Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

The freezing point of equimolal aqueous solution will be highest for (a) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\)(aniline hydrochloride) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{La}\left(\mathrm{NO}_{3}\right)_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose)

Step-by-Step Solution

Answer

The freezing point of the solution is highest for glucose (option d).

Step 1: Understand the Concept of Freezing Point Depression

The freezing point depression of a solution is 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 cryoscopic constant of the solvent, and \( m \) is the molality of the solution. For equimolal solutions, \( K_f \) and \( m \) remain constant, so \( \Delta T_f \) depends primarily on \( i \).

Step 2: Calculate Van't Hoff Factor for Each Solute

The van't Hoff factor \( i \) represents the number of particles the solute dissociates into in solution:- For \( \mathrm{C}_6 \mathrm{H}_3 \mathrm{NH}_3^+ \mathrm{Cl}^- \) (aniline hydrochloride), \( i = 2 \) (dissociates into \( \mathrm{C}_6 \mathrm{H}_3 \mathrm{NH}_3^+ \) and \( \mathrm{Cl}^-\)).- For \( \mathrm{Ca(NO_3)_2} \), \( i = 3 \) (dissociates into \( \mathrm{Ca^{2+}} \) and 2 \( \mathrm{NO_3^-} \)).- For \( \mathrm{La(NO_3)_3} \), \( i = 4 \) (dissociates into \( \mathrm{La^{3+}} \) and 3 \( \mathrm{NO_3^-} \)).- For \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \) (glucose), \( i = 1 \) because it does not dissociate in solution.

Step 3: Determine the Impact on Freezing Point

The higher the van't Hoff factor, the larger the freezing point depression. Therefore, the compound with the lowest \( i \) will have the highest freezing point because it causes the least depression of the freezing point.

Step 4: Identify the Compound with Highest Freezing Point

Comparing the van't Hoff factors, glucose \( (i = 1) \) has the lowest \( i \) value among the options, meaning it will cause the least freezing point depression, resulting in the highest freezing point.

Key Concepts

Freezing Point Depression

Freezing point depression refers to the process by which the freezing point of a liquid (solvent) is lowered by adding another substance. This process is significant in determining how solutions behave when a solute is introduced to a solvent.
Here are the key elements to understand it:- **Formula:** The change in freezing point \( \Delta T_f = i \cdot K_f \cdot m \) where: - \( \Delta T_f \) is the freezing point depression (how much the freezing point lowers). - \( i \) is the van't Hoff factor, representing the degree of particle dissociation in the solution. - \( K_f \) is the cryoscopic constant of the solvent, specific to each solvent used. - \( m \) is the molality of the solution.
In equimolal solutions, since \( K_f \) and \( m \) remain constant, the focus is primarily on the van't Hoff factor \( i \). This means that the extent of freezing point depression depends majorly on the number of particles the solute dissociates into.

van't Hoff Factor

The van't Hoff factor \( i \) is an integral part of understanding solutions' colligative properties. It measures how many particles a solute splits into when dissolved in a solution. The factor helps predict the effect of solutes on property changes, such as freezing point depression.- **Calculating \( i \):** The factor varies between different solutes, such as: - Aniline hydrochloride, \( \mathrm{C}_6 \mathrm{H}_3 \mathrm{NH}_3^+ \mathrm{Cl}^- \), dissociates into two ions, thus \( i = 2 \). - Calcium nitrate, \( \mathrm{Ca(NO_3)_2} \), produces three ions, resulting in an \( i = 3 \). - Lanthanum nitrate, \( \mathrm{La(NO_3)_3} \), results in four ions, so \( i = 4 \). - Glucose, \( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 \), does not dissociate, hence \( i = 1 \).
The van't Hoff factor highlights how different compounds will affect the freezing point of a solution, with higher \( i \) values leading to greater freezing point depression due to more dissociated particles.

Equimolal Solutions

Understanding equimolal solutions is vital when considering changes in colligative properties like freezing point depression. An equimolal solution is one where the molality (\( m \)) of different solute solutions is the same.- **Significance in Freezing Point Depression:** - Since molality remains constant, changes in properties are predominantly governed by the van't Hoff factor \( i \). This allows us to isolate the effect of different solutes' dissociation. - In such solutions, comparing solutes becomes easier as the innate properties of the solvent (represented by \( K_f \)) and the molality do not vary.
Equimolal solutions illustrate how colligative properties can be directly compared across different solutes, helping determine that glucose, with the lowest van't Hoff factor, maintains the highest freezing point because it contributes the least to the depression.