Problem 41 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Given that \(\Delta T_{f}\) is the depression in freezing point of the solvent in a solution of a non-volatile solute of molality, \(m\), the quantity \(\operatorname{Lim}_{m \rightarrow 0}\left(\Delta T_{f} / m\right)\) is equal to \(\ldots \ldots \ldots . .\)

Step-by-Step Solution

Answer

The limit is equal to the cryoscopic constant \( K_{f} \).

Step 1: Understanding the Freezing Point Depression Formula

The depression in freezing point \( \Delta T_{f} \) is given by the formula \( \Delta T_{f} = K_{f} \times m \), where \( K_{f} \) is the cryoscopic constant and \( m \) is the molality of the solution.

Step 2: Setting Up the Limit Expression

We need to evaluate the expression \( \operatorname{Lim}_{m \rightarrow 0} \left( \Delta T_{f} / m \right) \). Substitute \( \Delta T_{f} = K_{f} \times m \) into the limit expression.

Step 3: Simplifying the Limit Expression

By substituting, the limit expression becomes \( \operatorname{Lim}_{m \rightarrow 0} \left( \frac{K_{f} \times m}{m} \right) \). Simplify the fraction: \( \frac{K_{f} \times m}{m} = K_{f} \).

Step 4: Evaluating the Limit Expression

Because \( m \) cancels out, we have \( \operatorname{Lim}_{m \rightarrow 0} \left( \Delta T_{f} / m \right) = K_{f} \). Since there is no dependence on \( m \), the limit is equal to the cryoscopic constant \( K_{f} \).

Key Concepts

Freezing Point Depression

When a solute is added to a solvent, it usually leads to a lowering of the freezing point of the solvent. This phenomenon is known as freezing point depression. It occurs because the presence of solute particles interferes with the formation of a solid structure by the solvent molecules, effectively needing a lower temperature to freeze.
Freezing point depression is a type of colligative property, which depends on the number of solute particles in a solution and not on their identity. In simpler words, it doesn't matter what the solute is, just how much of it there is. This makes calculating changes in freezing points relatively straightforward once you know the concentration of the solution, often measured in molality.

Cryoscopic Constant

The cryoscopic constant, denoted by the symbol \( K_{f} \), is a proportionality constant used in calculating the freezing point depression of a solution. It is specific to a particular solvent and arises from the relationship between the depression of the freezing point and the molality of a solution.

• Each solvent has its own unique \( K_{f} \) value based on its molecular properties.
• The constant allows us to predict how much the freezing point will decrease based on the concentration of the solute in the solution.

The mathematical formula is given by: \( \Delta T_{f} = K_{f} \times m \),where \( \Delta T_{f} \) is the depression in freezing point, \( K_{f} \) is the cryoscopic constant, and \( m \) is the molality of the solution. In the context of the exercise, the solution demonstrates that when molality approaches zero, \( \Delta T_{f} / m \) equals \( K_{f} \), reflecting the property of \( K_{f} \) being intrinsic to the solvent and independent of solute concentration.

Molality

Molality is a way to express the concentration of a solution. It is defined as the number of moles of solute per kilogram of solvent. This measure is particularly useful in colligative properties because it doesn't change with temperature, unlike other concentration measures like molarity.

• To calculate molality (\( m \)), you need the number of moles of the solute and the mass of the solvent in kilograms.
• The formula is: \[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]

Using molality instead of molarity ensures that our calculations for colligative properties like freezing point depression remain accurate and consistent across different temperatures. Understanding molality is key to predicting how a solute will impact the properties of a solvent.