Chemical Principles
8th Edition
ISBN: 9781337247269
Author: Steven S. Zumdahl; Donald J. DeCoste
Publisher: Cengage Learning US
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 17, Problem 61E
Interpretation Introduction
Interpretation:
The boiling point of solution should be predicted.
Concept Introduction:
Colligative properties are the properties of the solution that depend on the number of moles of solute. These properties are different from the solute particles. These properties include boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure lowering.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionChapter 17 Solutions
Chemical Principles
Ch. 17 - Prob. 1DQCh. 17 - Consider Fig. 17.8. Suppose that instead of having...Ch. 17 - Prob. 3DQCh. 17 - Prob. 4DQCh. 17 - Prob. 5DQCh. 17 - Prob. 6DQCh. 17 - Prob. 7DQCh. 17 - Prob. 8DQCh. 17 - Prob. 9DQCh. 17 - Prob. 10DQ
Ch. 17 - Prob. 11DQCh. 17 - Prob. 12ECh. 17 - Prob. 13ECh. 17 - Prob. 14ECh. 17 - Prob. 15ECh. 17 - Prob. 16ECh. 17 - Prob. 17ECh. 17 - Prob. 18ECh. 17 - Prob. 19ECh. 17 - Prob. 20ECh. 17 - Prob. 21ECh. 17 - Prob. 22ECh. 17 - Prob. 23ECh. 17 - Prob. 24ECh. 17 - Prob. 25ECh. 17 - Prob. 26ECh. 17 - Prob. 27ECh. 17 - Prob. 28ECh. 17 - Prob. 29ECh. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Prob. 33ECh. 17 - Prob. 34ECh. 17 - Prob. 35ECh. 17 - Prob. 36ECh. 17 - Prob. 37ECh. 17 - Prob. 38ECh. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Rationalize the temperature dependence of the...Ch. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Prob. 44ECh. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - Prob. 50ECh. 17 - Prob. 51ECh. 17 - Prob. 52ECh. 17 - Prob. 53ECh. 17 - Prob. 54ECh. 17 - Prob. 55ECh. 17 - Prob. 56ECh. 17 - The following plot shows the vapor pressure of...Ch. 17 - Prob. 58ECh. 17 - Prob. 59ECh. 17 - Prob. 60ECh. 17 - Prob. 61ECh. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - Prob. 67ECh. 17 - An aqueous solution of 10.00 g of catalase, an...Ch. 17 - Prob. 69ECh. 17 - What volume of ethylene glycol (C2H6O2) , a...Ch. 17 - Prob. 71ECh. 17 - Erythrocytes are red blood cells containing...Ch. 17 - Prob. 73ECh. 17 - Prob. 74ECh. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - Prob. 80ECh. 17 - Consider the following solutions: 0.010 m Na3PO4...Ch. 17 - From the following: pure water solution of...Ch. 17 - Prob. 83ECh. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Prob. 87ECh. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - Prob. 92ECh. 17 - Prob. 93AECh. 17 - Prob. 94AECh. 17 - Prob. 95AECh. 17 - Prob. 96AECh. 17 - The term proof is defined as twice the percent by...Ch. 17 - Prob. 98AECh. 17 - Prob. 99AECh. 17 - Prob. 100AECh. 17 - Prob. 101AECh. 17 - Prob. 102AECh. 17 - Prob. 103AECh. 17 - Prob. 104AECh. 17 - Prob. 105AECh. 17 - Prob. 106AECh. 17 - Prob. 107AECh. 17 - Prob. 108AECh. 17 - Prob. 109AECh. 17 - Prob. 110AECh. 17 - Prob. 111AECh. 17 - Prob. 112AECh. 17 - Prob. 113AECh. 17 - Prob. 114AECh. 17 - Formic acid (HCO2H) is a monoprotic acid that...Ch. 17 - Prob. 116AECh. 17 - Prob. 117AECh. 17 - Prob. 118AECh. 17 - Prob. 119AECh. 17 - Prob. 120AECh. 17 - Prob. 121AECh. 17 - Prob. 122AECh. 17 - Prob. 123AECh. 17 - Prob. 124AECh. 17 - Prob. 125AECh. 17 - Prob. 126AECh. 17 - Prob. 127CPCh. 17 - Prob. 128CPCh. 17 - Prob. 129CPCh. 17 - Plants that thrive in salt water must have...Ch. 17 - Prob. 131CPCh. 17 - Prob. 132CPCh. 17 - Prob. 133CPCh. 17 - Prob. 134CPCh. 17 - Prob. 135CPCh. 17 - Prob. 136CP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- A 1.00 mol/kg aqueous sulfuric acid solution, H2SO4,freezes at 4.04 C. Calculate i, the vant Hoff factor,for sulfuric acid in this solution.arrow_forwardWhat is the freezing point of a solution of dibromobenzene, C6H4Br2, in 0.250 kg of benzene, if the solution boils at 83.5 C?arrow_forwardCalcium chloride, CaCl2, has been used to melt ice from roadways. Given that the saturated solution is 32% CaCl2 by mass, estimate the freezing point.arrow_forward
- What quantity of ethylene glycol, HOCH2CH2OH, must be added to 125 g of water to raise the boiling point by 1.0 C? Express the answer in grams.arrow_forwardA forensic chemist is given a white solid that is suspected of being pure cocaine (C17H21NO4, molar mass = 303.35 g/mol). She dissolves 1.22 0.01 g of the solid in 15.60 0.01 g benzene. The freezing point is lowered by 1.32 0.04C. a. What is the molar mass of the substance? Assuming that the percent uncertainty in the calculated molar mass is the same as the percent uncertainty in the temperature change, calculate the uncertainty in the molar mass. b. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine (C18H21NO3, molar mass = 299.36 g/mol)? c. Assuming that the absolute uncertainties in the measurements of temperature and mass remain unchanged, how could the chemist improve the precision of her results?arrow_forwardFreezing point depression is one means of determining the molar mass of a compound. The freezing point depression constant of benzene is 5.12 C/m. a. When a 0.503 g sample of the white crystalline dimer is dissolved in 10.0 g benzene, the freezing point of benzene is decreased by 0542 C. Verify that the molar mass of the dimer is 475 g/mol when determined by freezing point depression. Assume no dissociation of the dimer occurs. b. The correct molar mass of the dimer is 487 g/mol. Explain why the dissociation equilibrium causes the freezing point depression calculation to yield a lower molar mass for the dimer.arrow_forward
- The organic salt [(C4H9)4N][ClO4] consists of the ions (C4H9)4N+ and ClO4. The salt dissolves in chloroform. What mass (in grams) of the salt must have been dissolved if the boiling point of a solution of the salt in 25.0 g chloroform is 63.20 C? The normal boiling point of chloroform is 61.70 C and Kb = 3.63 C kg mol1. Assume that the salt dissociates completely into its ions in solution.arrow_forwardWhat would be the freezing point of a solution formed by adding 1.0 mole of glucose (a molecular compound) to the following amounts of water? a. 250 g (0.25 kg) b. 500 g (0.500 kg) c. 1000 g (1.000 kg) d. 2000 g (2.000 kg)arrow_forwardCalculate the molality of a solution made by dissolving 115.0 g ethylene glycol, HOCH2CH2OH, in 500. mL water. The density of water at this temperature is 0.978 g/mL. Calculate the molarity of the solution.arrow_forward
- A solution was prepared by dissolving 0.800 g of sulfur, Sg, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and boiling point of the solution.arrow_forwardSodium chloride (NaCl) is commonly used to melt ice on roads during the winter. Calcium chloride (CaCl2) is sometimes used for this purpose too. Let us compare the effectiveness of equal masses of these two compounds in lowering the freezing point of water, by calculating the freezing point depression of solutions containing 200. g of each salt in 1.00 kg of water. (An advantage of CaCl2 is that it acts more quickly because it is hygroscopic, that is. it absorbs moisture from the air to give a solution and begin the process. A disadvantage is that this compound is more costly.)arrow_forwardCalculate the freezing point and normal boiling points of each of the following aqueous solutions. (a) 2.63 m acetic acid (b) 33.0 % by mass lactose, C12H22O11 (c) 32.15 mL of ethylene glycol, C2H6O2(d=1.113g/mL) in 624 mL of water (d=1.00g/mL)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
- Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY