Chapter 15, Problem 28A

Interpretation Introduction

Interpretation: The mass of the solid sodium carbonate is to be determined.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V\left(\text{L}\right)}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

Answer to Problem 28A

The mass of the sodium carbonate is $63.\text{6 g}$ .

Explanation of Solution

Given information:

The volume of the solution, $V=\text{300}\text{.0 mL=300}{\text{.0×10}}^{\text{-3}}\text{ L}$

The molarity of the solution, $M=2.0\text{M}$

The molar mass of sodium carbonate is, $\text{molar}\text{mass=}\text{106 g/mol}$

Calculation:

The molarity formula is,

  $M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}$

Firstly, find the number of moles of the sodium carbonate,

  $\begin{array}{l}n=\frac{\text{mass}}{\text{molar}\text{mass}}\\ n=\frac{\text{mass}}{\text{106 g/mol}}\end{array}$

Now, use this value of number of moles in the molarity formula, and the result is,

  $\begin{array}{l}M=\frac{\left(\frac{\text{mass}}{\text{106g/mol}}\right)}{\text{300}{\text{.0×10}}^{\text{-3}}\text{L}}\\ \Rightarrow \text{mass}=\left(\text{2}\text{.0}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\right)\text{×300}{\text{.0×10}}^{\text{-3}}\text{ L×106 g/mol}\\ \text{mass}\text{=}\text{63}\text{.6 g}\end{array}$

Conclusion

The mass is $\text{63}\text{.6 g}$ .