Chapter 4, Problem 4.12P

(a)

Interpretation Introduction

Interpretation:

Using Raoult’s law the partial pressure of ethanol in the vapour phase has to be determined.

Concept introduction:

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

Partial pressure of a gas in terms of its mole fraction and total pressure is,

`            ${\text{P}}_{\text{A}}\text{=}{\text{χ}}_{\text{A}}\text{×}{\text{P}}_{\text{TOTAL}}$

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

$\text{Mass}=\text{Density}\text{×}\text{Volume}$

Explanation of Solution

Given,

An ideal-dilute solution of chloroform and ethanol with a mole fraction of ethanol in the liquid phase $x_{eth}$ is $0.9900$ has a total vapour pressure of $p=177.95Torr$. The vapour pressure of pure ethanol at this temperature is ${\text{p}}_{\text{eth}}^{\text{*}}\text{=}\text{172}\text{.76}\text{Torr}$.

Using Raoult’s law the partial pressure of ethanol in the vapour phase can be determined as follows,

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

Herein,

$\begin{array}{c}{\text{ x}}_{\text{eth}}\text{=}\text{0}\text{.9900}\\ {\text{p}}_{\text{eth}}^{\text{*}}\text{=}\text{172}\text{.76}\text{Torr}\end{array}$

Thus,

The partial pressure of ethanol is,

$\begin{array}{c}{P}_{eth}=0.9900\times 172.76Torr\\ =171.23Torr\end{array}$

(b)

Interpretation Introduction

Interpretation:

The partial pressure of chloroform in the vapour phase has to be determined.

Concept introduction:

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

Partial pressure of a gas in terms of its mole fraction and total pressure is,

`            ${\text{P}}_{\text{A}}\text{=}{\text{χ}}_{\text{A}}\text{×}{\text{P}}_{\text{TOTAL}}$

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

$\text{Mass}=\text{Density}\text{×}\text{Volume}$

Explanation of Solution

Given,

An ideal-dilute solution of chloroform and ethanol with a mole fraction of ethanol in the liquid phase $x_{eth}$ is $0.9900$ has a total vapour pressure of $p=177.95Torr$. The vapour pressure of pure ethanol at this temperature is ${\text{p}}_{\text{eth}}^{\text{*}}\text{=}\text{172}\text{.76}\text{Torr}$.

Using Raoult’s law the partial pressure of ethanol in the vapour phase can be determined as follows,

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

Herein,

$\begin{array}{c}{\text{ x}}_{\text{eth}}\text{=}\text{0}\text{.9900}\\ {\text{p}}_{\text{eth}}^{\text{*}}\text{=}\text{172}\text{.76}\text{Torr}\end{array}$

Thus,

The partial pressure of ethanol is,

$\begin{array}{c}{P}_{eth}=0.9900\times 172.76Torr\\ =171.23Torr\end{array}$

Total pressure is $P=177.95Torr$

Thus,

$\begin{array}{c}{\text{P}}_{\text{Chloroform}}\text{=}{\text{P}}_{\text{Total}}\text{-}{\text{P}}_{\text{Ethanol}}\\ \text{=}\text{177}\text{.95-}\text{171}\text{.23}\text{Torr}\\ \text{=}6.72Torr\end{array}$

(c)

Interpretation Introduction

Interpretation:

The mole fractions of chloroform and ethanol in the vapour phase has to be determined.

Concept introduction:

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

Partial pressure of a gas in terms of its mole fraction and total pressure is,

`            ${\text{P}}_{\text{A}}\text{=}{\text{χ}}_{\text{A}}\text{×}{\text{P}}_{\text{TOTAL}}$

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

Explanation of Solution

Given,

An ideal-dilute solution of chloroform and ethanol with a mole fraction of ethanol in the liquid phase $x_{eth}$ is $0.9900$ has a total vapour pressure of $p=177.95Torr$. The vapour pressure of pure ethanol at this temperature is ${\text{p}}_{\text{eth}}^{\text{*}}\text{=}\text{172}\text{.76}\text{Torr}$.

Using Raoult’s law the partial pressure of ethanol in the vapour phase can be determined as follows,

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

Herein,

Total pressure is $P=177.95Torr$

Thus,

$\begin{array}{c}{\text{P}}_{\text{Chloroform}}\text{=}{\text{P}}_{\text{Total}}\text{-}{\text{P}}_{\text{Ethanol}}\\ \text{=}\text{177}\text{.95-}\text{171}\text{.23}\text{Torr}\\ \text{=}6.72Torr\end{array}$

The mole fraction of chloroform in the vapour phase can be determined as follows,

$\begin{array}{c}{P}_{Chloroform}=x_{Chloroform}\times P_{total}\\ x_{Chloroform}=\frac{P_{Chloroform}}{P_{total}}\\ =\frac{6.72Torr}{177.85Torr}\\ =0.0378\end{array}$

Thus,

The mole fraction of ethanol in the vapour phase is, $\text{1-}\text{0}\text{.0378}\text{=}0.9622$

(d)

Interpretation Introduction

Interpretation:

Henry’s law constant for chloroform, $K_{chl}$ has to be determined.

Concept introduction:

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

Partial pressure of a gas in terms of its mole fraction and total pressure is,

`            ${\text{P}}_{\text{A}}\text{=}{\text{χ}}_{\text{A}}\text{×}{\text{P}}_{\text{TOTAL}}$

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture. Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

Henry’s law states that the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

$\text{p}={\text{K}}_{\text{H}}\text{x}\boxed{\begin{array}{l}\text{where,}\\ {\text{K}}_{\text{H}}\text{:}\text{Henry's}\text{constant}\\ \text{x}\text{:}\text{mole}\text{fraction}\\ \text{p}\text{:}\text{partial}\text{pressure}\end{array}}$

Explanation of Solution

Given,

An ideal-dilute solution of chloroform and ethanol with a mole fraction of ethanol in the liquid phase $x_{eth}$ is $0.9900$ has a total vapour pressure of $p=177.95Torr$. The vapour pressure of pure ethanol at this temperature is ${\text{p}}_{\text{eth}}^{\text{*}}\text{=}\text{172}\text{.76}\text{Torr}$.

Using Raoult’s law the partial pressure of ethanol in the vapour phase can be determined as follows,

Raoult’s law: The vapour pressure of the solvent over the solution ${\text{P}}_{\text{solvent}}$ is proportional to the mole fraction of the solvent. The ideal solution obeys the Raoult’s law.

$\begin{array}{l}{\text{P}}_{solvent}\text{= }{\chi }_{solvent}{\text{P}}_{solvent}^{\circ }\text{,}\\ \text{where,}{\text{P}}_{solvent}\text{is}\text{the}\text{vapor pressure of the solvent over the solution,}\\ {\chi }_{solvent}\text{is mole fraction of solvent in solution, }\\ {\text{P}}_{solvent}^{\circ }\text{ is the vapor pressure of the pure solvent}\text{.}\end{array}$

Herein,

Total pressure is $P=177.95Torr$

Thus,

$\begin{array}{c}{\text{P}}_{\text{Chloroform}}\text{=}{\text{P}}_{\text{Total}}\text{-}{\text{P}}_{\text{Ethanol}}\\ \text{=}\text{177}\text{.95-}\text{171}\text{.23}\text{Torr}\\ \text{=}6.72Torr\end{array}$

The mole fraction of chloroform in the vapour phase can be determined as follows,

$\begin{array}{c}{P}_{Chloroform}=x_{Chloroform}\times P_{total}\\ x_{Chloroform}=\frac{P_{Chloroform}}{P_{total}}\\ =\frac{6.72Torr}{177.85Torr}\\ =0.0378\end{array}$

If mole fraction of each component in the mixture is known then its respective Henry’s constant can be determined as shown below,

 Henry’s law states that the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

$\text{p}={\text{K}}_{\text{H}}\text{x}\boxed{\begin{array}{l}\text{where,}\\ {\text{K}}_{\text{H}}\text{:}\text{Henry's}\text{constant}\\ \text{x}\text{:}\text{mole}\text{fraction}\\ \text{p}\text{:}\text{partial}\text{pressure}\end{array}}$

Herein,

$\begin{array}{c}Partialpressureofchloroform=6.72Torr\\ Molefractionofchloroform=0.0378\end{array}$

$\begin{array}{c}Henry\text{'}slawcons\tan tforchloroform{K}_{chl}=\frac{Partialpressureofchloroform}{Molefractionofchloroform}\\ =\frac{6.72Torr}{0.0378}\\ =1.77\times 10^2\end{array}$