Problem 75 - Chemistry The Science in Context

Question

Which aqueous solution has the lowest freezing point: \(0.5 \mathrm{m}\) glucose, \(0.5 \mathrm{m} \mathrm{NaCl},\) or \(0.5 \mathrm{m} \mathrm{CaCl}_{2} ?\)

Step-by-Step Solution

Answer

Answer: The aqueous solution with the lowest freezing point is 0.5 m glucose.

Step 1: Recall the freezing point depression formula

The formula for freezing point depression is given by \(\Delta T_{f} = K_{f} \cdot i \cdot m\), where \(\Delta T_{f}\) is the freezing point depression, \(K_{f}\) is the cryoscopic constant for water, \(i\) is the van't Hoff factor, and \(m\) is the molality of the solute.

Step 2: Determine the van't Hoff factors for each solute

The van't Hoff factor is the number of particles formed by a solute when dissolved in solution. Since glucose is a non-electrolyte, it does not dissociate and thus has a van't Hoff factor of 1. Sodium chloride (\(\mathrm{NaCl}\)) dissociates into 2 ions (\(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\)), so it has a van't Hoff factor of 2. Calcium chloride (\(\mathrm{CaCl}_{2}\)) dissociates into 3 ions (\(\mathrm{Ca^{2+}}\) and \(2\mathrm{Cl^{-}}\)), so it has a van't Hoff factor of 3.

Step 3: Calculate the freezing point depression for each solution

Since all solutions have the same molality (\(0.5 \mathrm{m}\)) and cryoscopic constant \(K_{f}\) (for water), we can calculate the freezing point depression for each solution. For glucose: \(\Delta T_{f} = K_{f} \cdot 1 \cdot 0.5 \implies \Delta T_{f} = 0.5K_{f}\) For \(\mathrm{NaCl}\): \(\Delta T_{f} = K_{f} \cdot 2 \cdot 0.5 \implies \Delta T_{f} = 1K_{f}\) For \(\mathrm{CaCl}_{2}\): \(\Delta T_{f} = K_{f} \cdot 3 \cdot 0.5 \implies \Delta T_{f} = 1.5K_{f}\)

Step 4: Compare the freezing point depressions and determine the lowest

Comparing the calculated freezing point depressions, we have: Glucose: \(0.5K_{f}\) \(\mathrm{NaCl}\): \(1K_{f}\) \(\mathrm{CaCl}_{2}\): \(1.5K_{f}\) Since the freezing point depression is the lowest for glucose (\(0.5K_{f}\)), the aqueous solution with the lowest freezing point is \(0.5 \mathrm{m}\) glucose.

Key Concepts

van't Hoff factor

The van't Hoff factor, denoted as \(i\), is a key element in understanding how solutes affect the freezing point of a solution. It represents the number of particles a solute breaks into when dissolved in a solvent. The more parts the solute breaks into, the greater the impact on properties like freezing point depression.

Let's look at the examples from the exercise:

• Glucose: As a non-electrolyte, it doesn't dissociate in water, so its van't Hoff factor is 1.
• Sodium Chloride (\(\mathrm{NaCl}\)): Dissociates into 2 ions (\(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\)), with a van't Hoff factor of 2.
• Calcium Chloride (\(\mathrm{CaCl}_{2}\)): Breaks into 3 ions (\(\mathrm{Ca^{2+}}\) and 2\(\mathrm{Cl^{-}}\)), resulting in a van't Hoff factor of 3.

Understanding the van't Hoff factor helps predict how much the freezing point will lower, as the formula \(\Delta T_{f} = K_{f} \cdot i \cdot m\) depends significantly on this factor.

cryoscopic constant

The cryoscopic constant, \(K_{f}\), is a property of the solvent and measures how much the freezing point of the pure solvent decreases when a solute is added. Each solvent has its unique cryoscopic constant, which quantifies its sensitivity to the presence of solute particles.

For water, which is the most common solvent, the value of \(K_{f}\) is particularly important. When calculating the freezing point depression, this constant provides a critical link between the number of solute particles and the impact on the freezing point.

The formula used in the exercise, \(\Delta T_{f} = K_{f} \cdot i \cdot m\), reveals how \(K_{f}\) works together with the van't Hoff factor and molality to determine the new freezing point of the solution. Simply put, \(K_{f}\) helps gauge how much effect the solute has on lowering the freezing point of the solvent.

molality

Molality, represented by \(m\), is a measure of the concentration of a solute in a solution. It's different from molarity because it focuses on the moles of solute per kilogram of solvent, rather than per liter of solution.

This distinction makes molality especially useful in calculations involving temperature changes.

• Why Use Molality? - Unlike volume, mass is not affected by temperature, making molality a stable and reliable concentration measurement under varying conditions.
• Example: In the exercise, each solution has a molality of \(0.5 \mathrm{m}\). This means there are 0.5 moles of solute for every kilogram of water.

In the freezing point depression formula, \(\Delta T_{f} = K_{f} \cdot i \cdot m\), molality directly influences how much the freezing point will drop, working in tandem with the van't Hoff factor and cryoscopic constant.