MCGRAW: CHEMISTRY THE MOLECULAR NATURE
MCGRAW: CHEMISTRY THE MOLECULAR NATURE
8th Edition
ISBN: 9781264330430
Author: VALUE EDITION
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 13, Problem 13.77P

(a)

Interpretation Introduction

Interpretation:

The volume percent of ethylene glycol is to be calculated.

Concept introduction:

The volume percent is defined as the volume of the component divided by the volume of solution, multiplied by 100.

The formula to calculate the volume percent of the solution is as follows:

  Volume percent=(Volume of componentVolume of solution)(100 %)        (1)

The formula to calculate the density is as follows:

  Density(ρ)=Mass (M)Volume (V)        (2)

(a)

Expert Solution
Check Mark

Answer to Problem 13.77P

The volume percent of ethylene glycol is 50.61%.

Explanation of Solution

Rearrange equation (2) to calculate the mass as follows:

  Mass=(Density)(Volume)        (3)

Consider the volumes of ethylene glycol and water to be 1L.

Substitute 1.114g/mL for the density and 1L for the volume in equation (3) to calculate the mass of ethylene glycol.

  Mass of ethylene glycol=(1L)(103mL1L)(1.114g1mL)=1114g

Substitute 1g/mL for the density and 1L for the volume in equation (3) to calculate the mass of water.

  Mass of water=(1L)(103mL1L)(1g1mL)=1×103g

The formula to calculate the moles of the compound is as follows:

  Moles of compound=Given massMolar mass        (4)

Substitute 1114g for the given mass and 62.07g/mol for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  Moles of ethylene glycol=(1114g)(1mol62.07g)=17.94747865mol

Substitute 1×103g for the given mass and 18.02g/mol for the molar mass in equation (4) to calculate the moles of water.

  Moles of water=(1×103g)(1mol18.02g)=55.49389567mol

Rearrange equation (2) to calculate the volume as follows:

  Volume=MassDensity        (5)

The formula to calculate the mass of the solution is as follows:

  Mass of solution=(Mass of solute+Mass of solvent)        (6)

Substitute 1114g for the mass of solute and 1×103g for the mass of solvent in equation (6).

  Mass of solution=(1114g)+(1×103g)=2114g

Substitute 2114g for the mass and 1.070g/mL for the density in equation (5) to calculate the volume of the solution.

  Volume of solution=(2114g)(1mL1.070g)(1L103mL)=1.97570L

Substitute 1L for the volume of component and 1.97570L for the volume of solution in equation (1) to calculate the volume percent of ethylene glycol.

  Volume % of ethylene glycol=(1L1.97570L)(100 %)=50.61497%50.61%.

Conclusion

The volume percent is a concentration term that changes with the change in temperature as it includes the volumes of component and volume of solution.

(b)

Interpretation Introduction

Interpretation:

The mass percent of ethylene glycol is to be calculated.

Concept introduction:

Mass percent is defined as the mass of a component divided by the total mass of the mixture, multiplied by 100.

The formula to calculate the mass percent is as follows:

  Masspercent=(massofsolutemassofsolution)(100 %)        (7)

The formula to calculate the density is as follows:

  Density(ρ)=Mass (M)Volume (V)        (2)

(b)

Expert Solution
Check Mark

Answer to Problem 13.77P

52.7% is the volume percent of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be 1L.

Substitute 1.114g/mL for the density and 1L for the volume in equation (3) to calculate the mass of ethylene glycol.

  Mass of ethylene glycol=(1L)(103mL1L)(1.114g1mL)=1114g

Substitute 1g/mL for the density and 1L for the volume in equation (3) to calculate the mass of water.

  Mass of water=(1L)(103mL1L)(1g1mL)=1×103g

Substitute 1114g for the mass of solute and 1×103g for the mass of solvent in equation (6).

  Mass of solution=1114g+1×103g=2114g

Substitute 1114g for the mass of solute and 2114g for the mass of solution in equation (7).

  Mass % of ethylene glycol=(1114g2114g)(100 %)=52.6963%52.7%.

Conclusion

The mass percent is a concentration term that does not depend on the temperature as it depends only on the masses of solute and solution.

(c)

Interpretation Introduction

Interpretation:

The molarity of the ethylene glycol is to be calculated.

Concept introduction:

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by M and its unit is mol/L.

The formula to calculate the molarity of the solution is as follows:

  Molarity(M)=amount(mol)ofsolutevolume (L)ofsolution        (8)

The formula to calculate the density is as follows:

  Density(ρ)=Mass (M)Volume (V)        (2)

(c)

Expert Solution
Check Mark

Answer to Problem 13.77P

9.08M is the molarity of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be 1L.

Substitute 1.114g/mL for the density and 1L for the volume in equation (3) to calculate the mass of ethylene glycol.

  Mass of ethylene glycol=(1L)(103mL1L)(1.114g1mL)=1114g

Substitute 1g/mL for the density and 1L for the volume in equation (3) to calculate the mass of water.

  Mass of water=(1L)(103mL1L)(1g1mL)=1×103g

Substitute 1114g for the given mass and 62.07g/mol for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  Moles of ethylene glycol=(1114g)(1mol62.07g)=17.94747865mol

Substitute 1114g for the mass of solute and 1×103g for the mass of solvent in equation (6).

  Mass of solution=(1114g)+(1×103g)=2114g

Substitute 2114g for the mass and 1.070g/mL for the density in equation (5) to calculate the volume of the solution.

  Volume of solution=(2114g)(1mL1.070g)(1L103mL)=1.97570L

Substitute 17.94747865mol for the amount of solute and 1.97570L for the volume of solution in equation (8) to calculate the molarity of ethylene glycol.

  Molarity of ethylene glycol=17.94747865mol1.97570L=9.08411M9.08M.

Conclusion

Molarity is a concentration term that changes with the change in temperature as it includes a volume term which is temperature dependent.

(d)

Interpretation Introduction

Interpretation:

The molality of ethylene glycol is to be calculated.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by m and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the molality of the solution is as follows:

  Molality=amount(mol)ofsolutemass(kg)ofsolvent        (9)

The formula to calculate the density is as follows:

  Density(ρ)=Mass (M)Volume (V)        (2)

(d)

Expert Solution
Check Mark

Answer to Problem 13.77P

17.9m is the molality of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be 1L.

Substitute 1.114g/mL for the density and 1L for the volume in equation (3) to calculate the mass of ethylene glycol.

  Mass of ethylene glycol=(1L)(103mL1L)(1.114g1mL)=1114g

Substitute 1g/mL for the density and 1L for the volume in equation (3) to calculate the mass of water.

  Mass of water=(1L)(103mL1L)(1g1mL)=1×103g

Substitute 1114g for the given mass and 62.07g/mol for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  Moles of ethylene glycol=(1114g)(1mol62.07g)=17.94747865mol

Substitute 17.94747865mol for the amount of solute and 1×103g for the mass of solvent in equation (9) to calculate the molality of ethylene glycol.

  Molarity of ethylene glycol=(17.94747865mol1×103g)(103g1kg)=17.94747865m17.9m.

Conclusion

Molality is a concentration term that does not change with the change in temperature as it depends on the amount of solute and mass of the solvent.

(e)

Interpretation Introduction

Interpretation:

The mole fraction of ethylene glycol is to be determined.

Concept introduction:

The mole fraction is defined as the ratio of the number of moles of solute to the total number of moles in the mixture. It is represented by X.

The formula to calculate the mole fraction is as follows:

  Molefraction(X)=amount(mol)ofsolute(amount(mol)ofsolute+amount(mol)ofsolvent)        (10)

The formula to calculate the density is as follows:

  Density(ρ)=Mass (M)Volume (V)        (2)

(e)

Expert Solution
Check Mark

Answer to Problem 13.77P

0.244 is the mole fraction of ethylene glycol.

Explanation of Solution

Consider the volumes of ethylene glycol and water to be 1L.

Substitute 1.114g/mL for the density and 1L for the volume in equation (3) to calculate the mass of ethylene glycol.

  Mass of ethylene glycol=(1L)(103mL1L)(1.114g1mL)=1114g

Substitute 1g/mL for the density and 1L for the volume in equation (3) to calculate the mass of water.

  Mass of water=(1L)(103mL1L)(1g1mL)=1×103g

Substitute 1114g for the given mass and 62.07g/mol for the molar mass in equation (4) to calculate the moles of ethylene glycol.

  Moles of ethylene glycol=(1114g)(1mol62.07g)=17.94747865mol

Substitute 1×103g for the given mass and 18.02g/mol for the molar mass in equation (4) to calculate the moles of water.

  Moles of water=(1×103g)(1mol18.02g)=55.49389567mol

Substitute 17.94747865mol for the amount of solute and 55.49389567mol for the amount of solvent in equation (10) to calculate the mole fraction of ethylene glycol.

  Mole fraction of ethylene glycol=17.94747865mol(17.94747865mol+55.49389567mol)=0.2443780.244.

Conclusion

Mole fraction is a concentration term that does not depend on the temperature as it depends only on the amounts of solute and solvent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 13 Solutions

MCGRAW: CHEMISTRY THE MOLECULAR NATURE

Ch. 13.5 - Prob. 13.6AFPCh. 13.5 - Prob. 13.6BFPCh. 13.6 - Calculate the vapor pressure lowering of a...Ch. 13.6 - Prob. 13.7BFPCh. 13.6 - Prob. 13.8AFPCh. 13.6 - Prob. 13.8BFPCh. 13.6 - Prob. 13.9AFPCh. 13.6 - Prob. 13.9BFPCh. 13.6 - A solution is made by dissolving 31.2 g of...Ch. 13.6 - Prob. 13.10BFPCh. 13.7 - Prob. B13.1PCh. 13.7 - Prob. B13.2PCh. 13 - Prob. 13.1PCh. 13 - Prob. 13.2PCh. 13 - Prob. 13.3PCh. 13 - Which would you expect to be more effective as a...Ch. 13 - Prob. 13.5PCh. 13 - Prob. 13.6PCh. 13 - Prob. 13.7PCh. 13 - Prob. 13.8PCh. 13 - Prob. 13.9PCh. 13 - Prob. 13.10PCh. 13 - Prob. 13.11PCh. 13 - What is the strongest type of intermolecular force...Ch. 13 - Prob. 13.13PCh. 13 - Prob. 13.14PCh. 13 - Prob. 13.15PCh. 13 - Prob. 13.16PCh. 13 - Prob. 13.17PCh. 13 - Prob. 13.18PCh. 13 - Prob. 13.19PCh. 13 - Prob. 13.20PCh. 13 - Prob. 13.21PCh. 13 - Prob. 13.22PCh. 13 - Prob. 13.23PCh. 13 - What is the relationship between solvation and...Ch. 13 - Prob. 13.25PCh. 13 - Prob. 13.26PCh. 13 - Prob. 13.27PCh. 13 - Prob. 13.28PCh. 13 - Prob. 13.29PCh. 13 - Prob. 13.30PCh. 13 - Prob. 13.31PCh. 13 - Prob. 13.32PCh. 13 - Prob. 13.33PCh. 13 - Prob. 13.34PCh. 13 - Prob. 13.35PCh. 13 - Use the following data to calculate the combined...Ch. 13 - Use the following data to calculate the combined...Ch. 13 - State whether the entropy of the system increases...Ch. 13 - Prob. 13.39PCh. 13 - Prob. 13.40PCh. 13 - Prob. 13.41PCh. 13 - Prob. 13.42PCh. 13 - Prob. 13.43PCh. 13 - Prob. 13.44PCh. 13 - For a saturated aqueous solution of each of the...Ch. 13 - Prob. 13.46PCh. 13 - Prob. 13.47PCh. 13 - Prob. 13.48PCh. 13 - Prob. 13.49PCh. 13 - Prob. 13.50PCh. 13 - Prob. 13.51PCh. 13 - Prob. 13.52PCh. 13 - Prob. 13.53PCh. 13 - Prob. 13.54PCh. 13 - Prob. 13.55PCh. 13 - Calculate the molarity of each aqueous...Ch. 13 - Calculate the molarity of each aqueous...Ch. 13 - Prob. 13.58PCh. 13 - Calculate the molarity of each aqueous...Ch. 13 - How would you prepare the following aqueous...Ch. 13 - Prob. 13.61PCh. 13 - Prob. 13.62PCh. 13 - Prob. 13.63PCh. 13 - Prob. 13.64PCh. 13 - Prob. 13.65PCh. 13 - Prob. 13.66PCh. 13 - Prob. 13.67PCh. 13 - Prob. 13.68PCh. 13 - Prob. 13.69PCh. 13 - Prob. 13.70PCh. 13 - Prob. 13.71PCh. 13 - Prob. 13.72PCh. 13 - Prob. 13.73PCh. 13 - Prob. 13.74PCh. 13 - Prob. 13.75PCh. 13 - Prob. 13.76PCh. 13 - Prob. 13.77PCh. 13 - Prob. 13.78PCh. 13 - Prob. 13.79PCh. 13 - Prob. 13.80PCh. 13 - Prob. 13.81PCh. 13 - What are the most important differences between...Ch. 13 - Prob. 13.83PCh. 13 - Prob. 13.84PCh. 13 - Prob. 13.85PCh. 13 - Prob. 13.86PCh. 13 - Prob. 13.87PCh. 13 - Prob. 13.88PCh. 13 - Classify each substance as a strong electrolyte,...Ch. 13 - Prob. 13.90PCh. 13 - Prob. 13.91PCh. 13 - Which solution has the lower freezing point? 11.0...Ch. 13 - Prob. 13.93PCh. 13 - Prob. 13.94PCh. 13 - Prob. 13.95PCh. 13 - Prob. 13.96PCh. 13 - Prob. 13.97PCh. 13 - Prob. 13.98PCh. 13 - Prob. 13.99PCh. 13 - The boiling point of ethanol (C2H5OH) is 78.5°C....Ch. 13 - Prob. 13.101PCh. 13 - Prob. 13.102PCh. 13 - Prob. 13.103PCh. 13 - Prob. 13.104PCh. 13 - Prob. 13.105PCh. 13 - Prob. 13.106PCh. 13 - Prob. 13.107PCh. 13 - Prob. 13.108PCh. 13 - Prob. 13.109PCh. 13 - Prob. 13.110PCh. 13 - Prob. 13.111PCh. 13 - In a study designed to prepare new...Ch. 13 - The U.S. Food and Drug Administration lists...Ch. 13 - Prob. 13.114PCh. 13 - Prob. 13.115PCh. 13 - Prob. 13.116PCh. 13 - In a movie theater, you can see the beam of...Ch. 13 - Prob. 13.118PCh. 13 - Prob. 13.119PCh. 13 - Prob. 13.120PCh. 13 - Prob. 13.121PCh. 13 - Gold occurs in seawater at an average...Ch. 13 - Prob. 13.123PCh. 13 - Prob. 13.124PCh. 13 - Prob. 13.125PCh. 13 - Prob. 13.126PCh. 13 - Pyridine (right) is an essential portion of many...Ch. 13 - Prob. 13.128PCh. 13 - Prob. 13.129PCh. 13 - Prob. 13.130PCh. 13 - Prob. 13.131PCh. 13 - Prob. 13.132PCh. 13 - Prob. 13.133PCh. 13 - Prob. 13.134PCh. 13 - Prob. 13.135PCh. 13 - Prob. 13.136PCh. 13 - Prob. 13.137PCh. 13 - Prob. 13.138PCh. 13 - Prob. 13.139PCh. 13 - Prob. 13.140PCh. 13 - Prob. 13.141PCh. 13 - Prob. 13.142PCh. 13 - Prob. 13.143PCh. 13 - The release of volatile organic compounds into the...Ch. 13 - Although other solvents are available,...Ch. 13 - Prob. 13.146PCh. 13 - Prob. 13.147PCh. 13 - Prob. 13.148PCh. 13 - Prob. 13.149PCh. 13 - Prob. 13.150PCh. 13 - Prob. 13.151PCh. 13 - Suppose coal-fired power plants used water in...Ch. 13 - Urea is a white crystalline solid used as a...Ch. 13 - Prob. 13.154PCh. 13 - Prob. 13.155PCh. 13 - Prob. 13.156PCh. 13 - Prob. 13.157PCh. 13 - Prob. 13.158PCh. 13 - Prob. 13.159PCh. 13 - Prob. 13.160PCh. 13 - Prob. 13.161PCh. 13 - Prob. 13.162PCh. 13 - Figure 12.11 shows the phase changes of pure...Ch. 13 - KNO3, KClO3, KCl, and NaCl are recrystallized as...Ch. 13 - Prob. 13.165PCh. 13 - Prob. 13.166PCh. 13 - Prob. 13.167P
Knowledge Booster
Background pattern image
Chemistry
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.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY