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 7.67E
Interpretation Introduction

(a)

Interpretation:

The boiling and freezing points of a solution containing 50.0g of H2SO4, a strong electrolyte (both H s dissociate), 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 7.67E

The boiling and freezing points of a solution containing 50.0g of H2SO4, a strong electrolyte (both H s dissociate), per 250.mL are 103.18°C and 11.38°C respectively.

Explanation of Solution

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

Molesofsolute=GivenmassofsoluteMolarmassofsolute

The molar mass of H2SO4 can be calculated as follows.

H2SO4=(1×S)+(2×H)+(4×O)=(1×32)+(2×1)+(4×16)=98g/mol

Substitute the values in the above equation as follows.

MolesofH2SO4=GivenmassofH2SO4MolarmassofH2SO4=50g98g/mol=0.51mol

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=molesofH2SO4litersofH2SO4solution=0.51mol250mL×1L1000mL=2.04M

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 H2SO4 is a strong electrolyte, it will dissociate in the solution completely and the value of n for H2SO4 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×2.04M=3.18°C

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

Boiling point=100°C+3.18°C=103.18°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 H2SO4 is a strong electrolyte, it will dissociate in the solution completely and the value of n for urea 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×2.04M=11.38°C

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

Freezing point=0°C11.38°C=11.38°C

Conclusion

The boiling and freezing points of a solution containing 50.0g of H2SO4, a strong electrolyte (both H s dissociate), per 250.mL are 103.18°C and 11.38°C respectively.

Interpretation Introduction

(b)

Interpretation:

The boiling and freezing points of a solution containing 200.g of table sugar (C12H22O11), 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 7.67E

The boiling and freezing points of a solution containing 200.g of table sugar (C12H22O11), a nonelectrolyte, per 250.mL are 101.21°C and 4.33°C respectively.

Explanation of Solution

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

Molesofsolute=GivenmassofsoluteMolarmassofsolute

The molar mass of sugar can be calculated as follows.

C12H22O11=(12×C)+(22×H)+(11×O)=(12×12)+(22×1)+(11×16)=342g/mol

Substitute the values in the above equation as follows.

MolesofC12H22O11=GivenmassofC12H22O11MolarmassofC12H22O11=200g342g/mol=0.5847mol

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=molesofC12H22O12litersofC12H22O12solution=0.5847mol250mL×1L1000mL=2.33M

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 sugar is a nonelectrolyte, it will not dissociate in the solution and the value of n for sugar 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×2.33M=1.21°C

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

Boiling point=100°C+1.21°C=101.21°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 sugar is a nonelectrolyte, it will not dissociate in the solution and the value of n for sugar 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×2.33M=4.33°C

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

Freezing point=0°C4.33°C=4.33°C

Conclusion

The boiling and freezing points of a solution containing 200.g of table sugar (C12H22O11), a nonelectrolyte, per 250.mL are 101.21°C and 4.33°C respectively.

Interpretation Introduction

(c)

Interpretation:

The boiling and freezing points of a solution containing 75.0g of octanoic acid (C8H16O2), a nonelectrolyte, in enough benzene to give 250.mL of solution 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 7.67E

The boiling and freezing points of a solution containing 75.0g of octanoic acid (C8H16O2), a nonelectrolyte, in enough benzene to give 250.mL of solution are 85.36°C and 4.69°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.

C8H16O2=(8×C)+(16×H)+(2×O)=(8×12)+(16×1)+(2×16)=144g/mol

Substitute the values in the above equation as follows.

MolesofC8H16O2=GivenmassofC8H16O2MolarmassofC8H16O2=75g144g/mol=0.5208mol

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=molesofC8H16O2litersofC8H16O2solution=0.5208mol250mL×1L1000mL=2.08M

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 octanoic acid is a nonelectrolyte, it will not dissociate in the solution and the value of n for octanoic acid is 1. The value of Kb for benzene is 2.53°C/M.

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

ΔTb=nKbM=1×2.53°C/M×2.08M=5.26°C

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

Boiling point=80.1°C+5.26°C=85.36°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 octanoic acid is a nonelectrolyte, it will not dissociate in the solution and the value of n for octanoic acid is 1. The value of Kf for benzene is 4.90°C/M.

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

ΔTf=nKfM=1×4.90°C/M×2.08M=10.19°C

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

Freezing point=5.5°C10.19°C=4.69°C

Conclusion

The boiling and freezing points of a solution containing 75.0g of octanoic acid (C8H16O2), a nonelectrolyte, in enough benzene to give 250.mL of solution are 85.36°C and 4.69°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