Chapter 5, Problem 89E

a)

Interpretation Introduction

Interpretation: The total number of moles in the mixture, the mole fractions and the number of grams of each gas in the given mixture is needed to be determined if the partial pressure of ${\text{CH}}_4$ is $\text{0}\text{.175}\text{atm}$ and ${\text{O}}_{\text{2}}$ is $\text{0}\text{.250}\text{atm}$.

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,

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

• 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.

• The number of moles of molecule can be find out by using its mole fraction,

• $\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}\text{=}\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\times \text{total number of moles}$

• Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

To determine: the total number of moles in the mixture, the mole fractions and the number of grams of each gas in the given mixture.

Answer to Problem 89E

Mole fraction of ${\text{CH}}_4$ is $0.412$.

Mole fraction of ${\text{O}}_{\text{2}}$ is $0.588$.

Explanation of Solution

To find: the mole fractions of ${\text{CH}}_4$ and ${\text{O}}_{\text{2}}$ in the given mixture if the partial pressure of ${\text{CH}}_4$ is $\text{0}\text{.175}\text{atm}$ and ${\text{O}}_{\text{2}}$ is $\text{0}\text{.250}\text{atm}$.

Mole fraction of ${\text{CH}}_4$ is $0.412$.

Mole fraction of ${\text{O}}_{\text{2}}$ is $0.588$.

The partial pressure of ${\text{CH}}_4$ is given as $\text{0}\text{.175}\text{atm}$

The partial pressure of ${\text{O}}_{\text{2}}$ is given as $\text{0}\text{.250}\text{atm}$.

Therefore, the total pressure of the mixture is,

$\text{0}\text{.175}\text{atm}\text{+}\text{0}\text{.250}\text{atm}\text{=}\text{0}\text{.425}\text{atm}$

Equation for finding mole fraction from partial pressure and total pressure is,

${\text{χ}}_{\text{A}}\text{=}\frac{{\text{P}}_{\text{A}}}{{\text{P}}_{\text{TOTAL}}}$

Therefore,

The mole fraction of ${\text{CH}}_4$ is,

$\begin{array}{l}{\text{χ}}_{{\text{CH}}_{\text{4}}}\text{=}\frac{\text{0}\text{.175}\text{atm}}{\text{0}\text{.425}\text{atm}}\\ \text{=}\text{0}\text{.412}\end{array}$

The mole fraction of ${\text{O}}_{\text{2}}$ is,

$\begin{array}{l}{\text{χ}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{0}\text{.250atm}}{\text{0}\text{.425}\text{atm}}\\ \text{=}\text{0}\text{.588}\end{array}$

b)

Interpretation Introduction

Interpretation: The total number of moles in the mixture, the mole fractions and the number of grams of each gas in the given mixture is needed to be determined if the partial pressure of ${\text{CH}}_4$ is $\text{0}\text{.175}\text{atm}$ and ${\text{O}}_{\text{2}}$ is $\text{0}\text{.250}\text{atm}$.

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,

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

• 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.

• The number of moles of molecule can be find out by using its mole fraction,

• $\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}\text{=}\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\times \text{total number of moles}$

• Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

To determine: the total number of moles in the mixture, the mole fractions and the number of grams of each gas in the given mixture.

Answer to Problem 89E

Total number of moles of gas in the mixture is $\text{0}\text{.161}\text{mol}$.

Explanation of Solution

To find: the total number of moles of gases in the given mixture.

The total number of moles of gases in the given mixture is $\text{0}\text{.161}\text{mol}$.

The total pressure of gases presented in the mixture is given as $\text{0}\text{.425}\text{atm}$.

The volume of flask is given as $\text{10}\text{.5}\text{L}$.

The temperature of flask is given as ${\text{65}}^{\text{o}}\text{C}\text{=}\text{(273+65)}\text{K}\text{=}\text{338}\text{K}$.

The total number of moles of gas in the mixture is determined by using ideal gas equation According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

Therefore,

The total number of moles of gas in the mixture is,

$\begin{array}{l}\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{0}\text{.425}\text{atm}\text{×}\text{10}\text{.5}\text{L}}{\text{0}\text{.08206}\text{L}\text{atm/}\text{K}\text{mol}\text{×}\text{338}\text{K}}\\ \text{=}\text{0}\text{.161}\text{mol}\end{array}$

c)

Interpretation Introduction

Interpretation: The total number of moles in the mixture, the mole fractions and the number of grams of each gas in the given mixture is needed to be determined if the partial pressure of ${\text{CH}}_4$ is $\text{0}\text{.175}\text{atm}$ and ${\text{O}}_{\text{2}}$ is $\text{0}\text{.250}\text{atm}$.

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,

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

• 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.

• The number of moles of molecule can be find out by using its mole fraction,

• $\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}\text{=}\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\times \text{total number of moles}$

• Number of grams of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

• Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

To determine: the total number of moles in the mixture, the mole fractions and the number of grams of each gas in the given mixture.

Answer to Problem 89E

The number of grams of ${\text{CH}}_4$ is $\text{1}\text{.06}\text{g}$.

The number of grams of ${\text{O}}_{\text{2}}$ is $3.03\text{g}$.

Explanation of Solution

To find: the number of grams of ${\text{CH}}_4$ and ${\text{O}}_{\text{2}}$ in the given mixture.

The number of grams of ${\text{CH}}_4$ is $\text{1}\text{.06}\text{g}$.

The number of grams of ${\text{O}}_{\text{2}}$ is $3.03\text{g}$.

The mole fraction of ${\text{CH}}_4$ is calculated as $0.412$.

The mole fraction of ${\text{O}}_{\text{2}}$ is calculated as $0.588$.

The total number of moles of gases in the given mixture is calculated as $\text{0}\text{.161}\text{mol}$.

• The number of moles of molecule can be find out by using its mole fraction,

  Equation for finding number of moles,

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

Therefore,

-The number of moles of ${\text{CH}}_4$ is,

$\begin{array}{l}{\text{n}}_{{\text{CH}}_{\text{4}}}\text{=}\text{0}\text{.412 ×}\text{0}\text{.161}\text{mol}\\ \text{=}\text{6}\text{.63}\text{×}{\text{10}}^{\text{-2}}\text{mol}\end{array}$

-The number of moles of ${\text{O}}_{\text{2}}$ is,

$\begin{array}{l}{\text{n}}_{{\text{CH}}_{\text{4}}}\text{=}\text{0}\text{.588 ×}\text{0}\text{.161}\text{mol}\\ \text{=}9.47\text{×}{\text{10}}^{\text{-2}}\text{mol}\end{array}$

• The number of grams of a substance is determined from its number of moles is,

  Equation for finding number of grams is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{Number}\text{of}\text{grams}$

-The number of grams of ${\text{CH}}_4$ is,

$\text{6}\text{.63}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{×}\text{16}\text{g}\text{=}\text{1}\text{.06}\text{g}$

-The number of grams of ${\text{O}}_{\text{2}}$ is,

$\text{9}\text{.47}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{×}\text{32}\text{g}\text{=}\text{3}\text{.03}\text{g}$