Problem 45 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The boiling point and freezing point of a solvent 'A' are \(90.0^{\circ} \mathrm{C}\) and \(3.5^{\circ} \mathrm{C}\), respectively. \(K_{\mathrm{f}}\) and \(K_{\mathrm{b}}\) values of the solvent are \(17.5\) and \(5.0 \mathrm{~K}-\mathrm{kg} / \mathrm{mol}\), respectively. What is the boiling point of a solution of 'B' (non-volatile, nonelectrolyte solute) in 'A', if the solution freezes at \(2.8^{\circ} \mathrm{C} ?\) (a) \(90.0^{\circ} \mathrm{C}\) (b) \(89.8^{\circ} \mathrm{C}\) (c) \(90.2^{\circ} \mathrm{C}\) (d) \(90.7^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The boiling point of the solution is 90.2°C.

Step 1: Determine freezing point depression

The freezing point depression (ΔTf) is calculated by the difference between the freezing point of the pure solvent (A) and the freezing point of the solution. ΔTf = (Freezing point of pure solvent A) - (Freezing point of solution) = 3.5°C - 2.8°C.

Step 2: Calculate the molality of the solution

Using the freezing point depression constant (Kf) and the calculated ΔTf, find the molality (m) of the solution using the formula ΔTf = Kf * m. Rearrange to find m = ΔTf / Kf.

Step 3: Determine boiling point elevation

The boiling point elevation (ΔTb) is calculated using the same molality (m) obtained from step 2 and the boiling point elevation constant (Kb) for solvent A. ΔTb = Kb * m.

Step 4: Calculate the boiling point of the solution

To find the solution's boiling point, add the calculated ΔTb to the boiling point of the pure solvent (A). The solution's boiling point = (Boiling point of pure solvent A) + ΔTb.

Key Concepts

Freezing Point Depression

Freezing point depression is one of the colligative properties of solutions, which means it depends on the number of solute particles in a solvent and not on the type of particles. When a solute is dissolved in a solvent, it disrupts the formation of the solvent's solid phase, making it more difficult to freeze, thereby lowering its freezing temperature. This effect is quantified as the difference in temperatures between the pure solvent's freezing point and the solution's freezing point, denoted as \( \Delta T_f \).

The mathematical formula to calculate the freezing point depression is \( \Delta T_f = K_f \times m \) where \( K_f \) is the freezing point depression constant and \( m \) is the molality of the solution. In the context of our exercise, the freezing point of pure solvent 'A' is \( 3.5^\circ \mathrm{C} \) and it decreases to \( 2.8^\circ \mathrm{C} \) upon addition of solute 'B'. This shows that the freezing point has been depressed by \( 0.7^\circ \mathrm{C} \) due to the solute.

Boiling Point Elevation

Boiling point elevation is another colligative property of a solution which occurs when a solute is added to a solvent. Similar to how freezing point is lowered, the boiling point of a solvent is raised when solute particles are introduced. This is because the solute particles disrupt the evaporative process, requiring more energy (in the form of heat) to bring the solvent to a boil.

The increase in the boiling point, known as boiling point elevation \( \Delta T_b \), can be calculated via \( \Delta T_b = K_b \times m \) where \( K_b \) is the boiling point elevation constant and \( m \) is the molality of the solution. For our exercise, we calculate the molality from the freezing point depression and apply it to find the boiling point elevation of solvent 'A' with solute 'B' dissolved in it.

Molality

Molality is a measure of the concentration of a solute in a solution, representing the moles of solute per kilogram of solvent, and is expressed as moles/kg. One of the reasons molality is used in colligative properties calculations, such as freezing point depression and boiling point elevation, is because it does not change with changes in temperature or pressure, unlike molarity which depends on the volume of solution.

The formula to calculate molality is \( m = \frac{mol\ of\ solute}{kg\ of\ solvent} \) and this value remains constant for a given solution, allowing it to be used across different calculations. In our exercise scenario, once the molality is determined from the freezing point depression, the same value is used to calculate the boiling point elevation.

Nonelectrolyte Solutions

Nonelectrolyte solutions are composed of a solvent and a nonelectrolyte solute, which does not dissociate into ions in solution. Examples of nonelectrolytes include sugar and urea. Because nonelectrolytes do not break into ions, each molecule will remain intact in the solution. This has important implications for colligative properties because the effect observed (be it freezing point depression or boiling point elevation) is directly proportional to the number of particles dissolved.

In nonelectrolyte solutions like the one mentioned in our exercise with solute 'B' and solvent 'A', the freezing point depression and the boiling point elevation can be calculated without concern for ion dissociation, making these calculations straightforward as only the molality and the respective constants (\( K_f \) and \( K_b \) in this case) need to be considered.