The freezing point of equimolal aqueous solutions will be highest for: (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{La}\left(\mathrm{NO}_{3}\right)_{2}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Short Answer

Expert verified
Compound (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) because it does not dissociate in solution and will therefore have the highest freezing point.

Step by step solution

01

Understanding Colligative Properties

The freezing point depression of a solution is a colligative property, which means it depends on the number of solute particles in the solution, not on the type of particles. Ionic compounds dissociate into ions in water, increasing the number of particles more significantly compared to covalent compounds that do not dissociate.
02

Determining Number of Particles in Solution

To find the solution with the highest freezing point, we must consider the number of particles produced upon dissociation for each compound. - For (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\), it dissociates into 2 particles: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^+\) and \(\mathrm{Cl}^-\).- For (b) \(\mathrm{Ca}(\mathrm{NO}_{3})_{2}\), it dissociates into 3 particles: \(\mathrm{Ca}^{2+}\) and 2 \(\mathrm{NO}_{3}^-\).- For (c) \(\mathrm{La}(\mathrm{NO}_{3})_{2}\), this is incorrect because \(\mathrm{La}^{3+}\) would typically bond with three nitrate ions, not two, so the correct formula is \(\mathrm{La}(\mathrm{NO}_{3})_{3}\). It dissociates into 4 particles: \(\mathrm{La}^{3+}\) and 3 \(\mathrm{NO}_{3}^-\).- For (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), a covalent compound, does not dissociate in solution and remains as one particle.
03

Comparing the Effect on Freezing Point

Since freezing point depression is directly proportional to the number of particles in solution, compounds that produce fewer particles cause less freezing point depression. Therefore, we look for the compound that produces the fewest number of particles when dissolved.
04

Identifying the Compound with Highest Freezing Point

Based on the number of particles each compound produces when dissolved, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) does not dissociate and therefore causes the least freezing point depression. As a result, an equimolal aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) will have the highest freezing point among the options given.

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Key Concepts

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

Freezing Point Depression
One intriguing aspect of chemistry is the colligative properties of solutions—characteristics that depend on the quantity, not the identity, of solute particles in a solution. Freezing point depression is one such property, reflecting how the addition of solutes lowers a solvent's freezing temperature. It's a fascinating phenomenon for students because it appears in everyday life, from salting icy roads to making homemade ice cream.

For example, when salt is added to ice, it dissolves and disperses into its constituent ions, interfering with the formation of ice crystals. This process disrupts the equilibrium between the solid and liquid phases, requiring a colder temperature to reach a new freezing point. Hence, the more solute particles present, the more the freezing point is lowered. The mathematical expression for freezing point depression is \( \Delta T_f = i \cdot K_f \cdot m \), where \(\Delta T_f\) is the change in freezing point, \(i\) is the van't Hoff factor which is the number of particles the solute dissociates into, \(K_f\) is the freezing point depression constant, and \(m\) is the molality of the solution.

This can explain why certain solutions freeze at lower temperatures than pure solvents and is essential for industries like cryopreservation and food processing, as control over the freezing point is critical.
Dissociation of Ionic Compounds
Dissociation of ionic compounds is a critical concept in understanding colligative properties, including freezing point depression. An ionic compound in water breaks apart into its ions; this separation process is what we call dissociation. Such compounds are made of anions and cations held together by electrostatic forces, and when they dissolve in water, these forces weaken, allowing the ions to spread out within the solution.

Take table salt (\(NaCl\)) as an example. When dissolved, it dissociates into \(Na^+\) and \(Cl^-\) ions. The degree of dissociation is quantified by the van't Hoff factor (\(i\)), representing the number of particles a compound generates in solution. For instance, \(NaCl\) has a van't Hoff factor of 2, since it produces two separate ions. Similarly, \(Ca(NO_3)_2\) dissociates into three ions: one \(Ca^{2+}\) and two \(NO_3^-\), giving it a van't Hoff factor of 3.

Understanding the dissociation process helps students predict the impact on a solution's colligative properties. Incorrect knowledge of a compound's dissociation can lead to errors, such as incorrectly assuming the number of ions formed, which directly influences the colligative properties like freezing point depression.
Equimolal Aqueous Solutions
Equimolal aqueous solutions are intriguing because they bring uniformity to the comparative study of the colligative properties. An equimolal solution contains equal amounts (in moles) of solute in a given amount of solvent. This is a useful standardization when comparing different substances because it focuses the comparison on the intrinsic properties of the molecules themselves, rather than their concentration in solution.

For instance, equimolal solutions of glucose (\(C_6H_{12}O_6\)) and sodium chloride (\(NaCl\)) will contain the same number of moles of each compound dissolved in the same amount of water. However, their effect on the freezing point will differ not due to how much there is of each compound, but due to how they interact with water - glucose molecules do not dissociate while sodium chloride does.

An important exercise improvement advice here is to recognize that dissociation affects equimolal solutions differently. Ionic compounds will increase the number of particles more significantly than covalent ones, and thus, in the context of equal mole quantities, they will induce greater changes in colligative properties such as freezing point depression. This highlights the necessity of understanding both molar amounts and the nature of solutes when predicting the effects they will have on a solution's properties.

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Most popular questions from this chapter

If \(M_{\text {normal }}\) is the normal molecular mass and \(\alpha\) is the degree of ionization of \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\), then the abnormal molecular mass of the complex in the solution will be: (a) \(M_{\text {normal }}(1+2 \alpha)^{-1}\) (b) \(M_{\text {normal }}(1+3 \alpha)^{-1}\) (c) \(M_{\text {normal }}(1+\alpha)^{-1}\) (d) equal to \(M_{\text {normal }}\)

When a liquid that is immiscible with water was steam distilled at \(95.2^{\circ} \mathrm{C}\) at a total pressure of 748 torr, the distillate contained \(1.25 \mathrm{~g}\) of the liquid per gram of water. The vapour pressure of water is 648 torr at \(95.2^{\circ} \mathrm{C}\), what is the molar mass of liquid? (a) \(7.975 \mathrm{~g} / \mathrm{mol}\) (b) \(166 \mathrm{~g} / \mathrm{mol}\) (c) \(145.8 \mathrm{~g} / \mathrm{mol}\) (d) None of these

If two liquids \(A\left(P_{A}^{\circ}=100\right.\) torr \()\) and \(B\left(P_{B}^{\circ}=200\right.\) torr \()\) are completely immiscible with each other (each one will behave independently of the other) are present in a closed vessel. The total vapour pressure of the system will be: (a) less than 100 torr (b) greater than 200 torr (c) between 100 to 200 torr (d) 300 torr

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Select correct statement? (a) Heats of vaporisation for a pure solvent and for a solution are similar because similar intermolecular forces between solvent molecules must be overcome in both cases (b) Entropy change between solution and vapoür is smaller than the entropy change between pure solvent and vapour (c) Boiling point of the solution is larger than that of the pure solvent (d) All are correct statements

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