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Chapter 2: Problem 92

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)

Short Answer

Expert verified
The freezing point will be highest for glucose \(C_{6}H_{12}O_{6}\), because it does not dissociate and has the smallest Van’t Hoff factor.

Step by step solution

01

Understand Colligative Properties

Colligative properties depend on the number of particles in solution, not their identity. The freezing point depression is one such property, and it’s directly proportional to the molal concentration of particles in solution.
02

Determine the Van’t Hoff Factor

Calculate the Van’t Hoff factor (i) for each solute, which represents the number of particles a compound dissociates into. For example, ionic compounds dissociate into their respective ions, while non-electrolytes do not dissociate.
03

Calculate Van’t Hoff Factor for Each Solute

(a) \(C_{6}H_{5}NH_{3}Cl\) dissociates into 2 particles: aniline ion \(C_{6}H_{5}NH_{3}^{+}\) and chloride \(Cl^{-}\), so i = 2. (b) \(Ca(NO_{3})_{2}\) dissociates into 3 particles: calcium \(Ca^{2+}\) and two nitrate ions \(NO_{3}^{-}\), so i = 3. (c) \(La(NO_{3})_{3}\) dissociates into 4 particles: Lanthanum \(La^{3+}\) and three nitrate ions \(NO_{3}^{-}\), so i = 4. (d) \(C_{6}H_{12}O_{6}\) (glucose) does not dissociate, so i = 1.
04

Compare Van’t Hoff Factors

The higher the value of i, the greater the freezing point depression. Therefore, the solution with the smallest value of i will have the highest freezing point.
05

Identify the Solute with Highest Freezing Point

Glucose (\(C_{6}H_{12}O_{6}\)) does not dissociate in solution, giving it a Van’t Hoff factor of 1, the smallest among the options, thus the solution of glucose will have the highest freezing point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Colligative Properties
Colligative properties are characteristics of solutions that depend on the number of solute particles, regardless of their nature. A classic example of a colligative property is freezing point depression. When a solute dissolves in a solvent, it disrupts the pure solvent's structure, making it more difficult for the solvent to solidify and hence lowering its freezing point.

Understanding these properties is crucial for applications ranging from everyday tasks, such as salting roads in the winter to lower ice formation, to complex scientific procedures, like determining molecular masses by measuring boiling point elevation or freezing point depression.
Van't Hoff Factor
The Van't Hoff factor, represented by the symbol 'i', quantifies the extent of dissociation of a compound when it dissolves in solution. It indicates the number of particles a substance generates in a solution. For non-electrolytes like sugar that do not dissociate in water, the Van't Hoff factor is 1. Conversely, electrolytes such as salts can divide into multiple ions, increasing the Van't Hoff factor proportionally with the number of ions produced.

In the context of freezing point depression, the Van't Hoff factor plays a pivotal role in predicting how much the freezing point will be lowered, given the same concentration of different substances in solution.
Equimolal Solutions
Equimolal solutions are those that contain an equal number of moles of solute per kilogram of solvent, making them an excellent framework for observing colligative properties. Since these properties depend on the number of particles present—and equimolal solutions have the same number of moles of solute—the observed effects can be directly compared.

Understanding equimolal solutions is essential because it allows scientists and educators to isolate the impact of the solute's nature on freezing point depression and eliminate the concentration variable from the equation.
Dissociation of Ionic Compounds
Ionic compounds dissociate into ions when they dissolve in a solvent such as water. This dissociation is a crucial part of understanding how these compounds affect colligative properties like freezing point depression.

For instance, a compound containing a 1:1 ratio of ions will have a Van’t Hoff factor of 2, because it produces two separate ions per formula unit. If more complex ions are involved, such as polyatomic ions, the Van’t Hoff factor increases accordingly. However, it is also important to consider that not all ionic compounds fully dissociate. The degree of dissociation can depend on the solute concentration and the specific solvent's nature.

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Most popular questions from this chapter

When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) Freezing point is raised (b) Freezing point does not change (c) Freezing point is lowered (d) Boiling point does not change

A solution is obtained by dissolving \(6 \mathrm{~g}\) of urea (mol. wt. \(=60\) ) in a litre solution, another solution is prepared by dissolving \(34.2 \mathrm{~g}\) of cane sugar (mol. wt. = 342 ) in a litre of solution at the same temperature The lowering of vapour pressure in the first solution is: (a) Same as that of second solution (b) Double that of second solution (c) Half that of second solution (d) Nearly one fifth of the second solution

KBr is \(80 \%\) dissociated in aqueous solution of \(0.5 \mathrm{M}\) concentration. (Given \(\mathrm{K}_{\mathrm{f}}\) for water \(=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) ). The solution freezes at: (a) \(271.326 \mathrm{~K}\) (b) \(272 \mathrm{~K}\) (c) \(270.5 \mathrm{~K}\) (d) \(268.5 \mathrm{~K}\)

By dissolving 5 g substance in \(50 \mathrm{~g}\) of water, the decrease in freezing point is \(1.2^{\circ} \mathrm{C}\). The molaldepression constant is \(1.85^{\circ} \mathrm{kg} \mathrm{mol}^{-1} .\) The molecular weight of substance is: (a) \(105.4\) (b) \(118.2\) (c) \(137.2\) (d) \(154.2\)

Among the following aqueous solutions, the correct order of increasing boiling point can be given as: (1) \(10^{-4} \mathrm{M} \mathrm{KCl}\) (2) \(10^{-3}\) urea (3) \(10^{-3} \mathrm{M} \mathrm{CaCl}_{2}\) (4) \(10^{-3} \mathrm{M} \mathrm{KCl}\) (a) \(1<4<2<3\) (b) \(1<2<4<3\) (c) \(4<2<1<3\) (d) \(1<2<3<4\)
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