) (\(0.30\))4. \(\mathrm{K}_2\mathrm{SO}_4\) (\(0.36\))

Key concepts

Van't Hoff Factor

The van't Hoff factor, often denoted by \(i\), plays a crucial role in determining how a solute affects the properties of a solution. This factor essentially reflects the number of particles that a solute breaks into when dissolved in a solvent. For non-electrolytes, such as ethylene glycol, the van't Hoff factor is simply 1, as these substances do not dissociate into ions.

Electrolytes, on the other hand, increase the number of particles when dissolved. For example, \(\mathrm{K}_2\mathrm{SO}_4\) dissociates into three ions: two \(\mathrm{K}^+\) ions and one \(\mathrm{SO}_4^{2-}\) ion. Thus, its van't Hoff factor is 3. Similarly, \(\mathrm{MgCl}_2\) dissociates into three ions (one \(\mathrm{Mg}^{2+}\) ion and two \(\mathrm{Cl}^-\) ions), giving it a van't Hoff factor of 3.

The factor for \(\mathrm{KBr}\) is 2, as it separates into one \(\mathrm{K}^+\) ion and one \(\mathrm{Br}^-\) ion.

• Understanding the van't Hoff factor is key to predicting colligative properties like boiling point elevation and freezing point depression.

Molality

Molality is a measure of the concentration of a solute in a solution, calculated as the number of moles of solute per kilogram of solvent. It is represented by the symbol \(m\). Unlike molarity, which is affected by temperature, molality remains constant because it depends purely on the mass, not the volume, of the solvent.

When calculating freezing point depression, molality provides a key value, as it helps determine how much the presence of a solute will lower the freezing point of a solution. In the context of our solutions:

• Ethylene glycol has a molality of \(0.20 \ m\)
• \(\mathrm{K}_2\mathrm{SO}_4\) has a molality of \(0.12 \ m\)
• \(\mathrm{MgCl}_2\) carries a molality of \(0.10 \ m\)
• \(\mathrm{KBr}\) is \(0.12 \ m\) as well.

Molality is particularly useful in laboratory settings where precise measurements are essential, especially in calculations like those for freezing point depression.

Aqueous Solutions

Aqueous solutions are solutions in which water acts as the solvent. Water, known for its excellent solvent properties, can dissolve a wide range of substances, whether they are ionic or molecular.

The substances that dissolve in water take advantage of water's polarity. The positive and negative charges within water molecules help separate ions or interact with polar molecules, efficiently dissolving them.

In the context of our problem, all mentioned solutions, including ethylene glycol, \(\mathrm{K}_2\mathrm{SO}_4\), \(\mathrm{MgCl}_2\), and \(\mathrm{KBr}\), are aqueous. This means that:

• The solvent is always water.
• The behavior of dissolved solutes, be they electrolytes or non-electrolytes, is consistent in aqueous solutions.
• Aqueous solutions make use of water’s high dielectric constant, making it easier to dissociate ionic compounds into ions.

Electrolytes and Nonelectrolytes

In solutions, solutes are generally categorized as electrolytes or nonelectrolytes, based on their ability to conduct electricity.

Electrolytes are substances that dissociate into ions when dissolved in a solvent, enhancing the solution's electrical conductivity. Examples include \(\mathrm{K}_2\mathrm{SO}_4\), \(\mathrm{MgCl}_2\), and \(\mathrm{KBr}\). When these compounds dissolve:

• They release multiple ions, contributing to higher van't Hoff factors.
• They significantly affect the colligative properties, such as lowering the freezing point.

Nonelectrolytes, like ethylene glycol, do not dissociate into ions in solution, which means they do not conduct electricity or create as pronounced an effect on freezing point depression. Because they remain as molecules, their impact on properties like freezing point is generally weaker compared to electrolytes. Understanding these distinctions is fundamental when analyzing the behavior of different aqueous solutions.