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

Glucose (\( \mathrm{C_6H_{12}O_6} \)) will have the highest freezing point.

Step 1: Understand the Problem

We are tasked with determining which equimolal aqueous solution has the highest freezing point. This involves understanding the concept of freezing point depression, a colligative property that depends on the number of particles in solution.

Step 2: Determine Particles Produced by Dissolution

We need to determine how many particles each compound produces when dissolved in water. For (a) \( \mathrm{C_{6}H_{5}NH_{3}^{+}Cl^{-}} \) (aniline hydrochloride), it dissociates into two ions: \( \mathrm{C_{6}H_{5}NH_{3}^{+}} \) and \( \mathrm{Cl^{-}} \). For (b) \( \mathrm{Ca(NO_3)_2} \), it dissociates into three ions: \( \mathrm{Ca^{2+}} \) and two \( \mathrm{NO_3^{-}} \). For (c) \( \mathrm{La(NO_3)_3} \), it dissociates into four ions: \( \mathrm{La^{3+}} \) and three \( \mathrm{NO_3^{-}} \). (d) \( \mathrm{C_6H_{12}O_6} \) (glucose) does not dissociate and remains as one particle.

Step 3: Apply Freezing Point Depression Concept

The freezing point depression of a solution is proportional to the number of solute particles in the solution: more particles result in a greater depression, meaning a lower freezing point. Therefore, the fewer the dissociated particles, the lesser the freezing point depression.

Step 4: Compare the Number of Particles

Based on the number of particles produced:- \( \mathrm{C_6H_{12}O_6} \) results in 1 particle, causing the least depression.- \( \mathrm{C_{6}H_{5}NH_{3}^{+}Cl^{-}} \) results in 2 particles.- \( \mathrm{Ca(NO_3)_2} \) results in 3 particles.- \( \mathrm{La(NO_3)_3} \) results in 4 particles.Therefore, \( \mathrm{C_6H_{12}O_6} \) will have the highest freezing point as it produces the fewest particles.

Step 5: Conclusion

Considering these points, glucose (option d) has the least effect on lowering the freezing point due to producing only one particle in solution. Hence, it will have the highest freezing point among the options.

Key Concepts

Colligative Properties

Colligative properties are fundamental concepts in chemistry and relate to how the properties of a solution are affected by the presence of solute particles. There are several colligative properties, including vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. These properties depend solely on the number of solute particles in a given solution, rather than the identity of the solute itself.

When discussing freezing point depression, which is the key focus in our exercise, the idea is that the addition of solute particles causes the solution to freeze at a lower temperature than the pure solvent. This is because solute particles disrupt the formation of a solid structure. The relationship between the number of particles and the freezing point is inverse; more particles lead to a greater freezing point depression, while fewer particles result in a smaller depression. Given this, the colligative properties directly influence the physical characteristics of solutions and are crucial for solving related problems.

Dissociation of Compounds

Understanding how compounds dissociate in a solution is critical for predicting colligative effects. Dissociation is the process where compounds split into ions or simpler molecules when mixed with a solvent, such as water.

For example, in our exercise,

• Aniline hydrochloride (\( \mathrm{C_{6}H_{5}NH_{3}^{+}Cl^{-}} \)) dissociates into two ions: \( \mathrm{C_{6}H_{5}NH_{3}^{+}} \) and \( \mathrm{Cl^{-}} \).
• Calcium nitrate (\( \mathrm{Ca(NO_3)_2} \)) breaks into three ions: \( \mathrm{Ca^{2+}} \) and two \( \mathrm{NO_3^{-}} \).
• Lanthanum nitrate (\( \mathrm{La(NO_3)_3} \)) splits into four ions: \( \mathrm{La^{3+}} \) and three \( \mathrm{NO_3^{-}} \).

Glucose (\( \mathrm{C_6H_{12}O_6} \)), however, does not dissociate when dissolved; it remains as a single entity in the solution. The degree of dissociation indicates the number of particles formed in the solution, thus influencing the solution’s colligative properties.

Number of Particles in Solution

In colligative property analysis, knowing how many solute particles are present in a solution is crucial, as this directly impacts phenomena such as freezing point depression. The number of particles, also known as the van 't Hoff factor (i), provides insight into the effect different solutes have on colligative properties.

Consider our solution options:

• Glucose (\( \mathrm{C_6H_{12}O_6} \)) provides just one particle per molecule because it does not dissociate.
• Aniline hydrochloride produces two particles per molecule.
• Calcium nitrate yields three particles.
• Lanthanum nitrate results in four particles.

As more particles are formed, there is greater interference with the solvent's freezing process, leading to a lower freezing point. Conversely, fewer particles—a characteristic of glucose in the given exercise—mean the least disruption and thus a higher freezing point. Understanding the number of particles formed during dissociation allows us to predict which solution will have minimal freezing point depression.