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
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Chapter 7, Problem 7.64E

Calculate the boiling and freezing points of water solutions that are 1.50 M in the following solutes:

a. KCl , a strong electrolyte

b. glycerol, a nonelectrolyte

c. ( NH 4 ) 2 SO 4 , strong electrolyte

d. Al ( NO 3 ) 3 , a strong electrolyte

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of 1.50M water solution of KCl, a strong electrolyte 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.

Answer to Problem 7.64E

The boiling and freezing points of 1.50M water solution of KCl, a strong electrolyte are 101.6°C and 5.58°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 KCl is a strong electrolyte it will completely dissociate in the solution and the value of n for KCl is 2. 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=2×0.52°C/M×1.50M=1.56°C

The boiling point of water solution can be calculated by adding value the of ΔTb to normal boiling point of water as shown below.

Boiling point=100°C+1.56°C=101.56°C101.6°C

Thus, the boiling point of water solution is 101.6°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 KCl is a strong electrolyte it will completely dissociate in the solution and the value of n for KCl is 2. 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=2×1.86°C/M×1.50M=5.58°C

The freezing point of water solution can be calculated by subtracting the value of ΔTf from freezing point of water as shown below.

Freezing point=0°C5.58°C=5.58°C

Thus, the freezing point of the water solution is 5.58°C.

Conclusion

The boiling and freezing points of 1.50M water solution of KCl, a strong electrolyte are 101.6°C and 5.58°C respectively.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of 1.50M water solution of glycerol, a nonelectrolyte is 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.

Answer to Problem 7.64E

The boiling and freezing points of 1.50M water solution of glycerol, a nonelectrolyte are 100.78°C and 2.79°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 glycerol 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×1.50M=0.78°C

The boiling point of water solution can be calculated by adding value of ΔTb to the normal boiling point of water as shown below.

Boiling point=100°C+0.78°C=100.78°C

Thus, the boiling point of water solution is 100.78°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 glyerol is a nonelectrolyte it will not dissociate in the solution and the value of n for glycerol 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×1.50M=2.79°C

The freezing point of water solution can be calculated by subtracting the value of ΔTf from normal freezing point of water as given below.

Freezing point=0°C2.79°C=2.79°C

Thus, the freezing point of water solution is 2.79°C.

Conclusion

The boiling and freezing points of 1.50M water solution of glycerol, a nonelectrolyte are 100.78°C and 2.79°C respectively.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of 1.50M water solution of (NH4)2SO4, a strong electrolyte 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.

Answer to Problem 7.64E

The boiling and freezing points of 1.50M water solution of (NH4)2SO4, strong electrolyte are 102.34°C and 8.37°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 (NH4)2SO4 is a strong electrolyte it will dissociate in the solution completely and the value of n for (NH4)2SO4 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×1.50M=2.34°C

The boiling point of water solution can be calculated by adding value of ΔTb to the normal boing point of water as shown below.

Boiling point=100°C+2.34°C=102.34°C

Thus, the boiling point of water solution is 102.34°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 (NH4)2SO4 is a strong electrolyte it will dissociate in the solution completely and the value of n for (NH4)2SO4 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×1.50M=8.37°C

The freezing point of water solution can be calculated by subtracting the value of ΔTf from normal freezing point of water as given below.

Freezing point=0°C8.37°C=8.37°C

Thus, the freezing point of water solution is 8.37°C.

Conclusion

The boiling and freezing points of 1.50M water solution of (NH4)2SO4, strong electrolyte are 102.34°C and 8.37°C respectively.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The boiling and freezing points of 1.50M water solution of Al(NO3)3, a strong electrolyte 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.

Answer to Problem 7.64E

The boiling and freezing points of 1.50M water solution of Al(NO3)3, a strong electrolyte are 103.12°C and 11.16°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 Al(NO3)3 is a strong electrolyte it will dissociate in the solution completely and the value of n for Al(NO3)3 is 4. 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=4×0.52°C/M×1.50M=3.12°C

The boiling point of water solution can be calculated by adding value of ΔTb to the normal boing point of water as shown below.

Boiling point=100°C+3.12°C=103.12°C

Thus, the boiling point of water solution is 103.12°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 Al(NO3)3 is a strong electrolyte it will dissociate in the solution completely and the value of n for Al(NO3)3 is 4. 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=4×1.86°C/M×1.50M=11.16°C

The freezing point of water solution can be calculated by subtracting the value of ΔTf from normal freezing point of water as given below.

Freezing point=0°C11.16°C=11.16°C

Thus, the freezing point of water solution is 11.16°C.

Conclusion

The boiling and freezing points of 1.50M water solution of Al(NO3)3, a strong electrolyte are 103.12°C and 11.16°C respectively.

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Chapter 7 Solutions

Chemistry for Today: General, Organic, and Biochemistry

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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY