Chemistry for Today: General, Organic, and Biochemistry
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN: 9781305960060
Author: Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher: Cengage Learning
Question
Book Icon
Chapter 7, Problem 7.66E
Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of a 0.50M solution of urea, a nonelectrolyte, are to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Expert Solution
Check Mark

Answer to Problem 7.66E

The boiling and freezing points of a 0.50M solution of urea, a nonelectrolyte are 100.26°C and 0.93°C respectively.

Explanation of Solution

The formula to calculate boiling point is given below as,

ΔTb=nKbM

Where,

n is the number of ions in the solution.

Kb is the boiling point constant which is defined for a particular solvent.

M is the molarity of solution.

Since urea is a nonelectrolyte, it will not dissociate in the solution and the value of n for urea is 1. The value of Kb for water is 0.52°C/M.

Substitute the value of n, Kb and M in the above equation as follows.

ΔTb=nKbM=1×0.52°C/M×0.50M=0.26°C

Now, the boiling point of water solution can be calculated by adding value of ΔTb as given below.

Boiling point=100°C+0.26°C=100.26°C

The formula to calculate freezing point is given below as,

ΔTf=nKfM

Where,

n is the number of ions in the solution.

Kf is the freezing point constant which is defined for a particular solvent.

M is the molarity of solution.

Since urea is a nonelectrolyte, it will not dissociate in the solution and the value of n for urea is 1. The value of Kf for water is 1.86°C/M.

Substitute the value of n, Kf and M in the above equation as follows.

ΔTf=nKfM=1×1.86°C/M×0.50M=0.93°C

Now, the freezing point of water solution can be calculated by subtracting value of ΔTf as given below.

Freezing point=0°C0.93°C=0.93°C

Conclusion

The boiling and freezing points of a 0.50M solution of urea, a nonelectrolyte are 100.26°C and 0.93°C respectively.

Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of a 0.250M solution of CaCl2, a strong electrolyteare to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Expert Solution
Check Mark

Answer to Problem 7.66E

The boiling and freezing points of a 0.250M solution of CaCl2, a strong electrolyte are 100.39°C and 1.4°C respectively.

Explanation of Solution

The formula to calculate boiling point is given below as,

ΔTb=nKbM

Where,

n is the number of ions in the solution.

Kb is the boiling point constant which is defined for a particular solvent.

M is the molarity of solution.

Since CaCl2 is a strong electrolyte it will dissociate in the solution completely and the value of n for CaCl2 is 3. The value of Kb for water is 0.52°C/M.

Substitute the value of n, Kb and M in the above equation as follows.

ΔTb=nKbM=3×0.52°C/M×0.250M=0.39°C

Now, the boiling point of water solution can be calculated by adding value of ΔTb as given below.

Boiling point=100°C+0.39°C=100.39°C

The formula to calculate freezing point is given below as,

ΔTf=nKfM

Where,

n is the number of ions in the solution.

Kf is the freezing point constant which is defined for a particular solvent.

M is the molarity of solution.

Since CaCl2 is a strong electrolyte it will dissociate in the solution completely and the value of n for CaCl2 is 3. The value of Kf for water is 1.86°C/M.

Substitute the value of n, Kf and M in the above equation as follows.

ΔTf=nKfM=3×1.86°C/M×0.250M=1.395°C

Now, the freezing point of water solution can be calculated by subtracting value of ΔTf as given below.

Freezing point=0°C1.395°C=1.395°C1.4°C

Conclusion

The boiling and freezing points of a 0.250M solution of CaCl2, a strong electrolyte are 100.39°C and 1.4°C respectively.

Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of a solution containing 100.g of ethylene glycol (C2H6O2), a nonelectrolyte, per 250.mL are to be calculated.

Concept introduction:

Solutes which give conducting solutions on dissolution are called electrolytes. Those which dissociate completely in the solution are known as strong electrolytes. Solutes which do not give conducting solutions are called nonelectrolytes. The properties which depend on the number of solute particles are known as colligative properties. Some of these properties are boiling point, freezing point and osmotic pressure.

Expert Solution
Check Mark

Answer to Problem 7.66E

The boiling and freezing points of a solution containing 100.g of ethylene glycol (C2H6O2), a nonelectrolyte, per 250.mL are 103.354°C and 12°C respectively.

Explanation of Solution

The formula to calculate number of moles of solutes is given below as,

Molesofsolute=GivenmassofsoluteMolarmassofsolute

The molar mass of ethylene glycol can be calculated as follows.

C2H6O2=(2×C)+(6×H)+(2×O)=(2×12)+(6×1)+(2×16)=62g/mol

Substitute the values in the above equation as follows.

MolesofC2H6O2=GivenmassofC2H6O2MolarmassofC2H6O2=100g62g/mol=1.6129mol

The formula to calculate molarity is given below as,

M=molesofsolutelitersofsolution

Substitute the value of number of moles of solute and volume of solution in the above equation as follows.

M=molesofC2H6O2litersofC2H6O2solution=1.6129mol250mL×1L1000mL=6.45M

The formula to calculate boiling point is given below as,

ΔTb=nKbM

Where,

n is the number of ions in the solution.

Kb is the boiling point constant which is defined for a particular solvent.

M is the molarity of solution.

Since ethylene glycol is a nonelectrolyte, it will not dissociate in the solution and the value of n for ethylene glycol is 1. The value of Kb for water is 0.52°C/M.

Substitute the value of n, Kb and M in the above equation as follows.

ΔTb=nKbM=1×0.52°C/M×6.45M=3.354°C

Now, the boiling point of water solution can be calculated by adding value of ΔTb as given below.

Boiling point=100°C+3.354°C=103.354°C

The formula to calculate freezing point is given below as,

ΔTf=nKfM

Where,

n is the number of ions in the solution.

Kf is the freezing point constant which is defined for a particular solvent.

M is the molarity of solution.

Since ethylene glycol is a nonelectrolyte, it will not dissociate in the solution and the value of n for ethylene glycol is 1. The value of Kf for water is 1.86°C/M.

Substitute the value of n, Kf and M in the above equation as follows.

ΔTf=nKfM=1×1.86°C/M×6.45M=11.997°C

Now, the freezing point of water solution can be calculated by subtracting value of ΔTf as given below.

Freezing point=0°C11.997°C=11.997°C12°C

Conclusion

The boiling and freezing points of a solution containing 100.g of ethylene glycol (C2H6O2), a nonelectrolyte, per 250.mL are 103.354°C and 12°C respectively.

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 7 Solutions

Chemistry for Today: General, Organic, and Biochemistry

Ch. 7 - Prob. 7.11ECh. 7 - Classify each of the following solutes into the...Ch. 7 - Prob. 7.13ECh. 7 - Prob. 7.14ECh. 7 - Prob. 7.15ECh. 7 - Prob. 7.16ECh. 7 - Prob. 7.17ECh. 7 - Prob. 7.18ECh. 7 - Prob. 7.19ECh. 7 - Prob. 7.20ECh. 7 - Prob. 7.21ECh. 7 - Prob. 7.22ECh. 7 - Calculate the molarity of the following solutions:...Ch. 7 - Prob. 7.24ECh. 7 - Prob. 7.25ECh. 7 - Calculate: a. How many grams of solid would be...Ch. 7 - Prob. 7.27ECh. 7 - Prob. 7.28ECh. 7 - Calculate the concentration in (w/w) of the...Ch. 7 - Calculate the concentration in (w/w) of the...Ch. 7 - Prob. 7.31ECh. 7 - Calculate the concentration in (w/w) of the...Ch. 7 - Prob. 7.33ECh. 7 - Calculate the concentration in (v/v) of the...Ch. 7 - Calculate the concentration in (v/v) of the...Ch. 7 - Consider the blood volume of an adult to be 5.0L....Ch. 7 - Prob. 7.37ECh. 7 - Calculate the concentration in (w/v) of the...Ch. 7 - Calculate the concentration in (w/v) of the...Ch. 7 - Prob. 7.40ECh. 7 - Prob. 7.41ECh. 7 - Prob. 7.42ECh. 7 - Explain how you would prepare the following...Ch. 7 - Prob. 7.44ECh. 7 - Prob. 7.45ECh. 7 - Calculate the following: a. The number of grams of...Ch. 7 - Prob. 7.47ECh. 7 - Explain how you would prepare the following dilute...Ch. 7 - Prob. 7.49ECh. 7 - Prob. 7.50ECh. 7 - Prob. 7.51ECh. 7 - How many grams of solid Na2CO3 will react with...Ch. 7 - Prob. 7.53ECh. 7 - Prob. 7.54ECh. 7 - Prob. 7.55ECh. 7 - Prob. 7.56ECh. 7 - How many milliliters of 0.124MNaOH solution will...Ch. 7 - How many milliliters of 0.124MNaOH solution will...Ch. 7 - How many milliliters of 0.115MNaOH solution will...Ch. 7 - Stomach acid is essentially 0.10MHCl. An active...Ch. 7 - Prob. 7.61ECh. 7 - Prob. 7.62ECh. 7 - Prob. 7.63ECh. 7 - Calculate the boiling and freezing points of water...Ch. 7 - Calculate the boiling and freezing points of water...Ch. 7 - Prob. 7.66ECh. 7 - Prob. 7.67ECh. 7 - Prob. 7.68ECh. 7 - Calculate the osmolarity for the following...Ch. 7 - Prob. 7.70ECh. 7 - Calculate the osmotic pressure of a 0.125M...Ch. 7 - Prob. 7.72ECh. 7 - Prob. 7.73ECh. 7 - Calculate the osmotic pressure of a solution that...Ch. 7 - Prob. 7.75ECh. 7 - Prob. 7.77ECh. 7 - Prob. 7.78ECh. 7 - Prob. 7.79ECh. 7 - Suppose an osmotic membrane separates a 5.00 sugar...Ch. 7 - Prob. 7.81ECh. 7 - Prob. 7.82ECh. 7 - Suppose you have a bag made of a membrane like...Ch. 7 - Prob. 7.84ECh. 7 - Prob. 7.85ECh. 7 - Prob. 7.86ECh. 7 - Prob. 7.87ECh. 7 - Prob. 7.88ECh. 7 - Prob. 7.89ECh. 7 - When a patient has blood cleansed by hemodialysis,...Ch. 7 - Prob. 7.91ECh. 7 - Prob. 7.92ECh. 7 - Prob. 7.93ECh. 7 - Prob. 7.94ECh. 7 - Prob. 7.95ECh. 7 - Strips of fresh meat can be preserved by drying....Ch. 7 - If a salt is added to water, which of the...Ch. 7 - Prob. 7.98ECh. 7 - Prob. 7.99ECh. 7 - Prob. 7.100ECh. 7 - Which one of the following compounds is a...Ch. 7 - Prob. 7.102ECh. 7 - Prob. 7.103ECh. 7 - Prob. 7.104ECh. 7 - Prob. 7.105ECh. 7 - Prob. 7.106ECh. 7 - Prob. 7.107ECh. 7 - Prob. 7.108ECh. 7 - Prob. 7.109ECh. 7 - Prob. 7.110ECh. 7 - In a dilute solution of sodium chloride in water,...Ch. 7 - A salt solution has a molarity of 1.5M. How many...Ch. 7 - Prob. 7.113ECh. 7 - Prob. 7.114ECh. 7 - Prob. 7.115ECh. 7 - Prob. 7.116ECh. 7 - Prob. 7.117ECh. 7 - Prob. 7.118ECh. 7 - Prob. 7.119ECh. 7 - Prob. 7.120ECh. 7 - Prob. 7.121ECh. 7 - Prob. 7.122E
Knowledge Booster
Background pattern image
Similar questions
Recommended textbooks for you
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781285199030
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Text book image
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