Chapter 11, Problem 69E

Match the vapor pressure diagrams with the solute-solvent combinations and explain your answers.

a.  and

b.  and

c.  and

d.  and

Interpretation Introduction

Interpretation:

The dissolution of the given following solute and solvent has to be explained using Raoult’s law.

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 69E

A Negative deviations from Raoult’s law is the right option for Acetone-Water solution.

Explanation of Solution

To find the second diagram match with Acetone and Water

The second diagram illustrates negative variation from Raoult's law. This occurs whilst the solute-solvent connections are stronger than the connections in pure solvent and pure solute.

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

These two molecules are named Acetone ( ${\text{CH}}_{\text{3}}{\text{COCH}}_{\text{3}}$) and Water. As discussed in non-ideal solutions, Acetone-Water solutions exhibit negative deviations from Raoult’s law. Acetone and Water have the capability to hydrogen bond with each other, which gives the solution stronger intermolecular forces as compared to the pure states of together solute and solvent. In the pure state, Acetone cannot H-bond through itself. Consequently the second diagram illustrating negative deviations from Raoult’s law is the right option for Acetone-Water solutions.

Interpretation Introduction

Interpretation:

The dissolution of the given following solute and solvent has to be explained using Raoult’s law.

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 69E

A Positive deviation from Raoult’s law is the right option for Ethanol-Water solution.

Explanation of Solution

To find the first diagram match with ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$ and Water

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

The first diagram shows positive deviation from Raoult's law. This occurs when the solute-solvent connections are weaker than the connections in pure solvent and pure solute.

These two molecules are named Ethanol ( ${\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH}$) and Water. Ethanol-water solutions demonstrate positive deviations from Raoult’s law. Equally substances can hydrogen bond in the pure state, and they can carry on this in solution. Yet, the solute-solvent interactions are rather weaker for Ethanol-Water solutions due to the important 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 weakens the intermolecular forces in solution. As a result the first diagram illustrating positive deviations from Raoult’s law is the right option for Ethanol-Water solutions.

Interpretation Introduction

Interpretation:

The dissolution of the given following solute and solvent has to be explained using Raoult’s law.

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 69E

No deviation from Raoult’s law is the correct choice for Heptane-Hexane. solution.

Explanation of Solution

To find the polarity of Heptane and Hexane

Heptane and Hexane

The third diagram illustrates an perfect solution with no difference from Raoult's law. This occurs what time the solute-solvent interactions are concerning equal to the pure solvent and pure solute interactions.

These two molecules are named Heptane ( ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}$) and Hexane ( ${\text{C}}_{\text{6}}{\text{H}}_{\text{14}}$). Heptane and hexane are very similar non-polar substances. Equally are collected totally of non-polar $\text{C-C}$ bonds and relatively non-polar $\text{C-H}$ bonds, with together have a similar size and form. Solutions of Heptane and Hexane should be ideal. So the third diagram illustrating no deviation from Raoult’s law is the correct choice for Heptane-Hexane solutions.

Interpretation Introduction

Interpretation:

The dissolution of the given following solute and solvent has to be explained using Raoult’s law.

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 69E

Heptane and Water results in positive deviations from Raoult’s law (the first diagram).

Explanation of Solution

To find: The Heptane Vs Water.

${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}$ and Water

These two molecules are named Heptane ( ${\text{C}}_{\text{7}}{\text{H}}_{\text{16}}$) and Water. The connections flanked by the non-polar Heptane molecules and the polar water molecules will surely be weaker in solution as compared to the pure solvent and pure solute interactions. These results in positive deviations from Raoult’s law (the first diagram).