Problem 149 - A Complete Resource Book in Chemistry for JEE Main

Question

For an aqueous solution, freezing point is \(-0.186^{\circ} \mathrm{C}\). Elevation of the boiling point of the same solution is $$ \left(\mathrm{K}_{\ell}=1.86^{\circ} \mathrm{mol}^{-1} \mathrm{~kg} \text { and } \mathrm{K}_{\mathrm{b}}=0.512^{\circ} \mathrm{mol}^{-1} \mathrm{~kg}\right) $$ (a) \(0.186^{\circ}\) (b) \(0.0512^{\circ}\) (c) \(1.86^{\circ}\) (d) \(5.12^{\circ}\)

Step-by-Step Solution

Answer

The boiling point elevation is 0.0512°C, or option (b).

Step 1: Understanding the Freezing Point Depression Formula

The formula for freezing point depression is given as: \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f \) is the decrease in freezing point, \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution. In this problem, \( \Delta T_f = 0.186^{\circ} \).

Step 2: Calculate Molality using Freezing Point Depression

We have \( \Delta T_f = 0.186 \) and \( K_f = 1.86 \). Rearranging the formula: \( m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} \). Calculate this to find the molality of the solution.

Step 3: Calculate Molality

\( m = \frac{0.186}{1.86} = 0.1 \text{ mol/kg} \). So, the molality of the solution is 0.1 mol/kg.

Step 4: Use Molality to Find Boiling Point Elevation

The formula for boiling point elevation is \( \Delta T_b = K_b \cdot m \), where \( \Delta T_b \) is the increase in boiling point, \( K_b \) is the boiling point elevation constant, and \( m \) is the molality of the solution, which is 0.1 mol/kg.

Step 5: Calculate Boiling Point Elevation

Plug in \( K_b = 0.512 \) and \( m = 0.1 \) into the formula: \( \Delta T_b = 0.512 \cdot 0.1 = 0.0512 \).

Step 6: Determine the Correct Answer

The boiling point elevation of the solution is \( 0.0512^{\circ} \). Thus, the correct answer is (b) \( 0.0512^{\circ} \).

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property of solutions. This occurs when a solute is added to a solvent, effectively lowering the freezing point of the resulting mixture. At a molecular level, the solute particles disrupt the orderly structure of the solvent, making it harder for the solvent to form a solid structure. As a result, the solution needs to be cooled to an even lower temperature than the pure solvent to freeze.

The formula to calculate freezing point depression is \( \Delta T_f = K_f \cdot m \). Here, \( \Delta T_f \) represents the freezing point depression, \( K_f \) is the freezing point depression constant (unique to each solvent), and \( m \) is the molality of the solution. Let's say you have an aqueous solution where the freezing point is found to be \(-0.186^{\circ} \mathrm{C}\). To figure out the change in freezing point, we need the value of \( K_f \) for water, which is \(1.86^{\circ} \mathrm{C} \cdot \mathrm{mol}^{-1} \mathrm{~kg}\).

By rearranging the formula and solving for molality: \( m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} \). This calculation is crucial in determining the concentration of dissolved particles in the solution, providing insight into various scientific and industrial processes.

Boiling Point Elevation

Boiling point elevation is another key colligative property and happens when a non-volatile solute is added to a solvent. The addition of solute particles raises the boiling point of the solution compared to the pure solvent. This is because additional energy is needed to overcome the attraction between the solvent and the solute particles to transition to the gas phase.

The formula for boiling point elevation is \( \Delta T_b = K_b \cdot m \), with \( \Delta T_b \) being the increase in boiling point, \( K_b \) the boiling point elevation constant, and \( m \) again represents the molality. In our given solution, with a known molality of \(0.1 \text{ mol/kg}\) and a \( K_b \) value of \(0.512^{\circ} \mathrm{C} \cdot \mathrm{mol}^{-1} \mathrm{~kg}\), we can calculate the elevation as follows: \( \Delta T_b = 0.512 \times 0.1 = 0.0512^{\circ} \).

Understanding this concept is important in everyday phenomena as well as industrial applications where precise boiling point changes affect product outcomes.

Molality

Molality is a concentration unit used to express the amount of solute present in a solution. It is different from molarity as it is defined as the moles of solute per kilogram of solvent, rather than per liter of solution. This makes molality particularly useful in colligative property calculations where temperature changes can affect the volume but not the mass.

Molality is calculated using the formula \( m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \). In our previous calculations, we found the molality of a solution by using colligative properties such as freezing point depression. Specifically, in our example, the molality was calculated to be \(0.1 \text{ mol/kg}\).

Some benefits of using molality include its constancy with temperature changes, providing reliable data for calculations involving boiling point elevation and freezing point depression. These properties make molality an essential concept in chemical thermodynamics and various practical applications where temperature fluctuations are common.