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

Question

The freezing point of equimolal aqueous solution will be highest for (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\)(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 highest freezing point is for glucose (d) \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \).

Step 1: Understanding Depression of Freezing Point

The freezing point of a solution decreases from that of the pure solvent. For solutions, this depression in freezing point is dependent on the number of solute particles, given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is the molality. Thus, the higher the \( i \), the greater the freezing point depression and the lower the freezing point of the solution.

Step 2: Calculating van't Hoff Factor

The van't Hoff factor \( i \) is determined by the number of ions each solute dissociates into: - (a) Aniline hydrochloride \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-} \): \( i = 2 \) - (b) \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \): \( i = 3 \) - (c) \( \mathrm{La}\left(\mathrm{NO}_{3}\right)_{3} \): \( i = 4 \) - (d) Glucose \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \): \( i = 1 \), as it does not dissociate.

Step 3: Identifying Solution with Highest Freezing Point

The solution with the lowest depression in freezing point (highest freezing point) will have the smallest \( i \). Among the options, glucose (d) with \( i = 1 \) undergoes no dissociation, leading to the least lowering of the freezing point.

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating colligative property that occurs when a solute is added to a solvent. In simpler terms, it means that the solution will freeze at a lower temperature than the pure solvent alone. The phenomenon can be explained with the formula: \[ \Delta T_f = i \cdot K_f \cdot m \] where \( \Delta T_f \) is the change in the freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant of the solvent, and \( m \) is the molality of the solution.To understand this further, let's break it down:

• The van't Hoff factor \( i \): This denotes the number of particles the solute produces in solution. More particles mean a greater effect on freezing point depression.
• The constant \( K_f \): This is specific to each solvent and indicates how much the freezing point depresses with the addition of solute. Pure water, for example, has a specific \( K_f \).
• Molality \( m \): This refers to moles of solute per kilogram of solvent. More solute generally means more depression of the freezing point.

Understanding freezing point depression helps chemists predict the degree of cooling required for a solution to freeze.

Van't Hoff Factor

The van't Hoff factor, labeled as \( i \), plays a crucial role in determining the extent of colligative properties like freezing point depression. It tells us the number of individual particles formed in a solution from the dissociation of solute molecules. Knowing \( i \) allows predictions about how much a solution's freezing point will drop. The concept becomes even clearer with an example:

• Consider **calcium nitrate**, \( \text{Ca(NO}_3\text{)}_2 \). It dissociates into 3 ions: 1 calcium ion \( \text{Ca}^{2+} \) and 2 nitrate ions \( \text{NO}_3^- \). Here, \( i = 3 \).
• Contrast this with **glucose**, \( \text{C}_6 \text{H}_{12} \text{O}_6 \), a non-electrolyte. It does not dissociate in water, so \( i = 1 \).

Given solutions of equal molality, the one with a higher \( i \) will show greater freezing point depression. Understanding \( i \) is fundamental when predicting solution behavior.

Aqueous Solution

An aqueous solution is a specific type of chemical solution where the solvent is water. Water's unique properties make it an excellent solvent for many substances, assisting processes such as dissolution and reaction. Here's a simple breakdown of why aqueous solutions are important in this context:

• Dissolution: Solutes dissolve in water due to its polar nature, enabling the solute molecules or ions to spread throughout the solution.
• Colligative Properties: In learning about freezing point depression, water's solvent capabilities mean it will lower its freezing point based on the solute added.
• Everyday relevance: Many everyday solutions, like saline or sugar water, demonstrate the principles of aqueous solutions and colligative properties.

The effects of solutes like glucose or ionic compounds can be studied in water as it provides a consistent and familiar medium to understand these properties.