Chapter 5, Problem 5.139QP

(a)

Interpretation Introduction

Interpretation:

The volume of container having $16.0\text{ g}$  of ${\text{O}}_{\text{2}}$ and $14.0\text{ g}$ of ${\text{N}}_{\text{2}}$ at STP has to be given.

Concept Introduction:

$\text{Mole = }\frac{\text{mass}}{\text{Molar mass}}$

Answer to Problem 5.139QP

At STP, the volume of container having $16.0\text{ g}$ of ${\text{O}}_{\text{2}}$ and $14.0\text{ g}$ of ${\text{N}}_{\text{2}}$ is $22.4\text{ L}$

Explanation of Solution

Given data:

A container is filled with $16.0\text{ g}$ of ${\text{O}}_{\text{2}}$ and $14.0\text{ g}$ of ${\text{N}}_{\text{2}}$

Moles of oxygen:

Molar mass of oxygen, ${\text{O}}_{\text{2}}$= $32\text{ g}$

The number of moles in $32\text{ g}$ of ${\text{O}}_{\text{2}}$ = $1\text{ mole}$

Therefore,

$\begin{array}{l}{\text{Number of moles in 16 g of O}}_{\text{2}}\text{=}\frac{1}{32}\times 16\\ =0.5\text{ mole}\end{array}$

Moles of nitrogen:

Molar mass of nitrogen, ${\text{N}}_{\text{2}}$= $28\text{ g}$

The number of moles in $28\text{ g}$ of ${\text{N}}_{\text{2}}$ = $1\text{ mole}$

Therefore,

$\begin{array}{l}{\text{Number of moles in 28 g of N}}_{\text{2}}\text{=}\frac{1}{28}\times 14\\ =0.5\text{ mole}\end{array}$

Total volume of the container:

Altogether, $1.00\text{ mole}$ of gas is present in the container.

At STP $1.00\text{ mole}$ of gas occupies $22.4\text{ L}$

Hence, the volume of the container is $22.4\text{ L}$

Conclusion

At STP, the volume of container having $16.0\text{ g}$ of ${\text{O}}_{\text{2}}$ and $14.0\text{ g}$ of ${\text{N}}_{\text{2}}$ is given as $22.4\text{ L}$

(b)

Interpretation Introduction

Interpretation:

The partial pressure of the ${\text{O}}_{\text{2}}$ gas has to be calculated

Concept Introduction:

Partial Pressure:

Dalton’s law of partial pressure states that “the total pressure (P) of the mixture is equal to the sum of the partial pressures ${\text{(P}}_{\text{A}}{\text{, P}}_{\text{B}}{\text{, P}}_{\text{C}}\mathrm{.....}\text{)}$ of all the component gases $\text{(A, B, C}\mathrm{......}\text{)}$  present in the mixture” and is given as,

${\text{P = P}}_{\text{A}}{\text{ + P}}_{\text{B}}{\text{ + P}}_{\text{C}}\text{ + }\mathrm{.......}$

The ideal gas law for the individual gas component A is given as,

${\text{P}}_{\text{A}}{\text{V = n}}_{\text{A}}\text{RT}$

Mole Fraction:

$\text{The mole fraction of A = }\frac{n_{\text{A}}}{n}=\frac{{\text{P}}_{\text{A}}}{\text{P}}$

Answer to Problem 5.139QP

The partial pressure of ${\text{O}}_{\text{2}}$ is $0.500\text{ atm}$

Explanation of Solution

Given data:

A container is filled with $16.0\text{ g}$ of ${\text{O}}_{\text{2}}$ and $14.0\text{ g}$ of ${\text{N}}_{\text{2}}$

Calculation of Partial Pressure:

On the rearranging the formula of mole fraction, the partial pressure of oxygen is calculated.

$\begin{array}{l}{\text{Partial pressure of O}}_{\text{2}}{\text{ = Mole fraction of O}}_{\text{2}}\times \text{Total pressure}\\ {\text{ P}}_{{\text{O}}_{\text{2}}}{\text{ = X}}_{{\text{O}}_{\text{2}}}\times {\text{P}}_{\text{tot}}\end{array}$

Here,

$\begin{array}{l}{\text{P}}_{{\text{O}}_{\text{2}}}{\text{= X}}_{{\text{O}}_{\text{2}}}\times {\text{P}}_{\text{tot}}\\ =\frac{\text{0}{\text{.500 mol O}}_2}{\text{0}{\text{.500 mol O}}_2+0.500{\text{ mol N}}_{\text{2}}}\times 1.00\text{ atm}\\ \text{=0}\text{.500 atm}\end{array}$

Therefore, the partial pressure of ${\text{O}}_{\text{2}}$ is $0.500\text{ atm}$

Conclusion

The partial pressure of ${\text{O}}_{\text{2}}$ is calculated as $0.500\text{ atm}$

(c)

Interpretation Introduction

Interpretation:

The mole fraction and the mole percent of ${\text{N }}_{\text{2}}$ in the mixture has to be calculated

Concept Introduction:

Partial Pressure:

Dalton’s law of partial pressure states that “the total pressure (P) of the mixture is equal to the sum of the partial pressures ${\text{(P}}_{\text{A}}{\text{, P}}_{\text{B}}{\text{, P}}_{\text{C}}\mathrm{.....}\text{)}$ of all the component gases $\text{(A, B, C}\mathrm{......}\text{)}$  present in the mixture” and is given as,

${\text{P = P}}_{\text{A}}{\text{ + P}}_{\text{B}}{\text{ + P}}_{\text{C}}\text{ + }\mathrm{.......}$

The ideal gas law for the individual gas component A is given as,

${\text{P}}_{\text{A}}{\text{V = n}}_{\text{A}}\text{RT}$

Mole Fraction:

$\text{The mole fraction of A = }\frac{n_{\text{A}}}{n}=\frac{{\text{P}}_{\text{A}}}{\text{P}}$

Answer to Problem 5.139QP

The mole fraction of ${\text{N }}_{\text{2}}$ in the mixture is $0.500\text{ mol fraction}$

The mole percent of ${\text{N }}_{\text{2}}$ in the mixture is $50.0\%\text{ mole percent}$

Explanation of Solution

Given data:

A container is filled with $16.0\text{ g}$  of ${\text{O}}_{\text{2}}$ and $14.0\text{ g}$ of ${\text{N}}_{\text{2}}$

Calculation of mole fraction and mole percent:

The mole fraction of nitrogen is calculated as follows,

$\begin{array}{l}{\text{Mole fraction of N}}_{\text{2}}\text{= }\frac{{\text{Mole of N}}_{\text{2}}}{{\text{Mole of N}}_{\text{2}}+{\text{Mole of O}}_{\text{2}}}\\ =\frac{\text{0}{\text{.500 mol N}}_{\text{2}}}{\text{0}{\text{.500 mol N}}_{\text{2}}+0.500{\text{ mol O}}_{\text{2}}}\\ =0.500\text{ mol fraction}\end{array}$

The mole percent of nitrogen is calculated as below,

$\begin{array}{l}\text{Mole percent }\text{= Mole fraction }\times \text{ 100 \%}\\ \text{=0}{\text{.500 of N}}_{\text{2}}\times \text{ 100 \%}\\ \text{=50}\text{.0 \% mole percent}\end{array}$

Conclusion

The mole fraction of ${\text{N }}_{\text{2}}$ in the mixture is calculated as $0.500\text{ mol fraction}$

The mole percent of ${\text{N }}_{\text{2}}$ in the mixture is calculated as $50.0\%\text{ mole percent}$