Chapter 13, Problem 25QP

Interpretation Introduction

Interpretation:

The mole fractions of oxygen and nitrogen in the air are to be calculated.

Concept Introduction:

According to Henry’s law, solubility of a gas in a liquid is directly related to the pressure of the gas, and its expression is

$c=kP.$

Here, $c$ is the molar concentration $\left(\text{mol}/\text{L}\right)$, $k$ is proportionality constant $\left(\mathrm{mol}/\text{L}\text{.atm}\right),$ and $P$ is pressure $\left(\text{atm}\right)$.

Mole fraction $\left(\chi \right)$ is the ratio of moles of solute to the total moles (solute + solvent). So, the formula to calculate the mole fraction of any component, say A, will be as

${\chi }_{\text{A}}=\frac{\text{moles of A}}{\text{sum of moles of all components}}.$

Answer to Problem 25QP

Solution: 0.677;0.323

Explanation of Solution

Given information: The partial pressures of oxygen and nitrogen are $0.20\text{ \& 0}\text{.80 atm,}$ respectively and their solubility in water are $1.3\times {10}^{-3}\text{ mol/L}\text{.atm \& 6}\text{.8}\times {10}^{-4}\text{ mol/L}\text{.atm}$.

First, assume that ${\text{N}}_2{\text{ \& O}}_2$ are the only dissolved gases. Multiply each partial pressure by the corresponding Henry’s law constant.

$c=kP.$ …… (1)

For ${\text{N}}_2$,

substitute $1.3\times {10}^{-3}\text{ mol/L}\text{.atm}$ for $k$ and $0.20\text{ atm}$ for $P$ in the above expression as

$\begin{array}{c}c=\left(1.3\times {10}^{-3}\text{ mol/L}\text{.atm}\right)\left(0.20\text{ atm}\right)\\ =2.6\times {10}^{-4}\text{ mol/L}\text{.}\end{array}$

For ${\text{O}}_2$,

substitute $6.8\times {10}^{-4}\text{ mol/L}\text{.atm}$ for $k$ and $0.80\text{ atm}$ for $P$ in equation (1) as:

$\begin{array}{c}c=\left(6.8\times {10}^{-4}\text{ mol/L}\text{.atm}\right)\left(0.80\text{ atm}\right)\\ =5.44\times {10}^{-4}\text{ mol/L}\text{.}\end{array}$

In one liter of water, moles of ${\text{N}}_2{\text{ \& O}}_2$ will be $2.6\times {10}^{-4}\text{ mol/L \& 5}\text{.44}\times {10}^{-4}\text{ mol/L}$.

The expression to calculate the mole fraction is

${\chi }_{\text{A}}=\frac{\text{moles of A}}{\text{sum of moles of all components}}.$ …… (2)

For ${\text{N}}_2$,

substitute $2.6\times {10}^{-4}\text{ mol/L}$ for moles of $\text{A}$ and $\left(2.6\times {10}^{-4}\text{ mol/L + 5}\text{.44}\times {10}^{-4}\text{ mol/L}\right)$ for sum of moles of all components in the above equation as

$\begin{array}{c}{\chi }_{{\text{N}}_2}=\frac{2.6\times {10}^{-4}\cancel{\text{mol/}}\text{L}}{\left(2.6\times {10}^{-4}\cancel{\text{mol/L}}\text{ + 5}\text{.44}\times {10}^{-4}\cancel{\text{mol/L}}\right)}\\ =0.677\end{array}$ $\begin{array}{c}{\chi }_{{\text{N}}_2}=\frac{2.6\times {10}^{-4}\text{ mol/L}}{\left(2.6\times {10}^{-4}\text{ mol/L + 5}\text{.44}\times {10}^{-4}\text{ mol/L}\right)}\\ =0.677\end{array}$

For ${\text{O}}_2$,

Substitute $\text{5}\text{.44}\times {10}^{-4}\text{ mol/L}$ for moles of $\text{A}$ and $\left(2.6\times {10}^{-4}\text{ mol/L + 5}\text{.44}\times {10}^{-4}\text{ mol/L}\right)$ for sum of moles of all components in equation (2) as:

$\begin{array}{c}{\chi }_{{\text{N}}_2}=\frac{2.6\times {10}^{-4}\cancel{\text{mol/}}\text{L}}{\left(2.6\times {10}^{-4}\cancel{\text{mol/L}}\text{ + 5}\text{.44}\times {10}^{-4}\cancel{\text{mol/L}}\right)}\\ =0.677\end{array}$ $\begin{array}{c}{\chi }_{{\text{N}}_2}=\frac{\text{5}\text{.44}\times {10}^{-4}\text{ mol/L}}{\left(2.6\times {10}^{-4}\text{ mol/L + 5}\text{.44}\times {10}^{-4}\text{ mol/L}\right)}\\ =0.323\end{array}$

Conclusion

The mole fractions of ${\text{N}}_2{\text{ \& O}}_2$ are 0.677 and 0.323, respectively.