Textbook Question
(a) Does a 0.10 m aqueous solution of NaCl have a higher boiling point, a lower boiling point, or the same boiling point as a 0.10 m aqueous solution of C6H12O6?
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Ch.13 - Properties of Solutions
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement177
- Ch.2 - Atoms, Molecules, and Ions231
- Ch.3 - Chemical Reactions and Reaction Stoichiometry216
- Ch.4 - Reactions in Aqueous Solution189
- Ch.5 - Thermochemistry175
- Ch.6 - Electronic Structure of Atoms168
- Ch.7 - Periodic Properties of the Elements150
- Ch.8 - Basic Concepts of Chemical Bonding177
- Ch.9 - Molecular Geometry and Bonding Theories200
- Ch.10 - Gases179
- Ch.11 - Liquids and Intermolecular Forces113
- Ch.12 - Solids and Modern Materials154
- Ch.13 - Properties of Solutions157
- Ch.14 - Chemical Kinetics199
- Ch.15 - Chemical Equilibrium128
- Ch.16 - Acid-Base Equilibria164
- Ch.17 - Additional Aspects of Aqueous Equilibria163
- Ch.18 - Chemistry of the Environment105
- Ch.19 - Chemical Thermodynamics164
- Ch.20 - Electrochemistry171
- Ch.21 - Nuclear Chemistry122
- Ch.22 - Chemistry of the Nonmetals82
- Ch.23 - Transition Metals and Coordination Chemistry79
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry16
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
Chapter 13, Problem 71
List the following aqueous solutions in order of increasing boiling point: 0.120 m glucose, 0.050 m LiBr, 0.050 m Zn(NO3)2.
Verified step by step guidance1
Identify the colligative property involved: Boiling point elevation is a colligative property, which depends on the number of solute particles in a solution.
Determine the van 't Hoff factor (i) for each solute: Glucose (C_6H_{12}O_6) is a non-electrolyte, so i = 1. LiBr is an electrolyte that dissociates into Li^+ and Br^-, so i = 2. Zn(NO_3)_2 dissociates into Zn^{2+} and 2 NO_3^-, so i = 3.
Calculate the effective molality for each solution: Multiply the molality (m) by the van 't Hoff factor (i) for each solution. For glucose, it's 0.120 m * 1. For LiBr, it's 0.050 m * 2. For Zn(NO_3)_2, it's 0.050 m * 3.
Compare the effective molalities: The solution with the highest effective molality will have the highest boiling point, and the one with the lowest effective molality will have the lowest boiling point.
List the solutions in order of increasing boiling point based on their effective molalities.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Colligative Properties
Colligative properties are physical properties of solutions that depend on the number of solute particles in a given amount of solvent, rather than the identity of the solute. These properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. In this context, the boiling point elevation is crucial for determining the order of the solutions based on their molal concentrations.
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01:26
Colligative Properties
Boiling Point Elevation
Boiling point elevation occurs when a non-volatile solute is added to a solvent, resulting in an increase in the boiling point of the solution compared to the pure solvent. The extent of this elevation is directly proportional to the molality of the solute and the van 't Hoff factor (i), which accounts for the number of particles the solute dissociates into. Thus, solutions with higher concentrations of solute particles will have higher boiling points.
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03:05Boiling Point Elevation
Van 't Hoff Factor (i)
The van 't Hoff factor (i) represents the number of particles into which a solute dissociates in solution. For example, glucose does not dissociate (i=1), while LiBr dissociates into two ions (Li+ and Br-, i=2), and Zn(NO3)2 dissociates into three ions (Zn2+ and two NO3-, i=3). This factor is essential for calculating the effective concentration of solute particles, which influences the boiling point elevation of the solution.
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Related Practice
Textbook Question
Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.22 m glycerol (C3H8O3) in ethanol, (b) 0.240 mol of naphthalene (C10H8) in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water, (d) 2.04 g KBr and 4.82 g of glucose (C6H12O6) in 188 g of water.
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Textbook Question
At 20 °C, the vapor pressure of benzene (C6H6) is 75 torr, and that of toluene (C7H8) is 22 torr. Assume that benzene and toluene form an ideal solution. (b) What is the mole fraction of benzene in the vapor above the solution described in part (a)?
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Textbook Question
Using data from Table 13.3, calculate the freezing and boiling points of each of the following solutions: (a) 0.25 m glucose in ethanol; (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3; (c) 3.50 g NaOH in 175 g of water, (d) 0.45 mol ethylene glycol and 0.15 mol KBr in 150 g H2O.
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