Problem 7 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The freezing point of benzene decreases by \(0.45^{\circ} \mathrm{C}\) when \(0.2 \mathrm{~g}\) of acetic acid is added to \(20 \mathrm{~g}\) of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : \(\left(K_{f}\right.\) for benzene \(\left.=5.12 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\) (a) \(64.6 \%\) (b) \(80.4 \%\) (c) \(74.6 \%\) (d) \(94.6 \%\)

Step-by-Step Solution

Answer

The percentage association of acetic acid in benzene is 94.6%, matching option (d).

Step 1: Understanding the Freezing Point Depression

The freezing point depression \[ \Delta T_f = i \cdot K_f \cdot m \] where \(i\) is the van’t Hoff factor, \(K_f\) is the cryoscopic constant, and \(m\) is the molality of the solution.

Step 2: Calculate Molality

The molality \(m\) is given by \[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]First, we calculate the moles of acetic acid:\[ \text{Moles of acetic acid} = \frac{0.2}{60} = 0.00333 \text{ mol} \]Now, calculate the molality:\[ m = \frac{0.00333}{0.020} = 0.1665 \text{ mol/kg} \]

Step 3: Determine the Van 't Hoff Factor (i)

Given \( \Delta T_f = 0.45^{\circ} \text{C} \), \( K_f = 5.12 \), and \( m = 0.1665 \), we rearrange the freezing point formula: \[ i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.45}{5.12 \times 0.1665} \approx 0.527 \]

Step 4: Relate Van 't Hoff Factor to Association

The van't Hoff factor for a dimer of acetic acid is \( i = \frac{1}{2} + \frac{x}{2} \), where \(x\) is the fraction of solute that remains unassociated.Given that \( i = 0.527 \), solve:\[ 0.527 = \frac{1 + x}{2} \rightarrow x = 2(0.527) - 1 = 0.054 \]

Step 5: Calculate Percentage Association

Now calculate the percentage association:\[ \text{Percentage Association} = (1 - x) \times 100 = (1 - 0.054) \times 100 \approx 94.6\% \]

Step 6: Verify the Options

Comparing with given options, \(94.6\%\) matches option (d).

Key Concepts

Van't Hoff Factor

The van't Hoff factor, symbolized as \( i \), plays a crucial role in understanding colligative properties, such as freezing point depression. It represents the number of particles the solute forms in a solution. For example, when acetic acid is added to benzene, it forms a dimer. This dimerization, or pairing, affects how the solute influences the solution's properties.

It is important to understand that \( i \) changes based on the solute's behavior in the solution:

• For non-associating and non-dissociating solutes, \( i \) equals 1.
• For fully dissociating solutes, \( i \) equals the number of particles formed.

In the case of associating solutes like acetic acid dimerizing in benzene, \( i \) becomes a fractional value, reflecting the dimer formation. In this exercise, the calculated \( i \) = 0.527 indicates significant association, influencing the degree to which the freezing point is lowered.

Understanding \( i \) helps us relate molecular behaviors to macroscopic properties, bridging molecular interactions with observable changes in solutions.

Molality

Molality, often denoted as \( m \), is another important concept in the study of colligative properties. Unlike molarity, molality is temperature-independent because it is calculated based on the mass of the solvent, rather than its volume. This makes molality particularly useful in scenarios involving temperature changes, like freezing point depression.

To calculate molality, use the formula: \[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \] This formula highlights two critical components: the moles of the solute and the mass of the solvent.

In the given exercise, the moles of acetic acid are calculated by dividing the mass of the solute by its molar mass, yielding 0.00333 mol. Then, dividing this by the 0.020 kg of benzene (the solvent) results in a molality of 0.1665 mol/kg.

Understanding molality is key to accurately determining how solutes affect the properties of a solution, especially when evaluating colligative properties like freezing point depression, boiling point elevation, and osmotic pressure.

Percentage Association

Percentage association is a concept that helps measure how much of a solute associates or combines in a solution. In the context of acetic acid in benzene, it focuses on how much acetic acid forms dimers.

The percentage association reflects the portion of solute molecules that participate in association reactions, compared to those that remain as individual molecules. Here's the formula for percentage association: \[ \text{Percentage Association} = (1 - x) \times 100 \] Where \( x \) is the fraction of unassociated molecules.

In the exercise, solving \( x \) from the van’t Hoff factor equation as 0.054 indicates that only 5.4% of the acetic acid molecules do not form dimers. Converting this into percentage association, we find that approximately 94.6% of acetic acid molecules dimerize.

This concept is vital in chemical thermodynamics and helps to interpret molecular interactions and solution behavior, which is pivotal for predicting and explaining the properties of solutions.