Problem 5 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Elevation in the boiling point for 1 molal solution of glucose is \(2 \mathrm{~K}\). The depression in the freezing point for 2 molal solution of glucose in the same solvent is \(2 \mathrm{~K}\). The relation between \(\mathrm{K}_{\mathrm{b}}\) and \(\mathrm{K}_{\mathrm{f}}\) is: (a) \(\mathrm{K}_{\mathrm{b}}=1.5 \mathrm{~K}_{\mathrm{f}}\) (b) \(\mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{f}}\) (c) \(\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K}_{\mathrm{f}}\) (d) \(\mathrm{K}_{\mathrm{b}}=2 \mathrm{~K}_{\mathrm{f}}\)

Step-by-Step Solution

Answer

\( K_b = 2 K_f \), option (d).

Step 1: Understanding the Problem

The problem involves the colligative properties of solutions: boiling point elevation and freezing point depression. Given data include the boiling point elevation for a 1 molal solution and the freezing point depression for a 2 molal solution, both being 2 K, and we are to find the relation between the boiling point elevation constant \( K_b \) and the freezing point depression constant \( K_f \).

Step 2: Boiling Point Elevation Equation

The boiling point elevation equation is given by \( \Delta T_b = K_b \cdot m \), where \( \Delta T_b = 2 \text{ K} \) and \( m = 1 \text{ molal} \). From this, we can calculate \( K_b \) as: \[ K_b = \frac{2}{1} = 2 \text{ K} \].

Step 3: Freezing Point Depression Equation

The freezing point depression equation is given by \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f = 2 \text{ K} \) and \( m = 2 \text{ molal} \). From this, we can calculate \( K_f \) as: \[ K_f = \frac{2}{2} = 1 \text{ K} \].

Step 4: Relating \(K_b\) and \(K_f\)

Now, we know \( K_b = 2 \text{ K} \) and \( K_f = 1 \text{ K} \). To find the relation, we compare the two constants: \( K_b = 2 \cdot K_f \). Thus, \( K_b = 2 K_f \).

Step 5: Conclusion

The correct relation between \( K_b \) and \( K_f \) is \( K_b = 2 K_f \), which matches option \( (d) \). Thus, option \( (d) \) is correct.

Key Concepts

Boiling Point Elevation

Boiling point elevation is a fascinating concept that explains how the boiling point of a solvent increases when a solute is added. This is one of the colligative properties, which depend only on the number of solute particles in the solution, not the type of particles.
Let's break it down:

• The increase in the boiling point is directly proportional to the molality of the solution, which is the number of moles of solute per kilogram of solvent.
• The formula for boiling point elevation is \( \Delta T_b = K_b \cdot m \), where \( \Delta T_b \) represents the change in boiling point, \( K_b \) is the ebullioscopic constant specific to each solvent, and \( m \) is the molality.

In the exercise, a 1 molal glucose solution causes a 2 K elevation. Therefore, \( \Delta T_b = 2 \); and, as calculated, \( K_b = 2 \ \, \text{K} \). Understanding how the boiling point increases with solute addition helps us grasp real-world applications like in cooking, where salt in water raises the boiling point to cook food more efficiently.

Freezing Point Depression

Freezing point depression is another intriguing colligative property, closely related to boiling point elevation. It describes how the freezing point of a solvent decreases when a solute is dissolved in it.
The basic principles are:

• The freezing point decreases in proportion to the molality of the solution.
• Its formula is given by \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f \) is the decrease in freezing point, \( K_f \) is the cryoscopic constant of the solvent, and \( m \) is the molality.

In the provided exercise, a 2 molal glucose solution lowers the freezing point by the same amount—2 K. By applying the formula, we find \( K_f = 1 \ \, \text{K} \). This concept is often applied in real-life scenarios like in adding antifreeze to car radiators to lower the freezing point of the engine's coolant, thus preventing damage in cold weather.

Molality

Molality is a measure of the concentration of a solution. It is used specifically in colligative properties due to its temperature-independent nature. Let's delve into its specifics:
Molality refers to:

• The number of moles of solute per kilogram of solvent, denoted by \( m \).
• Unlike molarity, which involves volume and can vary with temperature, molality is unaffected by temperature changes, making it reliable for studies involving boiling point elevation and freezing point depression.

In the solved exercise, molality helps describe how the given concentration of glucose solutions affects the boiling and freezing points.Thus, molality serves as a crucial parameter in understanding colligative properties, allowing for predictable and quantifiable changes in physical properties of the solution just by dissolving more or fewer particles.