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Consider the following aqueous solutions: (i) \(0.20 \mathrm{m}\) \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (nonvolatile, nonelectrolyte); (ii) \(0.10 \mathrm{m}\) \(\mathrm{CaCl}_{2} ;\) (iii) \(0.12 \mathrm{m} \mathrm{KBr} ;\) and (iv) \(0.12 \mathrm{m} \mathrm{Na}_{2} \mathrm{SO}_{4}.\) (a) Which solution has the highest boiling point? (b) Which solution has the lowest freezing point? (c) Which solution has the highest water vapor pressure?

Short Answer

Expert verified
(a) Na₂SO₄ has the highest boiling point; (b) Na₂SO₄ has the lowest freezing point; (c) HOCH₂CH₂OH has the highest water vapor pressure.

Step by step solution

01

Understand the Problem

The problem involves determining which of the given solutions has the highest boiling point, the lowest freezing point, and the highest water vapor pressure. These properties are influenced by colligative properties, which depend on the number of solute particles in a solution.
02

Calculate the Van't Hoff Factor

For each solution, calculate the effective concentration of particles. The van't Hoff factor, \(i\), represents the number of particles a compound dissociates into in solution. - \( ext{HOCH}_2 ext{CH}_2 ext{OH} \) does not dissociate (i = 1).- \( ext{CaCl}_2 \) dissociates into 3 ions: Ca and 2 Cl (i = 3).- \( ext{KBr} \) dissociates into 2 ions: K and Br (i = 2).- \( ext{Na}_2 ext{SO}_4 \) dissociates into 3 ions: 2 Na and SO\( _4 \) (i = 3).
03

Boiling Point Elevation Analysis

The boiling point of a solution increases as the number of solute particles increases, due to boiling point elevation. Calculate the effective molality (i \( imes \) m) for each solution:- HOCH\(_2\)CH\(_2\)OH: 0.20 m \( \times \) 1 = 0.20- CaCl\(_2\): 0.10 m \( \times \) 3 = 0.30- KBr: 0.12 m \( \times \) 2 = 0.24- Na\(_2\)SO\(_4\): 0.12 m \( \times \) 3 = 0.36The highest value indicates the highest boiling point.
04

Freezing Point Depression Analysis

The freezing point of a solution decreases as the number of solute particles increases, due to freezing point depression. Effective molality impacts this as well: - The effective molality values are the same as in Step 3. The highest effective molality (same as highest boiling point elevation) also shows the lowest freezing point.
05

Water Vapor Pressure Analysis

The vapor pressure of a solution decreases as the solute particles increase (Raoult's Law). Thus, a higher effective molality means a lower vapor pressure. The solution with the least effective molality will have the highest vapor pressure.- Lowest effective molality: HOCH\(_2\)CH\(_2\)OH at 0.20, hence, highest vapor pressure.

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Key Concepts

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

Boiling Point Elevation
The boiling point of a liquid is the temperature at which it transitions into a vapor state. When you add a solute to a solvent, the boiling point increases. This phenomenon is known as boiling point elevation. It occurs because solute particles disrupt the formation of vapor bubbles, requiring more heat (higher temperature) to reach boiling.

This property does not depend on the type of solute, but rather on the number of particles present in the solution. Thus, it is classified as a colligative property. The key to understanding boiling point elevation is calculating the increase in molality (concentration of solute) and considering the van't Hoff factor (i), which represents how many particles a solute forms in solution.

For example, in the exercise, Na extsubscript{2}SO extsubscript{4} has a high effective molality (0.36), because it dissociates into three ions, resulting in the highest boiling point among the solutions analyzed.
Freezing Point Depression
Just like the boiling point, the freezing point of a solution is also affected by the presence of a solute. When a solute is added to a solvent, it lowers the temperature at which the solution turns solid. This effect is referred to as freezing point depression.

Colligative properties, such as freezing point depression, depend on the number of solute particles. Solute particles interfere with the solidification of the solvent, necessitating a lower temperature for the solution to freeze. Similar to boiling point elevation, the impact on freezing point is determined by the effective molality and the van't Hoff factor.

In the given problem, Na extsubscript{2}SO extsubscript{4} showed the largest freezing point depression because of its high effective molality of 0.36. This makes it have the lowest freezing point among the examined solutions.
Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase. When a nonvolatile solute is added to a solvent, the vapor pressure of the solvent decreases. This is a consequence of Raoult's Law, which explains that the presence of solute particles blocks some surface area, reducing the number of solvent molecules that can escape into the vapor phase.

As more solute is added, fewer solvent molecules escape into the vapor phase, leading to a drop in vapor pressure. This is another example of a colligative property, emphasizing the importance of solute particle concentration rather than the solute type.

In the exercise, the solution containing HOCH extsubscript{2}CH extsubscript{2}OH has the highest vapor pressure, due to its low effective molality (0.20). This means it has the smallest reduction in vapor pressure compared to other solutions.
Van't Hoff Factor
The van't Hoff factor ( (i) ) is a crucial concept in understanding colligative properties. It represents the number of particles formed when a solute dissolves. For nonelectrolytes like HOCH extsubscript{2}CH extsubscript{2}OH, which do not dissociate, the van't Hoff factor is 1. However, electrolytes like CaCl extsubscript{2} can dissociate into multiple ions, increasing this factor.

The van't Hoff factor helps in calculating the true concentration of particles in a solution (effective molality). For instance, CaCl extsubscript{2} dissociates into three ions (Ca and two Cl ions), so its van't Hoff factor is 3. This translates into a greater impact on boiling point elevation and freezing point depression as compared to substances with a lower van't Hoff factor.

Understanding the van't Hoff factor allows you to predict how solutions will behave in terms of colligative properties, giving you insight into their boiling, freezing, and vapor pressure characteristics.

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Most popular questions from this chapter

The dispersed phase of a certain colloidal dispersion consists of spheres of diameter \(1.0 \times 10^{2} \mathrm{nm}\) (a) What is the volume \(\left(V=\frac{4}{3} \pi r^{3}\right)\) and surface area \((A=\) \(\left.4 \pi r^{2}\right)\) of each sphere? (b) How many spheres are required to give a total volume of \(1.0 \mathrm{cm}^{3} ?\) What is the total surface area of these spheres in square meters?

The Henry's law constant for \(\mathrm{O}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is 1.66 \(\times 10^{-6} \mathrm{M} / \mathrm{mm}\) Hg. Which of the following is a reasonable constant when the temperature is \(50^{\circ} \mathrm{C}\) ? Explain the reason for your choice. (a) \(8.80 \times 10^{-7} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\) (b) \(3.40 \times 10^{-6} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\) (c) \(1.66 \times 10^{-6} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\) (d) \(8.40 \times 10^{-5} \mathrm{M} / \mathrm{mm} \mathrm{Hg}\)

Some lithium chloride, LiCl, is dissolved in 100 mL of water in one beaker and some \(\mathrm{Li}_{2} \mathrm{SO}_{4}\) is dissolved in 100 mL of water in another beaker. Both are at \(10^{\circ} \mathrm{C}\) and both are saturated solutions; some solid remains undissolved in each beaker. Describe what you would observe as the temperature is raised. The following data are available to you from a handbook of chemistry: $$\begin{array}{lll} \hline & \text { Solubility }(\mathrm{g} / 100 \mathrm{mL}) \\ \hline \text { Compound } & 10^{\circ} \mathrm{C} & 40^{\circ} \mathrm{C} \\ \hline \mathrm{Li}_{2} \mathrm{SO}_{4} & 35.5 & 33.7 \\\ \mathrm{LiCl} & 74.5 & 89.8 \\ \hline \end{array}$$

If \(52.5 \mathrm{g}\) of \(\mathrm{LiF}\) is dissolved in \(306 \mathrm{g}\) of water, what is the expected freezing point of the solution? (Assume the van't Hoff factor, \(i\), for LiF is \(2 .\) )

If you dissolve equal molar amounts of \(\mathrm{NaCl}\) and \(\mathrm{CaCl}_{2}\) in water, the CaCl, lowers the freezing point of the water almost 1.5 times as much as the NaCl. Why?

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