Which of the following aqueous solutions has (a) the higher boiling point, (b) the higher freezing point, and (c) the lower vapor pressure: \(0.35 \mathrm{~m}\) \(\mathrm{CaCl}_{2}\) or \(0.90 \mathrm{~m}\) urea? Explain. Assume \(\mathrm{CaCl}_{2}\) to undergo complete dissociation.

Short Answer

Expert verified
The 0.35 m CaCl2 solution has a higher boiling point, a lower freezing point, and a lower vapor pressure than the 0.9 m urea solution because, upon complete dissociation, CaCl2 produces more particles in solution than urea.

Step by step solution

01

Calculate the total molality for each solution

First, for each solution, determine the total concentration of particles (in molality). For CaCl2, since it dissociates into three particles (one Ca2+ ion and two Cl- ions), the total molality is \(0.35 m * 3 = 1.05 m\). For urea, since it doesn't disassociate, the total molality remains as the given molality: \(0.9 m\).
02

Compare the boiling points

The boiling point elevation is a colligative property which depends on the number of particles in solution. The solution with more particles per kg of solvent has the higher boiling point. Thus, since the CaCl2 solution has a higher total molality, it has a higher boiling point than the urea solution.
03

Compare the freezing points

The decrease in freezing point also relies on the total number of particles in solution. The more particles, the lower the freezing point. So, the CaCl2 solution, having more particles per kg of solvent, also has a lower freezing point than the urea solution.
04

Compare the vapor pressures

Vapor pressure lowering is another colligative property. The vapor pressure of a solution is lower than that of the pure solvent because the presence of solute particles decreases the number of solvent molecules that can escape the liquid phase. Thus, the solution with more particles will have the lower vapor pressure. In this case, the CaCl2 solution will have a lower vapor pressure than the urea solution.

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Key Concepts

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

Boiling Point Elevation
Boiling point elevation is one of the key colligative properties which implies that a solution has a higher boiling point than the pure solvent. This is caused by the addition of solute particles which disrupt the solvent's normal escape tendency into the gas phase. As a result, more energy (in the form of heat) is required to allow the solvent particles to escape, raising the boiling point.

The degree of boiling point elevation can be calculated using the formula \( \Delta T_b = K_b \times m \) where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the boiling point elevation constant, and \( m \) is the molality of the solution. It's crucial to consider not just the concentration but also the number of particles derived from a solute, particularly when dealing with electrolytes that dissociate in solution.
Freezing Point Depression
Conversely, freezing point depression is the reduction of a solution's freezing point compared to the pure solvent due to the presence of solute particles. These solute particles effectively hinder the formation of a solid structure as the solvent tries to freeze, requiring a lower temperature to achieve the solid phase.

The calculation for freezing point depression follows a similar principle using the formula \( \Delta T_f = K_f \times m \) where \( \Delta T_f \) is the freezing point depression, \( K_f \) is the freezing point depression constant, and \( m \) is the molality of the solution. This property is hugely important in applications such as using salt to melt ice on roads.
Vapor Pressure Lowering
Vapor pressure lowering is the phenomenon where the vapor pressure of a solvent above a solution is lower than that of the pure solvent at the same temperature. This happens because solute particles occupy some of the surface area above the liquid, reducing the number of solvent molecules that can evaporate. The extent to which the vapor pressure is lowered depends on the number of solute particles present.

This property can be described with Raoult's law for ideal solutions, which states that the vapor pressure of the solvent in a solution is proportional to the mole fraction of the solvent in the mixture. Vapor pressure lowering is particularly important in understanding the behavior of solutions and their physical properties, such as boiling and freezing points.
Molality
Molality (represented by \( m \) ) is a concentration term for solutions that expresses the moles of solute per kilogram of solvent. Unlike molarity, which is affected by changes in temperature due to volume expansion or contraction, molality is temperature-independent since it is based on mass.

Molality is an essential factor in calculating colligative properties as it directly correlates with the extent of boiling point elevation and freezing point depression. It is defined as \( m = \frac{{\text{moles of solute}}}{{\text{kilograms of solvent}}} \) and is crucial for precision in scenarios where temperature fluctuation is a concern.
Dissociation of Electrolytes
The dissociation of electrolytes when dissolved in water introduces free ions into the solution. Electrolytes like \( \mathrm{CaCl}_{2} \) dissociate into multiple particles. For instance, calcium chloride dissociates into one calcium ion (\( \mathrm{Ca}^{2+} \) ) and two chloride ions (\( \mathrm{Cl}^{-} \) ).

This dissociation is significant because it increases the number of particles in the solution, intensifying colligative properties effects compared to non-electrolytes. When an electrolyte like \( \mathrm{CaCl}_{2} \) is said to undergo 'complete dissociation,' it means that each unit of the substance contributes more than one particle to the solution, which is vital when calculating colligative properties.

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Most popular questions from this chapter

A \(50-g\) sample of impure \(\mathrm{KClO}_{3}\) (solubility \(=7.1 \mathrm{~g}\) per \(100 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) ) is contaminated with 10 percent of KCl (solubility \(=25.5 \mathrm{~g}\) per \(100 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) ). Calculate the minimum quantity of \(20^{\circ} \mathrm{C}\) water needed to dissolve all the \(\mathrm{KCl}\) from the sample. How much \(\mathrm{KClO}_{3}\) will be left after this treatment? (Assume that the solubilities are unaffected by the presence of the other compound.)

The vapor pressure of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) at \(20^{\circ} \mathrm{C}\) is \(44 \mathrm{mmHg},\) and the vapor pressure of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) at the same temperature is \(94 \mathrm{mmHg} . \mathrm{A}\) mixture of \(30.0 \mathrm{~g}\) of methanol and \(45.0 \mathrm{~g}\) of ethanol is prepared (and can be assumed to behave as an ideal solution). (a) Calculate the vapor pressure of methanol and ethanol above this solution at \(20^{\circ} \mathrm{C}\). (b) Calculate the mole fraction of methanol and ethanol in the vapor above this solution at \(20^{\circ} \mathrm{C}\). (c) Suggest a method for separating the two components of the solution.

As you know, some solution processes are endothermic and others are exothermic. Provide a molecular interpretation for the difference.

How do the solubilities of most ionic compounds in water change with temperature? With pressure?

Describe the factors that affect the solubility of a solid in a liquid. What does it mean to say that two liquids are miscible?

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