Chapter 13, Problem 13.74P

Interpretation Introduction

Interpretation:

The molality, molarity and mole fraction of ${\text{NH}}_3$ is to be determined.

Concept introduction:

Molality is the measure of the concentration of solute in the solution. It is the amount of solute that is dissolved in one kilogram of the solvent. It is represented by $\text{m}$ and its unit is moles per kilograms. The solute is the substance that is present in a smaller amount and solvent is the substance that is present in a larger amount.

The formula to calculate the molality of the solution is as follows:

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{mass}\left(\text{kg}\right)\text{of}\text{}\text{solvent}}$ (1)

Molarity is defined as the number of moles of solute that are dissolved in one litre of solution. It is represented by $\text{M}$ and its unit is $\text{mol/L}$.

The formula to calculate the molarity of the solution is as follows:

$\text{Molarity}=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (2)

The mole fraction is defined as the ratio of the number of moles of solute to the total number of moles in the mixture. It is represented by $X$.

The formula to calculate the mole fraction is as follows:

$\text{Mole}\text{fraction}\left(X\right)=\frac{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}}{\text{amount}\left(\text{mol}\right)\text{of}\text{solute}+\text{amount}\left(\text{mol}\right)\text{of}\text{solvent}}$ (3)

Answer to Problem 13.74P

The molality, molarity and mole fraction of the given solution of ${\text{NH}}_3$ is $5.11m$, $4.53M$ and $0.0843$.

Explanation of Solution

The formula to calculate the molality of the aqueous solution of ${\text{NH}}_3$ is as follows:

$\text{Molality}=\frac{\text{amount}\left(\text{mol}\right)\text{of}{\text{NH}}_3}{\text{mass}\left(\text{kg}\right)\text{of}{\text{H}}_2\text{O}}$ (4)

The formula to calculate the moles of a compound is as follows:

$\text{Moles of compound}=\left(\frac{\text{given mass}}{\text{molar mass of compound}}\right)$ (5)

The formula to calculate the moles of ${\text{NH}}_3$ is as follows:

$\text{Moles}\text{of}{\text{NH}}_3=\frac{\text{given mass }}{\text{molar mass of}{\text{NH}}_3}\left(\frac{\text{mass}\%\text{of}{\text{NH}}_3}{100\%\text{of}\text{solution}}\right)$ (6)

Substitute $100\text{g}$ for the given mass, $8\%$ for the mass percent of ${\text{NH}}_3$ and $17.03\text{g/mol}$ for the molar mass of ${\text{NH}}_3$ in equation (6).

$\begin{array}{c}\text{Amount}\text{of}{\text{NH}}_3=\left(100\text{g}\text{solution}\right)\left(\frac{8\%{\text{NH}}_3}{100\%\text{solution}}\right)\left(\frac{1\text{}\text{mol}{\text{NH}}_3}{17.03\text{g}{\text{NH}}_3}\right)\\ =0.469759\text{mol}\end{array}$

The formula to calculate the mass of ${\text{H}}_2\text{O}$ is as follows:

$\text{Mass}\text{of}{\text{H}}_2\text{O}=\left(\text{Mass}\text{of}\text{solution}-\text{Mass}\text{of}{\text{NH}}_3\right)$ (7)

Substitute $100\text{}\text{g}$ for the mass of solution and $8\text{g}$ for the mass of ${\text{NH}}_3$ in equation (7).

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_2\text{O}=\left(100\text{}\text{g}-8\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\\ =0.092\text{kg}\end{array}$

Substitute $0.469759\text{mol}$ for the amount of ${\text{NH}}_3$ and $0.092\text{kg}$ for the mass of ${\text{H}}_2\text{O}$ in equation (4).

$\begin{array}{c}\text{Molality}\text{of}{\text{NH}}_3=\frac{\left(0.469759\text{mol}\right)}{\left(0.092\text{kg}\right)}\\ =5.106076m\\ =5.11m\end{array}$

The formula to calculate the molarity of the aqueous solution of ${\text{NH}}_3$ is as follows:

$\text{Molarity}\text{of}{\text{NH}}_3=\frac{\text{amount}\left(\text{mol}\right)\text{of}{\text{NH}}_3}{\text{volume }\left(\text{L}\right)\text{of}\text{}\text{solution}}$ (8)

The formula to calculate the volume of the solution is as follows:

$\text{Volume}\text{of}\text{solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}$ (9)

Substitute $100\text{g}$ for the mass of solution and $0.9651\text{}\text{g/mL}$ for the density of the solution in equation (9).

$\begin{array}{c}\text{Volume}\text{of}\text{solution}=\left(100\text{g}\text{solution}\right)\left(\frac{1\text{mL}}{0.9651\text{}\text{g}\text{solution}}\right)\left(\frac{{10}^{-3}\text{L}}{1\text{mL}}\right)\\ =0.103616\text{ L}\end{array}$

Substitute $0.469759\text{mol}$ for the amount of ${\text{NH}}_3$ and $0.103616\text{ L}$ for the volume of solution in equation (8).

$\begin{array}{c}\text{Molarity}\text{of}{\text{NH}}_3=\frac{0.469759\text{mol}}{0.103616\text{ L}}\\ =4.53365M\\ =\text{4}\text{.53}M\end{array}$

The formula to calculate the mole fraction of ${\text{NH}}_3$ is as follows:

$\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{\text{amount}\left(\text{mol}\right)\text{of}{\text{NH}}_3}{\text{amount}\left(\text{mol}\right)\text{of}{\text{NH}}_3+\text{amount}\left(\text{mol}\right)\text{of}{\text{H}}_2\text{O}}$ (10)

Substitute $92\text{g}$ for the given mass and $18.02\text{}\text{g/mol}$ for the molar mass in equation (5).

$\begin{array}{c}\text{Amount}\text{of}{\text{H}}_2\text{O}=\left(92\text{g}\right)\left(\frac{1\text{mol}{\text{H}}_2\text{O}}{18.02\text{}\text{g}{\text{H}}_2\text{O}}\right)\\ =5.1054\text{mol}\end{array}$

Substitute $0.469759\text{mol}$ for the amount of ${\text{NH}}_3$ and $5.1054\text{mol}$ for the amount of ${\text{H}}_2\text{O}$ in equation (10).

$\begin{array}{c}\text{Mole}\text{fraction}\text{of}{\text{NH}}_3=\frac{0.469759\text{mol}}{\left(0.469759\text{mol}\right)+\left(5.1054\text{mol}\right)}\\ =\frac{1.80723\times {10}^{-3}\text{mol}}{5.57516\text{}\text{mol}}\\ =0.084259\\ =0.0843\end{array}$

Conclusion

The molality, molarity and mole fraction of the given solution of ${\text{NH}}_3$ is $5.11m$, $4.53M$ and $0.0843$.