Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
15th Edition
ISBN: 9781119231318
Author: Morris Hein
Publisher: Wiley (WileyPLUS Products)
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 14, Problem 95AE
Interpretation Introduction
Interpretation:
Molality of salt in rivers if it freezes at
Concept Introduction:
Colligative properties are dependent on amount of solute and not on their natures. These are caused due to addition of nonvolatile solute in any solution. Below mentioned are colligative properties.
1. Freezing point depression
2. Boiling point elevation
3. Vapor pressure lowering
4. Osmotic pressure
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Chapter 14 Solutions
Foundations of College Chemistry 15e Binder Ready Version + WileyPLUS Registration Card
Ch. 14.1 - Prob. 14.1PCh. 14.2 - Prob. 14.2PCh. 14.3 - Prob. 14.3PCh. 14.4 - Prob. 14.4PCh. 14.4 - Prob. 14.5PCh. 14.4 - Prob. 14.6PCh. 14.4 - Prob. 14.7PCh. 14.4 - Prob. 14.8PCh. 14.4 - Prob. 14.9PCh. 14.4 - Prob. 14.10P
Ch. 14.5 - Prob. 14.11PCh. 14.5 - Prob. 14.12PCh. 14 - Prob. 1RQCh. 14 - Prob. 2RQCh. 14 - Prob. 3RQCh. 14 - Prob. 4RQCh. 14 - Prob. 5RQCh. 14 - Prob. 6RQCh. 14 - Prob. 7RQCh. 14 - Prob. 8RQCh. 14 - Prob. 9RQCh. 14 - Prob. 10RQCh. 14 - Prob. 11RQCh. 14 - Prob. 12RQCh. 14 - Prob. 13RQCh. 14 - Prob. 14RQCh. 14 - Prob. 15RQCh. 14 - Prob. 16RQCh. 14 - Prob. 17RQCh. 14 - Prob. 18RQCh. 14 - Prob. 19RQCh. 14 - Prob. 20RQCh. 14 - Prob. 21RQCh. 14 - Prob. 22RQCh. 14 - Prob. 23RQCh. 14 - Prob. 24RQCh. 14 - Prob. 25RQCh. 14 - Prob. 26RQCh. 14 - Prob. 27RQCh. 14 - Prob. 28RQCh. 14 - Prob. 29RQCh. 14 - Prob. 30RQCh. 14 - Prob. 31RQCh. 14 - Prob. 32RQCh. 14 - Prob. 33RQCh. 14 - Prob. 34RQCh. 14 - Prob. 35RQCh. 14 - Prob. 37RQCh. 14 - Prob. 38RQCh. 14 - Prob. 39RQCh. 14 - Prob. 40RQCh. 14 - Prob. 41RQCh. 14 - Prob. 42RQCh. 14 - Prob. 1PECh. 14 - Prob. 2PECh. 14 - Prob. 3PECh. 14 - Prob. 4PECh. 14 - Prob. 5PECh. 14 - Prob. 6PECh. 14 - Prob. 7PECh. 14 - Prob. 8PECh. 14 - Prob. 9PECh. 14 - Prob. 10PECh. 14 - Prob. 11PECh. 14 - Prob. 12PECh. 14 - Prob. 13PECh. 14 - Prob. 14PECh. 14 - Prob. 15PECh. 14 - Prob. 16PECh. 14 - Prob. 17PECh. 14 - Prob. 18PECh. 14 - Prob. 19PECh. 14 - Prob. 20PECh. 14 - Prob. 21PECh. 14 - Prob. 22PECh. 14 - Prob. 23PECh. 14 - Prob. 24PECh. 14 - Prob. 25PECh. 14 - Prob. 26PECh. 14 - Prob. 27PECh. 14 - Prob. 28PECh. 14 - Prob. 29PECh. 14 - Prob. 30PECh. 14 - Prob. 31PECh. 14 - Prob. 32PECh. 14 - Prob. 33PECh. 14 - Prob. 34PECh. 14 - Prob. 35PECh. 14 - Prob. 36PECh. 14 - Prob. 37PECh. 14 - Prob. 38PECh. 14 - Prob. 39PECh. 14 - Prob. 40PECh. 14 - Prob. 41PECh. 14 - Prob. 42PECh. 14 - Prob. 44PECh. 14 - Prob. 45PECh. 14 - Prob. 46PECh. 14 - Prob. 47PECh. 14 - Prob. 48PECh. 14 - Prob. 49PECh. 14 - Prob. 50PECh. 14 - Prob. 51PECh. 14 - Prob. 52PECh. 14 - Prob. 53AECh. 14 - Prob. 54AECh. 14 - Prob. 55AECh. 14 - Prob. 56AECh. 14 - Prob. 57AECh. 14 - Prob. 58AECh. 14 - Prob. 59AECh. 14 - Prob. 60AECh. 14 - Prob. 61AECh. 14 - Prob. 62AECh. 14 - Prob. 63AECh. 14 - Prob. 65AECh. 14 - Prob. 66AECh. 14 - Prob. 67AECh. 14 - Prob. 68AECh. 14 - Prob. 69AECh. 14 - Prob. 70AECh. 14 - Prob. 71AECh. 14 - Prob. 72AECh. 14 - Prob. 73AECh. 14 - Prob. 74AECh. 14 - Prob. 75AECh. 14 - Prob. 76AECh. 14 - Prob. 77AECh. 14 - Prob. 78AECh. 14 - Prob. 79AECh. 14 - Prob. 80AECh. 14 - Prob. 81AECh. 14 - Prob. 82AECh. 14 - Prob. 83AECh. 14 - Prob. 84AECh. 14 - Prob. 85AECh. 14 - Prob. 86AECh. 14 - Prob. 87AECh. 14 - Prob. 88AECh. 14 - Prob. 90AECh. 14 - Prob. 91AECh. 14 - Prob. 92AECh. 14 - Prob. 93AECh. 14 - Prob. 94AECh. 14 - Prob. 95AECh. 14 - Prob. 96AECh. 14 - Prob. 97AECh. 14 - Prob. 98AECh. 14 - Prob. 99CECh. 14 - Prob. 100CECh. 14 - Prob. 102CECh. 14 - Prob. 103CECh. 14 - Prob. 104CECh. 14 - Prob. 105CE
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
- Calculate 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_forward6-111 As noted in Section 6-8C, the amount of external pressure that must be applied to a more concentrated solution to stop the passage of solvent molecules across a semipermeable membrane is known as the osmotic pressure The osmotic pressure obeys a law similar in form to the ideal gas law (discussed in Section 5-4), where Substituting for pressure and solving for osmotic pressures gives the following equation: RT MRT, where M is the concentration or molarity of the solution. (a) Determine the osmotic pressure at 25°C of a 0.0020 M sucrose (C12H22O11) solution. (b) Seawater contains 3.4 g of salts for every liter of solution. Assuming the solute consists entirely of NaCl (and complete dissociation of the NaCI salt), calculate the osmotic pressure of seawater at 25°C. (c) The average osmotic pressure of blood is 7.7 atm at 25°C. What concentration of glucose (C6H12O6) will be isotonic with blood? (d) Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing 0.150 g of this enzyme in 210. mL of solution has an osmotic pressure of 0.953 torr at 25°C. What is the molar mass of lysozyme? (e) The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine the protein's molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein.arrow_forwardWater at 25 C has a density of 0.997 g/cm3. Calculate the molality and molarity of pure water at this temperature.arrow_forward
- Sodium 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_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_forwardA patient has a “cholesterol count” of 214. Like manyblood-chemistry measurements,this result is measured inunits of milligrams per deciliter (mgdL1). Determine the molar concentration of cholesterol inthis patient’s blood, taking the molar mass of cholesterolto be 386.64gmol1. Estimate the molality of cholesterol in the patient’sblood. If 214 is a typical cholesterol reading among men inthe United States, determine the volume of such bloodrequired to furnish 8.10 g of cholesterol.arrow_forward
- A compound has a solubility in water of 250 mg/L at 25C. Should this compound be characterized as a soluble or insoluble compound at 25C?arrow_forwardThe 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_forwardFor each of the following pairs of solutions, select the solution for which solute solubility is greatest. a. Oxygen gas in water with P = 1 atm and T = 10C Oxygen gas in water with P = 1 atm and T = 20C b. Nitrogen gas in water with P = 2 atm and T = 50C Nitrogen gas in water with P = 1 atm and T = 70C c. Table salt in water with P = 1 atm and T = 40C Table salt in water with P = 1 atm and T = 70C d. Table sugar in water with P = 3 atm and T = 30C Table sugar in water with P = 1 atm and T = 80Carrow_forward
- Simple acids such as formic acid, HCOOH, and acetic acid, CH3COOH, are very soluble in water; however, fatty acids such as stearic acid, CH3(CH2)16COOH, and palmitic acid, CH3(CH2)14COOH, are water-insoluble. Based on what you know about the solubility of alcohols, explain the solubility of these organic acids.arrow_forwardConsider three test tubes. Tube A has pure water. Tube B has an aqueous 1.0 m solution of ethanol, C2H5OH. Tube C has an aqueous 1.0 m solution of NaCl. Which of the following statements are true? (Assume that for these solutions 1.0m=1.0M.) (a) The vapor pressure of the solvent over tube A is greater than the solvent pressure over tube B. (b) The freezing point of the solution in tube B is higher than the freezing point of the solution in tube A. (c) The freezing point of the solution in tube B is higher than the freezing point of the solution in tube C. (d) The boiling point of the solution in tube B is higher than the boiling point of the solution in tube C. (e) The osmotic pressure of the solution in tube B is greater than the osmotic pressure of the solution in tube C.arrow_forwardA 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_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning
Introduction to General, Organic and BiochemistryChemistryISBN:9781285869759Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar TorresPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY