Problem 3 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Molal depression constant for a solvent is \(4.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\). The depression in the freezing point of the solvent for \(0.03 \mathrm{~mol} \mathrm{~kg}^{-1}\) solution \(\mathrm{K}_{2} \mathrm{SO}_{4}\) is: (Assume complete dissociation of the electrolyte) (a) \(0.18 \mathrm{~K}\) (b) \(0.24 \mathrm{~K}\) (c) \(0.12 \mathrm{~K}\) (d) \(0.36 \mathrm{~K}\)

Step-by-Step Solution

Answer

The depression in freezing point is 0.36 K; option (d) is correct.

Step 1: Understand Depression Formula

The depression in freezing point can be calculated using the formula \(\Delta T_f = i \cdot k_f \cdot m\), where \(\Delta T_f\) is the depression in freezing point, \(i\) is the van't Hoff factor, \(k_f\) is the molal depression constant, and \(m\) is the molality of the solution.

Step 2: Find the van't Hoff Factor

Since \(\text{K}_2\text{SO}_4\) completely dissociates in solution, we must consider the ionization. \(\text{K}_2\text{SO}_4\) dissociates into 2 K⁺ ions and 1 SO₄²⁻ ion, making a total of 3 ions. Therefore, the van't Hoff factor \(i = 3\).

Step 3: Substitute Values into the Formula

We know \(k_f = 4.0 \, \text{K kg mol}^{-1}\), \(m = 0.03 \, \text{mol kg}^{-1}\), and \(i = 3\). Substitute these values into the formula: \(\Delta T_f = 3 \cdot 4.0 \cdot 0.03\).

Step 4: Calculate Depression in Freezing Point

Perform the multiplication: \(\Delta T_f = 3 \times 4.0 \times 0.03 = 0.36\, \text{K}\).

Step 5: Choose the Correct Answer

The depression in the freezing point is \(0.36 \, \text{K}\). Hence, the correct answer is option (d) \(0.36 \, \text{K}\).

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property that occurs when a solute is added to a solvent, causing the solution to freeze at a lower temperature than the pure solvent. This happens because the solute particles disrupt the crystal formation of the solvent during freezing.
The basic equation to calculate the freezing point depression is:

• \( \Delta T_f = i \cdot k_f \cdot m \)

Here, \( \Delta T_f \) is the change in freezing point, \( i \) is the van't Hoff factor, \( k_f \) is the molal depression constant, and \( m \) is the molality of the solution.
This concept is applied in many real-world situations, such as making ice cream, where salt may be added to lower the freezing point of the ice, thus making it colder than ice alone. It's perhaps even more famously seen when salt is spread on icy roads to lower the freezing point of water and help melt the ice.

Van't Hoff Factor

The van't Hoff factor is a crucial component in calculating colligative properties. It represents the number of particles a solute splits into when dissolved in a solvent.
For non-electrolytes, which don’t dissociate in solution, the van't Hoff factor \( i \) is simply 1. However, electrolytes dissociate into ions, increasing the number of particles in solution, which affects the freezing point.

• For instance, when \( \text{K}_2\text{SO}_4 \) dissociates in water, it splits into 2 \( \text{K}^+ \) ions and 1 \( \text{SO}_4^2^- \) ion.
• This makes the van't Hoff factor \( i = 3 \) because there are 3 ions in total.

It's essential to know the degree of dissociation or ionization to accurately compute the impact on freezing point depression. This factor can also change with concentration and temperature, making it versatile yet complex in different conditions.

Molal Depression Constant

The molal depression constant \( k_f \) is a unique value for each solvent that provides essential information about how much the freezing point of a solution differs from that of the pure solvent per molal concentration of a solute. It is expressed in units of \( \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \).
Each solvent has its own specific molal depression constant:

• This constant depends on the interactions between the solute and solvent molecules.
• In our example exercise, \( k_f = 4.0 \, \text{K kg mol}^{-1} \) for the solvent in question.

Understanding \( k_f \) is vital because it helps predict how much a solution's freezing point will decrease when a known quantity of solute is added. Hence, it finds common usage in fields where precise temperature control is necessary, like chemical engineering and food science.