Problem 18 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

When \(20 \mathrm{~g}\) of naphthoic acid \(\left(\mathrm{C}_{11} \mathrm{H}_{8} \mathrm{O}_{2}\right)\) is dissolved in \(50 \mathrm{~g}\) of benzene \(\left(K_{f}=1.72 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\), a freezing point depression of \(2 \mathrm{~K}\) is observed. The van't Hoff factor \((i)\) is (a) \(0.5\) (b) 1 (c) 2 (d) 3

Step-by-Step Solution

Answer

The van't Hoff factor (i) is 0.5, answer (a).

Step 1: Understanding the Formula

To find the van't Hoff factor, we need to understand the formula for freezing point depression: \[ \Delta T_f = i \cdot K_f \cdot m \] where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality.

Step 2: Calculate Molality

Molality \( m \) is calculated using the formula: \[ m = \frac{n}{m_{solvent}} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]We first calculate the moles of naphthoic acid: \[ n = \frac{\text{mass of naphthoic acid}}{\text{molar mass of naphthoic acid}} \] The molar mass of naphthoic acid (\( \text{C}_{11} \text{H}_8 \text{O}_2 \)) is 172 g/mol.So, \[ n = \frac{20}{172} = 0.11627 \text{ mol} \] The mass of benzene is 50 g, which is 0.05 kg, thus \[ m = \frac{0.11627}{0.05} = 2.3254 \text{ mol/kg} \]

Step 3: Solving for Van't Hoff Factor (i)

Using the freezing point depression formula \[ 2 = i \cdot 1.72 \cdot 2.3254 \]We solve for \( i \):\[ i = \frac{2}{1.72 \times 2.3254} \approx 0.5 \]

Key Concepts

Freezing Point Depression

Freezing point depression is an interesting phenomenon that happens when a solute is added to a solvent. When a substance is dissolved, it generally lowers the freezing point of the solvent. This effect is primarily due to the solute particles interfering with the formation of the solvent's solid state structure. In mathematical terms, freezing point depression is expressed through the equation: \[ \Delta T_f = i \cdot K_f \cdot m \]where:

• \( \Delta T_f \) is the change in freezing point
• \( i \) is the van't Hoff factor
• \( K_f \) is the cryoscopic constant of the solvent
• \( m \) is the molality of the solution

This equation tells us how much the freezing point will decrease based on the amount and type of solute added. The decrease in freezing point is often a linearly dependent property of how many particles (ions or molecules) a solute produces in the solution. This is where the van't Hoff factor becomes crucial, as it accounts for the number of particles into which a compound dissociates or associates.

Molality

Molality is a measure of concentration that compares the number of moles of the solute to the mass of the solvent in kilograms. It is particularly useful in calculations involving colligative properties like freezing point depression, because it remains constant with temperature changes, unlike molarity. The formula for molality is: \[ m = \frac{n}{m_{\text{solvent}}} \]Here, \( n \) represents the moles of solute, and \( m_{\text{solvent}} \) is the mass of the solvent in kilograms. Let's consider an example. If you have 20 grams of naphthoic acid (a solute) dissolved in 50 grams of benzene (a solvent), you would first need to find the moles of naphthoic acid. By dividing the weight of the solid by its molar mass, we determine the moles of our solute. Once we have the moles, the next step is calculating the molality using the mass of the solvent in kilograms. In our example, we calculated it as 2.3254 mol/kg, which assisted us in solving for the van't Hoff factor later.

Cryoscopic Constant

The cryoscopic constant, often denoted as \( K_f \), is a property of solvents and is used in the formula for freezing point depression. It quantifies the ability of a particular solvent to have its freezing point lowered when a solute is present.For each solvent, the cryoscopic constant is unique and determined experimentally. In the context of our problem with benzene, the cryoscopic constant is given as 1.72 K kg mol\(^{-1}\). This constant helps us understand the relationship between the number of solute particles and the freezing point depression.Here’s a small breakdown:

• The unit of \( K_f \) is Kelvin kilogram per mole (K kg mol\(^{-1}\)), showing how much the freezing point will decrease per molal concentration of a solute.
• Using \( K_f \) in the freezing point depression formula allows calculation of the van't Hoff factor, which helps predict how many particles a solute will produce in a solvent.

Indeed, without knowing \( K_f \), calculating the expected freezing point depression would be difficult. It's the pivotal bridge between understanding basic freezing point changes and assessing more complex solution behaviors.