Chapter 7, Problem 7.44E

Interpretation Introduction

Interpretation:

The approximate molarity of the given solution and the value of Henry’s law constant for the given gas are to be calculated.

Concept introduction:

According to Henry’s law, the temperature and solubility of dissolved gas are inversely related. When temperature increases, kinetic energy increases which causes the molecules to break the intermolecular bonds and escape from the solution. Hence solubility decreases.

Answer to Problem 7.44E

The approximate molarity is $\text{2}\text{.32}\times {10}^{-3}\text{ M}$. The Henry’s law constant is $2.43\times {10}^9\text{ Pa}$.

Explanation of Solution

The mole fraction of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ was found to be $4.17\times {10}^{-5}$ at normal pressure.

The mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is calculated by the formula,

${\text{Mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}=\text{Number}\text{of}\text{moles}\times {\text{Molar mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}$

Molar mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is $120.91 \text{g/mol}$.

The number of moles of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is $4.17\times {10}^{-5}$.

The number of moles of ${\text{H}}_{\text{2}}\text{O}$ is $1-\left(4.17\times {10}^{-5}\right)=0.999$

Substitute the values of mass and molar mass in the above formula.

$\begin{array}{rll}{\text{Mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}& =& \text{Number}\text{of}\text{moles}\times {\text{Molar mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}\\ & =& 4.17\times {10}^{\text{-5}}\text{ mol}\times 120.91 \text{g/mol}\\ & =& \text{5}\text{.04}\times {10}^{\text{-3}}\text{ g}\end{array}$

The mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is calculated by the formula,

${\text{Mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}=\text{Number}\text{of}\text{moles}\times {\text{Molar mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}$

Molar mass of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is $120.91 \text{g/mol}$.

The number of moles of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ is $4.17\times {10}^{-5}$.

Substitute the values of mass and molar mass in the above formula.

$\begin{array}{rll}{\text{Mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}& =& \text{Number}\text{of}\text{moles}\times {\text{Molar mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}\\ & =& 4.17\times {10}^{\text{-5}}\text{ mol}\times 120.91 \text{g/mol}\\ & =& \text{5}\text{.04}\times {10}^{\text{-3}}\text{ g}\end{array}$

The mass of ${\text{H}}_{\text{2}}\text{O}$ is calculated by the formula,

${\text{Mass of H}}_{\text{2}}\text{O}=\text{Number}\text{of}\text{moles}\times {\text{Molar mass of H}}_{\text{2}}\text{O}$

Molar mass of ${\text{H}}_{\text{2}}\text{O}$ is $18 \text{g/mol}$.

Substitute the values of mass and molar mass of ${\text{H}}_{\text{2}}\text{O}$ in the above formula.

$\begin{array}{rll}{\text{Mass of H}}_{\text{2}}\text{O}& =& \text{Number}\text{of}\text{moles}\times {\text{Molar mass of H}}_{\text{2}}\text{O}\\ & =& 0.999\text{ mol}\times 18 \text{g/mol}\\ & =& 17.982\text{ g}\end{array}$

The mass of the solution will be,

$\begin{array}{rll}\text{Mass of solution}& =& {\text{Mass of CCl}}_{\text{2}}{\text{F}}_{\text{2}}+{\text{Mass of H}}_{\text{2}}\text{O}\\ & =& \text{5}\text{.04}\times {10}^{\text{-3}}\text{ g}+17.982\text{ g}\\ & =& \text{17}\text{.987 g}\end{array}$

The volume of solution is calculated by the formula,

$\text{Volume}=\frac{\text{Mass}\text{of}\text{solution}}{\text{Density of water}}$

The density of water is $\text{1}{\text{.00 g/cm}}^{\text{3}}$.

Substitute the value of mass and density in the above formula.

$\begin{array}{rll}\text{Volume}& =& \frac{17.987\text{g}}{{\text{1 g/cm}}^3}\\ & =& 17.98{\text{7 cm}}^{\text{3}}\end{array}$

Convert $17.98{\text{7 cm}}^{\text{3}}$ to $\text{L}$.

$\begin{array}{rll}17.98{\text{7 cm}}^{\text{3}}& =& \frac{17.98\text{7 }}{1000}\text{L}\\ & =& 0.0179\text{L}\end{array}$

Therefore, the volume of solution is $0.0179\text{L}$.

The molarity of the aqueous solution is calculated by the formula,

$\text{Molarity}=\frac{\text{Moles of Solute}}{\text{Liters of solution}}$

Substitute the values of number of moles of ${\text{CCl}}_{\text{2}}{\text{F}}_{\text{2}}$ and volume in the above equation.

$\begin{array}{rll}\text{Molarity}& =& \frac{\text{Moles of Solute}}{\text{Liters of solution}}\\ & =& \frac{4.17\times {10}^{-5}\text{ mol}}{\text{0}\text{.0179 L}}\\ & =& \text{2}\text{.32}\times {10}^{-3}\text{ mol/L}\\ & =& \text{2}\text{.32}\times {10}^{-3}\text{ M}\end{array}$

The expression for Henry’s Law is shown below.

$p_{\text{i}}=K_{\text{i}}\times x_{\text{i}}$

Where,

• $p_{\text{i}}$ is the pressure.

• $K_{\text{i}}$ is Henry constant.

• $x_{\text{i}}$ is mole fraction.

The standard pressure in Pascal’s is $101325\text{ Pa}$.

The value of mole fraction is $4.17\times {10}^{-5}$.

Substitute the value of pressure and value of mole fraction in the above expression.

$\begin{array}{rll}{K}_{{\text{CCl}}_2{\text{F}}_{\text{2}}}& =& \frac{p}{x_{{\text{CCl}}_2{\text{F}}_{\text{2}}}}\\ & =& \frac{101325\text{ Pa}}{4.17\times {10}^{-5}}\\ & =& 2.43\times {10}^9\text{ Pa}\end{array}$

Therefore, the Henry’s law constant is $2.43\times {10}^9\text{ Pa}$.

Conclusion

The Henry’s law constant is $2.43\times {10}^9\text{ Pa}$.