Problem 126 - A Complete Resource Book in Chemistry for JEE Main

Question

An aqueous solution of a solute which neither associates nor dissociates has a freezing point depression of \(\mathrm{X}^{\circ} \mathrm{C}\). An equimolar solution of a second has a freezing point depression of \(4 \mathrm{X}^{\circ} \mathrm{C}\). The second solution could be a salt of formula (assuming \(100 \%\) dissociation ) (a) \(\mathrm{AB}_{2}\) (b) \(\mathrm{A}_{2} \mathrm{~B}_{2}\) (c) \(\mathrm{A}_{3} \mathrm{~B}\) (d) \(\mathrm{AB}_{3}\)

Step-by-Step Solution

Answer

The second solution could be \( \mathrm{A}_2\mathrm{B}_2 \), \( \mathrm{A}_3\mathrm{B} \) or \( \mathrm{AB}_3 \).

Step 1: Understanding Freezing Point Depression

The freezing point depression is a colligative property that depends on the number of solute particles in a solution. It can be calculated using the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality of the solution. For the first solute, since it neither associates nor dissociates, \( i = 1 \).

Step 2: Comparing Freezing Point Depression

The equimolar solution of the second solute has a freezing point depression \( 4X^{\circ}C \) compared to \( X^{\circ}C \) for the first solute. This implies the van't Hoff factor for the second solute, \( i_2 \), is 4 times that of the first solute. Therefore, \( i_2 = 4 \).

Step 3: Analyzing Possible Solute Formulas

Since the second solute dissociates completely into ions, the formula must result in 4 moles of ions per mole of solute. Checking each provided option:- \( \mathrm{AB}_2 \) dissociates into \( 1A^+ + 2B^- \), total ions = 3.- \( \mathrm{A}_2\mathrm{B}_2 \) dissociates into \( 2A^+ + 2B^- \), total ions = 4.- \( \mathrm{A}_3\mathrm{B} \) dissociates into \( 3A^+ + 1B^- \), total ions = 4.- \( \mathrm{AB}_3 \) dissociates into \( 1A^+ + 3B^- \), total ions = 4.

Step 4: Determining the Correct Formula

Options \( \mathrm{A}_2\mathrm{B}_2 \), \( \mathrm{A}_3\mathrm{B} \), and \( \mathrm{AB}_3 \) each give 4 ions upon dissociation, matching the required van't Hoff factor of 4. Thus, these formulas are possible, but without any canonical order constraint, list them as equivalent possible options.

Key Concepts

Freezing Point Depression

Freezing point depression refers to the lowering of a liquid's freezing point by adding solute particles. When we dissolve a solute in a solvent, the presence of solute particles disturbs the formation of a structured solid, thereby lowering the temperature required for the liquid to solidify.
This phenomenon is explained by the colligative property, meaning it depends on the number of solute particles, rather than their nature.
Its magnitude can be calculated using the equation: \[ \Delta T_f = i \cdot K_f \cdot m \]where:

• \( \Delta T_f \) is the freezing point depression (the change in freezing point).
• \( i \) is the van't Hoff factor, indicating how many particles a solute produces in solution.
• \( K_f \) is the cryoscopic constant, specific to each solvent.
• \( m \) is the molality of the solution, measuring moles of solute per kilogram of solvent.

In this exercise, we learn that the second solution has a freezing point depression four times greater than the first. This highlights the importance of the number of particles in influencing colligative properties.

Van't Hoff Factor

The van't Hoff factor, denoted as \( i \), is crucial in calculating colligative properties like freezing point depression. It signifies the number of particles that a solute produces in solution.
For a solute that doesn't dissociate or associate in solution, \( i \) is typically 1. This means each solute molecule stays intact and forms one particle in the solution.

• Non-electrolytes, like sugar, have an \( i \) value of 1.
• Electrolytes, which dissociate into ions, have \( i > 1 \).

In the given problem, since the second solution's freezing point depression is four times greater than the first, \( i \) for the second solute is 4. Therefore, it suggests complete dissociation into multiple ions. Understanding the van't Hoff factor helps us predict how solutes will affect a solution's colligative properties.

Complete Dissociation

Complete dissociation refers to a solute entirely separating into individual ions when dissolved in a solution. This is common in strong electrolytes, such as ionic salts.

• For instance, NaCl dissociates fully into Na\(^+\) and Cl\(^-\) ions.
• Strong acids and bases also tend to completely dissociate in water.

In solutions exhibiting complete dissociation, the number of particles increases significantly, thus elevating the colligative property effects.
For example, a formula of \( \mathrm{A}_2\mathrm{B}_2 \) dissolving will yield 4 ions. This property is why solutions with complete dissociation have larger van't Hoff factors, as more particles mean a greater impact on properties like freezing point depression.

Equimolar Solutions

Equimolar solutions have equal molar concentrations of solutes, meaning they contain the same amount of moles of solute per volume.
These solutions enable direct comparison of colligative properties because differences in behavior are due purely to the nature of the solutes, rather than their amounts.
When encountering a scenario like the one described, where two equimolar solutions have drastically different freezing point depressions, it highlights the role of particle number in colligative effects.

• A non-dissociating solute leads to fewer particles, thus a smaller impact.
• Dissociating solutes increase particle count, amplifying properties like freezing point depression.

Thus, equimolar solutions are a useful tool for exploring and understanding the nuances of colligative properties and their reliance on particle count.