Chemistry for Today: General  Organic  and Biochemistry
Chemistry for Today: General Organic and Biochemistry
9th Edition
ISBN: 9781337514576
Author: Seager
Publisher: Cengage
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Chapter 7, Problem 7.65E

Calculate the boiling and freezing points of water solutions that are 1.15 M in the following solutes:

a. KBr , a strong electrolyte

b. ethylene glycol, a nonelectrolyte

c. ( NH 4 ) 2 CO 3 , strong electrolyte

d. Al 2 ( SO 4 ) 3 , a strong electrolyte

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of 1.15M water solution of KBr, 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 7.65E

The boiling and freezing points of 1.15M water solution of KBr, a strong electrolyte are 101.2°C and 4.3°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 KBr is a strong electrolyte it will completely dissociate in the solution and the value of n for KBr 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×1.15M=1.196°C

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

Boiling point=100°C+1.196°C=101.196°C101.2°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 KBr is a strong electrolyte it will completely dissociate in the solution and the value of n for KBr 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×1.15M=4.278°C

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

Freezing point=0°C4.278°C=4.278°C4.3°C

Conclusion

The boiling and freezing points of 1.15M water solution of KBr, a strong electrolyte are 101.2°C and 4.3°C respectively.

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of 1.15M water solution of ethylene glycol, 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.

Answer to Problem 7.65E

The boiling and freezing points of 1.15M water solution of ethylene glycol, a nonelectrolyte are 100.6°C and 2.1°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 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×1.15M=0.598°C

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

Boiling point=100°C+0.598°C=100.598°C100.6°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 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×1.15M=2.139°C

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

Freezing point=0°C2.139°C=2.139°C2.1°C

Conclusion

The boiling and freezing points of 1.15M water solution of ethylene glycol, a nonelectrolyte are 100.6°C and 2.1°C respectively.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of 1.15M water solution of (NH4)2CO3, 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 7.65E

The boiling and freezing points of 1.15M water solution of (NH4)2CO3, strong electrolyte are 101.8°C and 6.4°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 (NH4)2CO3 is a strong electrolyte it will dissociate in the solution completely and the value of n for (NH4)2CO3 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×1.15M=1.794°C

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

Boiling point=100°C+1.794°C=101.794°C101.8°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)2CO3 is a strong electrolyte it will dissociate in the solution completely and the value of n for (NH4)2CO3 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×1.15M=6.417°C

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

Freezing point=0°C6.417°C=6.417°C6.4°C

Conclusion

The boiling and freezing points of 1.15M water solution of (NH4)2CO3, strong electrolyte are 101.8°C and 6.4°C respectively.

Expert Solution
Check Mark
Interpretation Introduction

(d)

Interpretation:

The boiling and freezing points of 1.15M water solution of Al2(SO4)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 7.65E

The boiling and freezing points of 1.15M water solution of Al2(SO4)3, a strong electrolyte are 103°C and 10.7°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 Al2(SO4)3 is a strong electrolyte it will dissociate in the solution completely and the value of n for Al2(SO4)3 is 5. 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=5×0.52°C/M×1.15M=2.99°C

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

Boiling point=100°C+2.99°C=102.99°C103°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 Al2(SO4)3 is a strong electrolyte it will dissociate in the solution completely and the value of n for Al2(SO4)3 is 5. 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=5×1.86°C/M×1.15M=10.695°C

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

Freezing point=0°C10.695°C=10.695°C10.7°C

Conclusion

The boiling and freezing points of 1.15M water solution of Al2(SO4)3, a strong electrolyte are 103°C and 10.7°C respectively.

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Chapter 7 Solutions

Chemistry for Today: General Organic and Biochemistry

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