Question

The freezing point constant of toluene is \(3.33^{\circ} \mathrm{C}\) per mole per \(1000 \mathrm{~g}\). Calculate the freezing point of a solution prepared by dissolving \(0.4\) mole of solute in \(500 \mathrm{~g}\) of toluene. The freezing point of toluene is \(-95.0^{\circ} \mathrm{C}\).

Short answer

The freezing point of the solution prepared by dissolving 0.4 mole of solute in 500 g of toluene is -97.664°C.

Step-by-step solution

Identify the given values

Here, we are given: 1. Freezing point constant of toluene (Kf) = 3.33°C per mole per 1000 g 2. Amount of solute (n) = 0.4 mole 3. Mass of toluene (m) = 500 g 4. Freezing point of pure toluene (Tf_tol) = -95.0°C

Calculate the molality of the solute

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Thus, to calculate the molality of the solute, we can use the formula: \[m = \frac{n}{M}\] where n is the number of moles of solute (0.4 mole) and M is the mass of the solvent (toluene) in kg (0.5 kg, since we have 500 g of toluene). Therefore, the molality of the solute is: \[m = \frac{0.4 \: \text{mole}}{0.5 \: \text{kg}} = 0.8 \: \text{mol/kg}\]

Calculate the freezing point depression

The freezing point depression (ΔTf) can be calculated using the formula: \[ΔTf = Kf \times m\] where Kf is the freezing point constant of toluene (3.33°C per mole per 1000 g) and m is the molality of the solute (0.8 mol/kg). Thus, the freezing point depression is: \[ΔTf = 3.33 \: \text{°C / mol per 1000 g} \times 0.8 \: \text{mol/kg} = 2.664 °C\]

Calculate the final freezing point of the solution

Now that we have the freezing point depression (ΔTf), we can find the freezing point of the solution by subtracting ΔTf from the freezing point of pure toluene (Tf_tol): \[Tf_{sol} = Tf_{tol} - ΔTf\] \[Tf_{sol} = -95.0 °C - 2.664 °C = -97.664 °C\] So, the freezing point of the solution is -97.664°C.

Key concepts

Colligative Properties

Colligative properties are a collection of properties of solutions that depend only on the number of solute particles present, regardless of their nature. What makes colligative properties important is their ability to predict how adding a solute will affect the physical state of a solvent.

Freezing point depression, one of these colligative properties, helps us understand why throwing salt on icy sidewalks melts the ice. In the context of our exercise, the freezing point of toluene lowers when a solute is dissolved in it. This is because solute particles disrupt the formation of the solid phase, requiring the temperature to be lower for solidification to happen. For students, connecting real-world examples, like the salt-ice scenario, can help cement the concept of freezing point depression and its practical applications.

Molality Calculation

Molality is a measure of the concentration of a solution, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is dependent on the volume of the solution, molality is especially useful because it remains unaffected by changes in temperature and pressure.

To calculate molality, as shown in the exercise's solution, you use the formula \[m = \frac{n}{M}\], where \(m\) is the molality, \(n\) is the number of moles of the solute, and \(M\) is the mass of the solvent in kilograms. By keeping the steps to calculate molality distinct and straightforward, students can easily apply the formula to a variety of problems involving different substances and conditions, enhancing their problem-solving skills.

Solute-Solvent Interactions

In a solution, solute-solvent interactions play a critical role in determining the properties of the solution. When a solute dissolves, it disrupts the uniform structure of the solvent, and new interactions between solute and solvent particles occur. These interactions include forces like hydrogen bonding, dipole-dipole interactions, and dispersion forces.

Understanding solute-solvent interactions can explain why certain substances dissolve more readily in a particular solvent or why the solution process affects the freezing and boiling points of the solvent, which are both relevant in the exercise given. Exploring these interactions makes it easier for students to grasp why the presence of a solute changes the physical properties of a solution, such as vapor pressure, boiling point, and as mentioned in our exercise, the freezing point.