Question

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

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

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.

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.

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.

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.

Key concepts

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.