Chapter 12, Problem 12.96QP

Interpretation Introduction

Interpretation:

Mole fractions of Oxygen and Nitrogen air dissolved in water to be determined which then the answer has to be commented.

Concept Introduction:

Mole fraction: Concentration of the solution can also expressed by mole fraction. Mole fraction is equal to moles of the component divided by total moles of the mixture.

$\text{Mole fraction =}\frac{\text{moles of a component}}{\text{total moles}}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

$\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Henry’s law: At constant temperature, amount of dissolved gas and volume of liquid proportional to the partial pressure of that gas in equilibrium with that liquid. This law applies most precisely for dilute solutions of a gas that does not react or dissociate with the solvent.

Henry’s law states that dissolved gases in liquid are directly proportional to the partial pressure of the gas over the solution.

$\text{c = kP}$

Where,

$\text{c}$ is molar concentration of dissolved gas $\text{(mol/L)}$

$\text{k}$ is proportionality constant

$\text{P}$ is partial pressure of the gas over the solution $\text{(mol/L}\text{.atom)}$

Answer to Problem 12.96QP

The mole fraction of Oxygen is $\text{4}{\text{.7×10}}^{\text{-6}}$

The mole fraction of Nitrogen is $\text{9}{\text{.7×10}}^{\text{-6}}$

Explanation of Solution

Given:

Partial pressure of Oxygen in air $\text{=0}\text{.20}\text{atm}$

Partial pressure of Nitrogen in air $\text{=}\text{0}\text{.80}\text{atm}$

Temperature $=298K$

The concentration for oxygen and nitrogen is calculated as follows,

$\begin{array}{c}\text{c = kP}\\ {\text{c}}_{{\text{O}}_{\text{2}}}{\text{ = k}}_{{\text{O}}_{\text{2}}}{\text{P}}_{{\text{O}}_{\text{2}}}\text{=1}{\text{.3×10}}^{\text{-3}}\text{mol/L}\text{.atm×0}\text{.20}\text{atm}\text{=}\text{2}\text{.6}{\text{×10}}^{\text{-4}}\text{M}\\ {\text{c}}_{{\text{N}}_{\text{2}}}{\text{ = k}}_{{\text{N}}_{\text{2}}}{\text{P}}_{{\text{N}}_{\text{2}}}\text{=6}{\text{.8×10}}^{\text{-4}}\text{mol/L}\text{.atm×0}\text{.80}\text{atm}\text{=}5.4{\text{×10}}^{\text{-4}}\text{M}\\ {\text{Where, k}}_{{\text{O}}_{\text{2}}}=\text{1}{\text{.3×10}}^{\text{-3}}\text{mol/L}\text{.atm}\\ {\text{k}}_{{\text{N}}_{\text{2}}}=\text{6}{\text{.8×10}}^{\text{-4}}\text{mol/L}\text{.atm}\text{from}Example12.6\end{array}$

Assume the water density is $\text{1}\text{g/mL}$, the water moles in $\text{1L}$ is determined as follows,

$1000\cancel{g}{H}_{2}O\times \frac{1mol{H}_{2}O}{18.02\cancel{g}{H}_{2}O}=55.5mol{H}_{2}O$

Calculate the mole fractions of oxygen and nitrogen:

$\begin{array}{c}{\text{X}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{2}{\text{.6×10}}^{\text{-4}}\text{mol}{}_{{\text{O}}_{\text{2}}}}{\text{55}\text{.5}\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \text{= 4}{\text{.7×10}}^{\text{-6}}\\ {\text{X}}_{N_{\text{2}}}\text{=}\frac{\text{5}{\text{.4×10}}^{\text{-4}}\text{mol}{}_{N_{\text{2}}}}{\text{55}\text{.5}\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ \text{= 9}{\text{.7×10}}^{\text{-6}}\end{array}$

Comment about the obtained answer:

$\begin{array}{l}\text{In air}\Rightarrow \frac{{\text{X}}_{{\text{O}}_{\text{2}}}}{{\text{X}}_{N_{\text{2}}}}=\frac{0.20}{0.80}=0.25\\ \text{In water}\Rightarrow \frac{{\text{X}}_{{\text{O}}_{\text{2}}}}{{\text{X}}_{N_{\text{2}}}}=\frac{4.7\times {10}^{-6}}{9.7\times {10}^{-6}}=0.48\end{array}$

The ratios between oxygen and nitrogen mole fractions in air and water are represented above.

Comparing the oxygen and nitrogen mole fractions in air and water shows that in water their mole fraction are greater due to the fact that oxygen is more soluble in water than nitrogen. Hence, fish is more efficient in extracting oxygen which is in small concentration in water.

Conclusion

Mole fractions of Oxygen and Nitrogen air dissolved in water were determined which then the answer was commented.