Chapter 14.2, Problem 9PP

Interpretation Introduction

Interpretation:

The percent by mass of sodium hydrogen carbonate $\left(\text{N}\text{a}\text{H}\text{C}{\text{O}}_3\right)$ in a solution is to be calculated where $20\text{g}\text{m}$ sodium hydrogen carbonate $\left(\text{N}\text{a}\text{H}\text{C}{\text{O}}_3\right)$ is dissolved in $600\text{m}\text{L}$ water.

Concept Introduction:

The concentrations of solutions are expressed in different ways. All of them are involved in the quantity of solute and the quantity of solution or solvent. Among them, Mass percentage (w/W percent) is used to express the concentration of the solution.

Mass percentage (w/W percent):

It is expressed as the ratio of the mass of the solute to the mass of the solution multiplied by 100. (Mass of the solution = mass of the solute + mass of the solvent)

In expression, $\text{Mass}\text{p}\text{e}\text{r}\text{c}\text{e}\text{n}\text{t}\text{a}\text{g}\text{e}\left(\text{w}/\text{W}\right)\text{\%}= \frac{\text{Mass of solute}}{\text{Mass of}\text{solution}}\times 100\text{\%}$

Answer to Problem 9PP

The percent by mass of sodium hydrogen carbonate $\left(\text{N}\text{a}\text{H}\text{C}{\text{O}}_3\right)$ in a solution is $3.22\text{\%}$ .

Explanation of Solution

  $\text{Mass}\text{p}\text{e}\text{r}\text{c}\text{e}\text{n}\text{t}\text{a}\text{g}\text{e}\left(\text{w}/\text{W}\right)\text{\%}= \frac{\text{Mass of solute}}{\text{Mass of}\text{solution}}\times 100\text{\%}$

  $\begin{array}{l}\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{v}\text{e}\text{n}\text{t}\left(\text{W}\text{a}\text{t}\text{e}\text{r}\right)=\text{V}\text{o}\text{l}\text{u}\text{m}\text{e}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\times \text{d}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\hfill \\ =600\text{m}\text{L}\times 1.0\left(\text{g}\text{m}/\text{m}\text{L}\right)=600\text{g}\text{m}\hfill \end{array}$

Mass of NaHCO₃ in water = $20\text{g}\text{m}$

Mass of the solvent = $600\text{g}\text{m}$

  $\begin{array}{l}\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{t}\text{h}\text{e}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}=\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{N}\text{a}\text{H}\text{C}{\text{O}}_3+\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{w}\text{a}\text{t}\text{e}\text{r}\hfill \\ \begin{array}{l} =\left(20+600\right)\text{g}\text{m}\end{array}\hfill \end{array}$

           $\begin{array}{c}=620\text{g}\text{m}\end{array}$

  $\text{Percent}\text{b}\text{y}\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{N}\text{a}\text{H}\text{C}{\text{O}}_3\text{i}\text{n}\text{a}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}= \frac{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{N}\text{a}\text{H}\text{C}{\text{O}}_3}{\text{M}\text{a}\text{s}\text{s}\text{o}\text{f}\text{S}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}}\times 100\text{\%}$

  $\text{Percent}\text{b}\text{y}\text{m}\text{a}\text{s}\text{s}\text{o}\text{f}\text{N}\text{a}\text{H}\text{C}{\text{O}}_3\text{i}\text{n}\text{a}\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}= \frac{20}{620}\times 100\text{\%}$

                     = $3.22\text{\%}$

Conclusion

Percent by mass of sodium hydrogen carbonate $\left(\text{N}\text{a}\text{H}\text{C}{\text{O}}_3\right)$ in a solution is $3.22\text{\%}$ .