Problem 76 - Chemistry: The Central Science

Question

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

The freezing point of the aqueous solution is \(-18.186^{\circ}\mathrm{C}\).

Step 1: Calculate boiling point elevation

Since we know the boiling point of the aqueous solution (\(105.0^{\circ}\mathrm{C}\)) and the normal boiling point of water (\(100.0^{\circ}\mathrm{C}\)), we can calculate the boiling point elevation: \(\Delta T_b = T_{b, solution} - T_{b, pure \, water}\) \(\Delta T_b = 105.0 - 100.0\) \(\Delta T_b = 5.0^{\circ} \mathrm{C}\)

Step 2: Calculate the molality of the solution

Use the boiling point elevation formula to calculate the molality of the solution: \(\Delta T_b = K_b \cdot m\) \(5.0 = 0.512 \cdot m\) \(m = \frac{5.0}{0.512}\) \(m = 9.766\, \mathrm{mol/kg}\)

Step 3: Calculate the freezing point depression

Use the molality and the cryoscopic constant of water to calculate the freezing point depression: \(\Delta T_f = K_f \cdot m\) \(\Delta T_f = 1.86 \cdot 9.766\) \(\Delta T_f = 18.186^{\circ} \mathrm{C}\)

Step 4: Calculate the freezing point of the aqueous solution

Subtract the freezing point depression from the normal freezing point of water to determine the freezing point of the aqueous solution: \(T_{f, solution} = T_{f, pure \, water} - \Delta T_f\) \(T_{f, solution} = 0.0^{\circ}\mathrm{C} - 18.186^{\circ}\mathrm{C}\) \(T_{f, solution} = -18.186^{\circ}\mathrm{C}\) So the freezing point of the aqueous solution is \(-18.186^{\circ}\mathrm{C}\).

Key Concepts

Boiling Point Elevation

Boiling point elevation is a colligative property, meaning it depends on the number of solute particles in a solution and not on their identity. When you dissolve a nonvolatile solute into a solvent, the boiling point of the resulting solution is higher than that of the pure solvent. This happens because the solute particles disrupt the formation of vapor at the surface of the liquid, requiring more heat energy to reach the point of boiling.

This relationship can be quantified using the formula for boiling point elevation:

• \[\Delta T_b = K_b \cdot m\]
• Where \(\Delta T_b\) is the change in boiling point, \(K_b\) is the ebullioscopic constant specific to the solvent (for water it's 0.512 \(\text{°C/m}\)), and \(m\) is the molality of the solution.

In our exercise, the boiling point of the aqueous solution was observed at 105.0 °C, which is 5.0 °C above the normal boiling point of water at 100.0 °C. By using the formula, we calculated the molality, which helps to further determine changes in the freezing point, illustrating the connection between boiling point elevation and freezing point depression.

Freezing Point Depression

Freezing point depression, another important colligative property, occurs when the freezing point of a solution is lower than that of the pure solvent. Just like boiling point elevation, this effect is due to the presence of solute particles. These particles disrupt the orderly crystal formation of the solid phase, requiring a lower temperature to freeze the solution.

To model this concept, we use the formula:

• \[\Delta T_f = K_f \cdot m\]
• Here, \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant (for water it's 1.86 \(\text{°C/m}\)), and \(m\) is the molality of the solution.

In the given problem, after determining the molality from boiling point elevation, we used it to find the freezing point depression. The aqueous solution's freezing point was calculated to drop by 18.186 °C, leading to a new freezing point of -18.186 °C. This showcases how changes in solute concentration can significantly alter the physical properties of a solution.

Molality

Molality is a measure of the concentration of a solute in a solution, expressed in moles of solute per kilogram of solvent. Unlike molarity, molality remains unaffected by changes in temperature because it depends on the mass of the solvent, not the volume of the entire solution.

To find molality, you use the formula:

• \[m = \frac{\text{moles of solute}}{\text{kg of solvent}}\]

In the context of colligative properties like boiling point elevation and freezing point depression, molality is crucial because these effects depend on solute concentration.
In our exercise, the molality was calculated using the boiling point elevation data. It was found to be 9.766 mol/kg, indicating a relatively concentrated solution. This concentration directly affected both the increase in boiling point and the decrease in freezing point, underlining how molality is a key factor in manipulating the colligative properties of solutions.