Problem 46 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(1.22 \mathrm{~g}\) of benzoic acid is dissolved in \(100 \mathrm{~g}\) of acetone and \(100 \mathrm{~g}\) of benzene separately. Boiling point of the solution in acetone increases by \(0.17^{\circ} \mathrm{C}\), while that in the benzene increases by \(0.13^{\circ} \mathrm{C}\); \(K_{b}\) for acetone and benzene is \(1.7 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) and \(2.6 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) respectively. Find molecular weight of benzoic acid in two cases and justify your answer.

Step-by-Step Solution

Answer

The molecular weight of benzoic acid is 122 g/mol in acetone and 244 g/mol in benzene, indicating association in benzene.

Step 1: Understand the Problem

In this exercise, we are given that benzoic acid is dissolved in two different solvents, acetone and benzene, each with a given boiling point elevation. We are also provided with the ebullioscopic constants \(K_b\) for these solvents. Our goal is to determine the molecular weight of benzoic acid in each solvent by using the boiling point elevation formula.

Step 2: Identify the Formula

The boiling point elevation \(\Delta T_b\) can be calculated using the formula \(\Delta T_b = K_b \cdot m\), where \(m\) is the molality of the solution. Molality \(m\) is given by \(m = \frac{n_{solute}}{m_{solvent\, in\, kg}} = \frac{w}{M \cdot m_{solvent\, in\, kg}}\). Here, \(n_{solute}\) is the number of moles of solute, \(w\) is the weight of solute in grams, \(M\) is the molar mass of solute, and \(m_{solvent\, in\, kg}\) is the weight of the solvent in kilograms.

Step 3: Calculate Molality in Acetone

For the solution in acetone:\[\Delta T_b = 0.17 \, ^{\circ}C, \quad K_b = 1.7 \, K \, kg \, mol^{-1}\]\[m_{acetone} = \frac{100 \, g}{1000} = 0.1 \, kg\]Using the formula, we rearrange to find \(M\):\[M = \frac{w \cdot K_b}{\Delta T_b \cdot m_{solvent\, in\, kg}} = \frac{1.22 \cdot 1.7}{0.17 \cdot 0.1}\]Calculating gives:\[M \approx 122 \, g/mol\]

Step 4: Calculate Molality in Benzene

For the solution in benzene:\[\Delta T_b = 0.13 \, ^{\circ}C, \quad K_b = 2.6 \, K \, kg \, mol^{-1}\]\[m_{benzene} = \frac{100 \, g}{1000} = 0.1 \, kg\]Using the formula, we rearrange to find \(M\):\[M = \frac{w \cdot K_b}{\Delta T_b \cdot m_{solvent\, in\, kg}} = \frac{1.22 \cdot 2.6}{0.13 \cdot 0.1}\]Calculating gives:\[M \approx 244 \, g/mol\]

Step 5: Conclusion Analysis

The calculated molecular weight of benzoic acid is different in the two solvents: \(122 \, g/mol\) in acetone and \(244 \, g/mol\) in benzene. This discrepancy suggests that benzoic acid associates or forms dimers in benzene, thereby effectively doubling its apparent molecular weight.

Key Concepts

Molecular Weight Determination

Molecular weight, also known as molar mass, is a critical property of a compound, reflecting the mass of one mole of its molecules. To determine molecular weight using boiling point elevation (a colligative property depending only on the number of particles in a solution), we use the formula: \( \Delta T_b = K_b \cdot m \). Here, \( \Delta T_b \) represents the change in boiling point, \( K_b \) is the ebullioscopic constant specific to the solvent, and \( m \) denotes molality, the moles of solute per kilogram of solvent.

To find the molecular weight \( M \), we rearrange the equation:

• Express molality: \( m = \frac{w}{M \cdot m_{solvent\, in\, kg}} \)
• Substitute it back: \( M = \frac{w \cdot K_b}{\Delta T_b \cdot m_{solvent\, in\, kg}} \)

This calculation provides a direct method to derive molecular weights, especially when standard methods are not feasible. However, it relies heavily on the uniqueness of the solute's behavior in each solvent, which could lead to different results due to solvent interactions.

Ebullioscopic Constant

The ebullioscopic constant \( K_b \) is a distinct physical property of each solvent, analogous to its fingerprint in the realm of boiling point elevation. This constant communicates how much the boiling point of the solvent will be elevated per molal concentration of the solute.

For example, in our exercise, acetone and benzene have distinct ebullioscopic constants:

• Acetone: \( K_b = 1.7 \, K \, kg \, mol^{-1} \)
• Benzene: \( K_b = 2.6 \, K \, kg \, mol^{-1} \)

These values mean that benzene is more responsive to solute addition than acetone in terms of its boiling point rise for every mole of solute, which is a crucial factor in determining the solute's effective molecular weight.

In practical applications, \( K_b \) is used in conjunction with experimental boiling point changes to calculate solute properties like molecular weight and to infer potential solute association or dissociation within the solvent.

Solvent Interactions

Solvent interactions are at the heart of understanding the behavior of dissolved solutes. These interactions define how a solute behaves in solution, often altering its effective properties. In the case presented, benzoic acid behaves differently in acetone and benzene, leading to variations in its apparent molecular weight.

Here’s why:

• In acetone, the calculated molecular weight of benzoic acid is closer to its actual value (122 g/mol), suggesting minimal association or dissociation.
• In benzene, however, the molecular weight is calculated as 244 g/mol. This is double the acetone value, indicating the possibility of dimerization where two benzoic acid molecules pair up, effectively doubling the mass per mole.

Such phenomena occur due to hydrogen bonding, van der Waals forces, or other intermolecular forces that can be significant in certain solvent environments. Recognizing these interactions is crucial for accurately interpreting experiments and adjusting methodologies accordingly.