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(a) Why does a \(0.10 m\) aqueous solution of NaCl have a higher boiling point than a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?\) (b) Calculate the boiling point of each solution. (c) The experimental boiling point of the \(\mathrm{NaCl}\) solution is lower than that calculated, assuming that \(\mathrm{NaCl}\) is completely dissociated in solution. Why is this the case?

Short Answer

Expert verified
The boiling point of a 0.10 m NaCl solution is higher than a 0.10 m C鈧咹鈧佲倐O鈧 solution because NaCl dissociates into two ions (Na鈦 and Cl鈦) in water, while C鈧咹鈧佲倐O鈧 doesn't dissociate and remains as a single molecule. The boiling points of the NaCl and C鈧咹鈧佲倐O鈧 solutions are 100.1024 掳C and 100.0512 掳C, respectively. The experimental boiling point of the NaCl solution is lower than the calculated value because not all NaCl molecules dissociate completely, leading to a lower boiling point elevation.

Step by step solution

01

(a) Understanding the boiling point elevation due to solutes

Boiling point elevation occurs when a non-volatile solute is added to a solvent, causing the boiling point of the solution to be higher than that of the pure solvent. In this case, both NaCl and C鈧咹鈧佲倐O鈧 are non-volatile solutes being added to water. The main reason why the boiling point of a 0.10 molarity (m) NaCl solution is higher than a 0.10 m C鈧咹鈧佲倐O鈧 solution is due to the number of particles in the solution. NaCl is an ionic compound and dissociates into two ions (Na鈦 and Cl鈦) when dissolved in water, while C鈧咹鈧佲倐O鈧, a sugar molecule, does not dissociate and remains as a single molecule in solution.
02

(b) Calculating the boiling points of NaCl and C鈧咹鈧佲倐O鈧 solutions

We can use the formula for boiling point elevation to calculate the boiling points of the NaCl and C鈧咹鈧佲倐O鈧 solutions: 螖T = i 脳 Kb 脳 m Where 螖T is the boiling point elevation, i is the van't Hoff factor (number of particles produced per formula unit in the solution), Kb is the molal boiling point constant of water (0.512 掳C/m), and m is the molality. For NaCl: i = 2 (since NaCl dissociates into two ions) For C鈧咹鈧佲倐O鈧: i = 1 (since it doesn't dissociate) Now we can calculate the boiling point elevation for each solution: 螖T_NaCl = 2 脳 0.512 脳 0.10 = 0.1024 掳C 螖T_C鈧咹鈧佲倐O鈧 = 1 脳 0.512 脳 0.10 = 0.0512 掳C The boiling point of pure water is 100 掳C, so we can find the boiling points of both solutions by adding the calculated boiling point elevation to the boiling point of water: Boiling point of NaCl solution = 100 掳C + 0.1024 掳C = 100.1024 掳C Boiling point of C鈧咹鈧佲倐O鈧 solution = 100 掳C + 0.0512 掳C = 100.0512 掳C
03

(c) Explaining the discrepancy between experimental NaCl boiling point and calculated boiling point

The calculated boiling point of the NaCl solution assumes complete dissociation of NaCl into Na鈦 and Cl鈦 ions. In reality, however, not all NaCl molecules dissociate completely in solution. Some Na鈦 and Cl鈦 ions may associate with each other, forming ion pairs. These ion pairs effectively reduce the number of particles in the solution, leading to a lower boiling point elevation than expected from the calculation. Consequently, the experimental boiling point of the NaCl solution is lower than the calculated boiling 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 fascinating aspects of chemistry that only depend on the number of solute particles in a solution, not the nature of the solute itself. This means that characteristics like boiling point elevation, freezing point depression, and osmotic pressure are affected by how many particles are present, regardless of their specific type. This property is crucial when considering solutions like saltwater, where adding more particles changes how the solution behaves.
For instance, in a salt solution compared to pure water, the introduction of salt affects the boiling point. After adding the solute, the solution requires more energy (heat) to convert into vapor, resulting in a higher boiling point. This is why the boiling point of a saline solution is higher than that of pure water. Understanding colligative properties helps us predict and control these changes in solution behavior for various practical applications.
Van't Hoff Factor
The van't Hoff factor, denoted as \(i\), is a critical concept when discussing colligative properties. It represents the number of particles formed in a solution from a single solute molecule. For instance, when sodium chloride (NaCl) dissolves in water, it dissociates into two ions: Na鈦 and Cl鈦. Therefore, the van't Hoff factor for NaCl is 2.
In contrast, glucose (C鈧咹鈧佲倐O鈧) does not dissociate into separate particles when it dissolves. It stays as a single intact molecule, giving it a van鈥檛 Hoff factor of 1. This difference significantly impacts the boiling point elevation between solutions of NaCl and C鈧咹鈧佲倐O鈧, even if they have the same concentration. The NaCl solution generates more particles, enhancing the colligative effect and increasing the boiling point more than the glucose solution.
Solute Dissociation
Solute dissociation refers to how a solute breaks down into its individual ions when dissolved in a solvent. This process is pivotal in determining the van't Hoff factor and thus the colligative properties. Sodium chloride is a classic example of a solute that dissociates. In water, it separates into two ions, Na鈦 and Cl鈦, contributing to the higher boiling point elevation compared to non-dissociating solutes like glucose.
This breakdown directly impacts how solutions behave under various conditions. However, solute dissociation is not always complete. Factors like concentration and intermolecular forces can cause some ions to reassociate or form ion pairs, reducing the number of free particles. This incomplete dissociation can lead to experimental boiling points being lower than theoretical predictions, as seen with NaCl solutions in reality.

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Most popular questions from this chapter

A sulfuric acid solution containing \(571.6 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per liter of solution has a density of \(1.329 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate (a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in this solution.

(a) Why is there no colloid in which both the dispersed substance and the dispersing substance are gases? (b) Michael Faraday first prepared ruby-red colloids of gold particles in water that were stable indefinitely. To the unaided eye these brightly colored colloids are not distinguishable from solutions. How could you determine whether a given colored preparation is a solution or colloid?

In general, the attractive intermolecular forces between solvent and solute particles must be comparable or greater than solute-solute interactions for significant solubility to occur. Explain this statement in terms of the overall energetics of solution formation.

Water and glycerol, \(\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH},\) are miscible in all proportions. What does this mean? How do the OH groups of the alcohol molecule contribute to this miscibility?

Indicate the principal type of solute-solvent interaction in each of the following solutions and rank the solutions from weakest to strongest solute- solvent interaction: (a) \(\mathrm{KCl}\) in water, (b) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right),(\mathrm{c})\) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water.

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