Question

In each pair, identify the solution that will have a lower freezing point. Explain. a. \(1.0\) mole of glycerol (nonelectrolyte) or \(2.0\) moles of ethylene glycol (nonelectrolyte) each in \(1.0 \mathrm{~L}\) of water. b. \(0.50\) mole of \(\mathrm{KCl}\) (strong electrolyte) or \(0.50\) mole of \(\mathrm{MgCl}_{2}\) (strong electrolyte) each in \(2.0 \mathrm{~L}\) of water.

Short answer

a. Ethylene glycol solution. b. MgCl_2 solution.

Step-by-step solution

- Understanding Freezing Point Depression

The freezing point decreases when a solute is added to a solvent. This phenomenon is called freezing point depression. It is calculated using the formula \[ \text{Freezing Point Depression} (\triangle T_f) = iK_f m \] where - \(i\) is the van 't Hoff factor (number of particles the solute breaks into) - \(K_f\) is the freezing point depression constant of the solvent - \(m\) is the molality of the solution.

- Calculate for Glycerol and Ethylene Glycol (a)

Both glycerol and ethylene glycol are nonelectrolytes, meaning \(i\) = 1 for both. For glycerol: - Moles: 1.0 mole - Volume of water: 1.0 L Therefore, molality \( m = \frac{{1.0 \text{ mole}}}{{1.0 \text{ kg}}} = 1.0 \text{ m} \) For ethylene glycol: - Moles: 2.0 moles - Volume of water: 1.0 L Therefore, molality \( m = \frac{{2.0 \text{ moles}}}{{1.0 \text{ kg}}} = 2.0 \text{ m} \) Comparing the two values, the solution with ethylene glycol has a higher molality.

- Calculate for KCl and MgCl_2 (b)

Both KCl and MgCl_2 are strong electrolytes. For KCl: - Van 't Hoff factor \( i \) = 2 (K^+ and Cl^-) - Moles: 0.50 mole - Volume of water: 2.0 L Therefore, molality \( m = \frac{{0.50 \text{ mole}}}{{2.0 \text{ kg}}} = 0.25 \text{ m} \) For MgCl_2: - Van 't Hoff factor \( i \) = 3 (Mg^{2+} and 2Cl^-) - Moles: 0.50 mole - Volume of water: 2.0 L Therefore, molality \( m = \frac{{0.50 \text{ mole}}}{{2.0 \text{ kg}}} = 0.25 \text{ m} \) Since the freezing point depression depends on \( i \times m \), for KCl, \( \triangle T_f = 2 \times 0.25 = 0.5 \text{ } \). For MgCl_2, \( \triangle T_f = 3 \times 0.25 = 0.75 \text{ } \). The solution with MgCl_2 will have a lower freezing point.

Key concepts

van 't Hoff factor

The van 't Hoff factor, represented as \( i \), is a crucial element in understanding freezing point depression. It indicates the number of particles a solute splits into when it dissolves in a solvent. For instance, nonelectrolytes like glycerol and ethylene glycol do not dissociate in solution; thus, their van 't Hoff factor \( i \) is 1. On the other hand, electrolytes dissociate into multiple ions. For example, potassium chloride (KCl) dissociates into two ions (K\(^+\) and Cl\(^-\)), giving it an \( i \) of 2. Magnesium chloride (MgCl\(_2\)) dissociates into three ions (Mg\(^{2+}\) and 2Cl\(^-\)), resulting in an \( i \) of 3. The van 't Hoff factor plays a significant role in calculating the effect of a solute on the freezing point of a solution.

nonelectrolyte

Nonelectrolytes are substances that do not dissociate into ions when dissolved in a solvent. This means they don't conduct electricity in solution. Common examples include organic compounds like glycerol and ethylene glycol. When considering freezing point depression, nonelectrolytes have a van 't Hoff factor \( i \) of 1 because they do not break apart. In the given exercise, both glycerol and ethylene glycol are nonelectrolytes. Hence, their effect on freezing point depression is determined solely by their molality.

molality

Molality (\( m \)) is a measure of the concentration of a solution. It is defined as the number of moles of solute per kilogram of solvent. Mathematically, it is expressed as: \[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \] In the context of freezing point depression, molality is crucial because it appears in the formula \( \Delta T_f = iK_f m \). As seen in the exercise, the molality of glycerol is calculated with 1.0 mole in 1.0 kg of water, resulting in 1.0 \( m \), while the molality of ethylene glycol is 2.0 moles in 1.0 kg of water, equating to 2.0 \( m \). Higher molality contributes to a greater depression of the freezing point.

electrolyte

Electrolytes are substances that dissociate into ions when dissolved in water, making the solution capable of conducting electricity. They can be strong or weak electrolytes. In this exercise, potassium chloride (KCl) and magnesium chloride (MgCl\(_2\)) are strong electrolytes, meaning they completely dissociate in water. KCl dissociates into two ions, so it has an \( i \) of 2. MgCl\(_2\) dissociates into three ions, giving it an \( i \) of 3. The more ions an electrolyte produces, the greater its effect on the freezing point depression, as demonstrated in the calculations.

solute

A solute is the substance that is dissolved in a solvent to form a solution. In the context of freezing point depression, the solute is the key factor in determining how much the freezing point of the solvent will be lowered. For example, in the given exercise, glycerol and ethylene glycol are solutes in water (solvent). The nature of the solute, whether it is an electrolyte or nonelectrolyte, and its quantity (molality) impact the extent of freezing point depression. More solute particles lead to a greater disruption of the solvent's ability to form a solid structure, thus lowering the freezing point of the solution.