Chapter 11, Problem 106AE

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,

                                 $\begin{array}{l}{\text{P}}_{\text{solution}}{\text{=P°}}_{\text{solvent}}{\text{X}}_{\text{solvent}}\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{vapor pressure}\text{of the solution}\\ {\text{P°}}_{\text{solvent}}\text{pressure}\text{of the solvent}\\ {\text{X}}_{\text{solvent}}\text{ mole fraction of}\text{solvent}\end{array}$

Answer to Problem 106AE

Answer

${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ refer to the breaking of intermolecular forces in pure solute and in pure solvent. ${\text{ΔH}}_{\text{3}}$ refers to the formation of the intermolecular forces in solution between the solute and solvent. ${\text{ΔH}}_{\text{soln}}$ is the sum ${\text{ΔH}}_{\text{3}}$

Explanation of Solution

To find the Acetone and water polarity

${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}$ and ${\text{H}}_{\text{2}}\text{O}$

The electrostatic possible drawing illustrates that acetone ( ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}$), 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, ${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ will be large and positive, whilst ${\text{ΔH}}_{\text{3}}$ 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 ${\text{ΔH}}_{\text{soln}}$ to be slightly negative. Here ${\text{ΔH}}_{\text{3}}$ will be more than negative enough to overcome the large positive value from the ${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ 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,

                                 $\begin{array}{l}{\text{P}}_{\text{solution}}{\text{=P°}}_{\text{solvent}}{\text{X}}_{\text{solvent}}\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{vapor pressure}\text{of the solution}\\ {\text{P°}}_{\text{solvent}}\text{pressure}\text{of the solvent}\\ {\text{X}}_{\text{solvent}}\text{ mole fraction of}\text{solvent}\end{array}$

Answer to Problem 106AE

Answer

${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ refer to the breaking of intermolecular forces in pure solute and in pure solvent. ${\text{ΔH}}_{\text{3}}$ refers to the formation of the intermolecular forces in solution between the solute and solvent. ${\text{ΔH}}_{\text{soln}}$ is the sum ${\text{ΔH}}_{\text{3}}$

Explanation of Solution

To find the polarity of ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ and water

${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ and water

These two molecules are named Ethanol ( ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$) 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 ( ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}$ 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, ${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ are large and positive, while ${\text{ΔH}}_{\text{3}}$ 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, ${\text{ΔH}}_{\text{soln}}$ will be slightly positive because the ${\text{ΔH}}_{\text{3}}$ term is not negative enough to overcome the large, positive ${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ 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,

                                

$\begin{array}{l}{\text{P}}_{\text{solution}}{\text{=P°}}_{\text{solvent}}{\text{X}}_{\text{solvent}}\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{vapor pressure}\text{of the solution}\\ {\text{P°}}_{\text{solvent}}\text{pressure}\text{of the solvent}\\ {\text{X}}_{\text{solvent}}\text{ mole fraction of}\text{solvent}\end{array}$

Answer to Problem 106AE

Answer

${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ refer to the breaking of intermolecular forces in pure solute and in pure solvent. ${\text{ΔH}}_{\text{3}}$ refers to the formation of the intermolecular forces in solution between the solute and solvent. ${\text{ΔH}}_{\text{soln}}$ is the sum ${\text{ΔH}}_{\text{3}}$

Explanation of Solution

To find the polarity of Heptane and Hexane

Heptane and Hexane

Explanation: Because the electrostatic possible diagrams specify, together heptane ( ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}$) and hexane ( ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$) are non-polar substances. For a non-polar solute dissolved in a non-polar solvent, ${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ are little and positive, while the ${\text{ΔH}}_{\text{3}}$ 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 ${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$, and ${\text{ΔH}}_{\text{3}}$ are all small values, the ${\text{ΔH}}_{\text{soln}}$ 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, ${\text{ΔH}}_{\text{soln}}$ = 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,

                                 $\begin{array}{l}{\text{P}}_{\text{solution}}{\text{=P°}}_{\text{solvent}}{\text{X}}_{\text{solvent}}\mathrm{......}\text{(1)}\\ \text{P}\text{is}\text{vapor pressure}\text{of the solution}\\ {\text{P°}}_{\text{solvent}}\text{pressure}\text{of the solvent}\\ {\text{X}}_{\text{solvent}}\text{ mole fraction of}\text{solvent}\end{array}$

Answer to Problem 106AE

Answer

${\text{ΔH}}_{\text{1}}{\text{ and ΔH}}_{\text{2}}$ refer to the breaking of intermolecular forces in pure solute and in pure solvent. ${\text{ΔH}}_{\text{3}}$ refers to the formation of the intermolecular forces in solution between the solute and solvent. ${\text{ΔH}}_{\text{soln}}$ is the sum ${\text{ΔH}}_{\text{3}}$

Explanation of Solution

To find the polarity of Methane

${\text{CH}}_{\text{4}}$

This combination represents a non-polar solute in a polar solvent. ${\text{ΔH}}_{\text{1}}$ will be small due to the relative weak London dispersion forces which are broken when the solute     ( ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}$) expands. ${\text{ΔH}}_{\text{2}}$ 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 ${\text{ΔH}}_{\text{3}}$ term will be small because the non-polar solute and polar solvent do not interact with each other. The end result is that ${\text{ΔH}}_{\text{soln}}$ is a large positive value