Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
Question
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Chapter 10, Problem 98AE

a)

Interpretation Introduction

Interpretation: The dissolution of the given following solute and solvent has to be explained.

Concept Introduction: Concept introduction:

Raoult's law:

The mole fraction of a solute is related to the vapor pressure of the solution thus,

                                 Psolution=P°solventXsolvent......(1)Pisvapor pressureof the solutionsolventpressureof the solventXsolvent mole fraction ofsolvent

a)

Expert Solution
Check Mark

Answer to Problem 98AE

Answer

ΔH1 and ΔH2 refer to the breaking of intermolecular forces in pure solute and in pure solvent. ΔH3 refers to the formation of the intermolecular forces in solution between the solute and solvent. ΔHsoln is the sum ΔH3

Explanation of Solution

To find the Acetone and water polarity

CH3COCH3 and H2O

The electrostatic possible drawing illustrates that acetone ( CH3COCH3 ), like water, is a polar substance (each has a red end indicating the partial negative end of the dipole moment and a blue end indicating the partial positive end). For a polar solute in a polar solvent, ΔH1 and ΔH2 will be large and positive, whilst ΔH3 will be a large negative value. As on non-ideal solutions, acetone-water solutions exhibit negative deviations from Raoult’s law. Acetone and water have the ability to hydrogen bond with every previous, which gives the solution stronger intermolecular forces as compared to the pure states of together solute and solvent. In the clean state, acetone cannot H-bond among itself. Since acetone and water prove negative deviations from Raoult’s law, one would wait for ΔHsoln to be slightly negative. Here ΔH3 will be more than negative enough to overcome the large positive value from the ΔH1 and ΔH2 terms combined.

b)

Interpretation Introduction

Interpretation: The dissolution of the given following solute and solvent has to be explained.

Concept Introduction: Concept introduction:

Raoult's law:

The mole fraction of a solute is related to the vapor pressure of the solution thus,

                                 Psolution=P°solventXsolvent......(1)Pisvapor pressureof the solutionsolventpressureof the solventXsolvent mole fraction ofsolvent

b)

Expert Solution
Check Mark

Answer to Problem 98AE

Answer

ΔH1 and ΔH2 refer to the breaking of intermolecular forces in pure solute and in pure solvent. ΔH3 refers to the formation of the intermolecular forces in solution between the solute and solvent. ΔHsoln is the sum ΔH3

Explanation of Solution

To find the polarity of CH3CH2OH and water

CH3CH2OH and water

These two molecules are named Ethanol ( CH3CH2OH ) and water. Ethanol-water solutions demonstrate affirmative deviations from Raoult’s law. Together substances can hydrogen bond in the pure state, and they can carry on this in solution. Still, the solute-solvent interactions are rather weaker for ethanol-water solutions due to the small non-polar part of ethanol ( CH3CH2 is the non-polar part of ethanol). This non-polar part of ethanol slightly weakens the intermolecular forces in solution. So as in part a, when a polar solute and polar solvent are present, ΔH1 and ΔH2 are large and positive, while ΔH3 is large and negative. For positive deviations from Raoult’s law, the interactions in solution are weaker than the interactions in pure solute and pure solvent. Here, ΔHsoln will be slightly positive because the ΔH3 term is not negative enough to overcome the large, positive ΔH1 and ΔH2 terms combined.

c)

Interpretation Introduction

Interpretation: The dissolution of the given following solute and solvent has to be explained.

Concept Introduction: Concept introduction:

Raoult's law:

The mole fraction of a solute is related to the vapor pressure of the solution thus,

                                

Psolution=P°solventXsolvent......(1)Pisvapor pressureof the solutionsolventpressureof the solventXsolvent mole fraction ofsolvent

c)

Expert Solution
Check Mark

Answer to Problem 98AE

Answer

ΔH1 and ΔH2 refer to the breaking of intermolecular forces in pure solute and in pure solvent. ΔH3 refers to the formation of the intermolecular forces in solution between the solute and solvent. ΔHsoln is the sum ΔH3

Explanation of Solution

To find the polarity of Heptane and Hexane

Heptane and Hexane

Explanation: Because the electrostatic possible diagrams specify, together heptane ( C7H16 ) and hexane ( C6H14 ) are non-polar substances. For a non-polar solute dissolved in a non-polar solvent, ΔH1 and ΔH2 are little and positive, while the ΔH3 term is small and negative. These three provisos have small values due to the comparatively weak London dispersion forces that are broken and formed for solutions consisting of a non-polar solute in a non-polar solvent.  Because ΔH1 and ΔH2 , and ΔH3 are all small values, the ΔHsoln value will be small.  Heptane and Hexane would form an ideal solution because the relative strengths of the London dispersal armed forces are about equal in pure solute and pure solvent as compared to those LD forces in solution. For ideal solutions, ΔHsoln = 0.

d)

Interpretation Introduction

Interpretation: The dissolution of the given following solute and solvent has to be explained.

Concept Introduction: Concept introduction:

Raoult's law:

The mole fraction of a solute is related to the vapor pressure of the solution thus,

                                 Psolution=P°solventXsolvent......(1)Pisvapor pressureof the solutionsolventpressureof the solventXsolvent mole fraction ofsolvent

d)

Expert Solution
Check Mark

Answer to Problem 98AE

Answer

ΔH1 and ΔH2 refer to the breaking of intermolecular forces in pure solute and in pure solvent. ΔH3 refers to the formation of the intermolecular forces in solution between the solute and solvent. ΔHsoln is the sum ΔH3

Explanation of Solution

To find the polarity of Methane

CH4

This combination represents a non-polar solute in a polar solvent. ΔH1  will be small due to the relative weak London dispersion forces which are broken when the solute     ( C7H16 ) expands. ΔH2 will be large and positive because of the relatively strong hydrogen bonding interactions that must be broken when the polar solvent (water) is expanded. And finally, the ΔH3 term will be small because the non-polar solute and polar solvent do not interact with each other. The end result is that ΔHsoln is a large positive value

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Chapter 10 Solutions

Chemistry: An Atoms First Approach

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