Textbook Question
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.
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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 69
(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?
Verified step by step guidance1
Identify the colligative property involved: Boiling point elevation is a colligative property, which means it depends on the number of solute particles in a solution, not their identity.
Determine the van't Hoff factor (i) for each solute: NaCl dissociates into two ions (Na^+ and Cl^-), so i = 2. C6H12O6 (glucose) does not dissociate in solution, so i = 1.
Use the boiling point elevation formula: \( \Delta T_b = i \cdot K_b \cdot m \), where \( \Delta T_b \) is the boiling point elevation, \( K_b \) is the ebullioscopic constant of the solvent, and \( m \) is the molality of the solution.
Compare the boiling point elevations: Since the molality (m) and the solvent (water) are the same for both solutions, the difference in boiling point elevation will depend on the van't Hoff factor (i).
Conclude which solution has a higher boiling point: The solution with the higher van't Hoff factor (NaCl) will have a greater boiling point elevation, thus a higher boiling point than the glucose solution.
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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 particularly relevant, as it indicates how the presence of solute particles affects the boiling point of the solvent.
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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. This phenomenon is quantified by the formula ΔT_b = i * K_b * m, where ΔT_b is the boiling point elevation, i is the van 't Hoff factor (number of particles the solute dissociates into), K_b is the ebullioscopic constant of the solvent, and m is the molality of the solution. The greater the number of solute particles, the higher the boiling point.
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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, NaCl dissociates into two ions (Na+ and Cl-), giving it an i value of 2, while glucose (C6H12O6) does not dissociate and has an i value of 1. This factor is crucial for calculating colligative properties, as it directly influences the extent of boiling point elevation in solutions.
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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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