Chapter 15.2, Problem 5RQ

Interpretation Introduction

Interpretation: The molarity of $10\text{g}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ solution in $450\text{mL}$ has to be determined.

Concept introduction: Molarity is a concentration term that is defined as the number of moles of solute dissolved in one litre of the solution. It is denoted by M, and its unit is mol/L.

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

Answer to Problem 5RQ

Molarity of $10\text{g}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ solution in $450\text{mL}$ is $0.2776\text{M}$ .

Explanation of Solution

Given information: Mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $10\text{g}$ and its volume is $450\text{mL}$ .

molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ can be calculated as follows:

  $\begin{array}{c}{\text{Molar mass of NH}}_{\text{4}}{\text{NO}}_{\text{3}}=\left[\begin{array}{l}\left(\text{number}\text{of}\text{}\text{N}\right)\left(\text{Atomic}\text{mass}\text{of}\text{N}\right)\text{+}\\ \left(\text{number}\text{of}\text{}\text{H}\right)\left(\text{Atomic}\text{mass}\text{of}\text{H}\right)+\\ \left(\text{number}\text{of}\text{}\text{O}\right)\left(\text{Atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(\text{2}\right)\left(\text{14}\text{.0067}\text{g/mol}\right)\text{+}\\ \left(\text{4}\right)\left(\text{1}\text{.00784}\text{g/mol}\right)+\\ \left(\text{3}\right)\left(15.999\text{g/mol}\right)\end{array}\right]\\ =80.04176\text{g/mol}\\ \end{array}$

Number of moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ can be calculated by formula as follows:

  ${\text{Moles of NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{= }\frac{\text{mass}{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{molar mass}{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}$

Substitute $10\text{g}$ for mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ and $80.04176\text{g/mol}$ for molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in above equation.

  $\begin{array}{c}{\text{Moles of NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{= }\frac{\text{mass}{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{molar mass}{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\\ \text{=}\frac{10\text{g}}{80.04176\text{g/mol}}\\ =0.12493\text{mol}\end{array}$

Now, ${\text{Molarity of NH}}_{\text{4}}{\text{NO}}_{\text{3}}=\frac{{\text{moles of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{volume}{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}$

Substitute $0.12493\text{mol}$ for moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ and $450\text{mL}$ for volume of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ in above equation.

  $\begin{array}{c}{\text{Molarity of NH}}_{\text{4}}{\text{NO}}_{\text{3}}=\frac{{\text{moles of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{volume}{\text{of NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\\ =\frac{0.12493\text{mol}}{450\text{mL}\left(\frac{1\text{L}}{1000\text{mL}}\right)}\\ =0.2776\text{M}\end{array}$

Conclusion

Molarity of $10\text{g}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ solution in $450\text{mL}$ is $0.2776\text{M}$ .