Chapter 15.2, Problem 9SP

Interpretation Introduction

Interpretation: The mass of $N{a}_{2}C{O}_3.10H_{2}O$ should be calculated that is required to prepare the $\text{5}\text{.00 g}$ solution with anhydrous $N{a}_{2}C{O}_3$ .

Concept Introduction: Mole concept is used to calculate the moles, mass, number of atoms, and moles of a compound. The relation between these values can be shown as:

  $\begin{array}{l}\text{Moles = }\frac{\text{mass}}{\text{molar mass}}\\ {\text{1 mole = N}}_{\text{a}}\text{ atoms}\\ {\text{N}}_{\text{a}}\text{ = 6}\text{.022 }\times {\text{ 10}}^{\text{23}}\end{array}$

A chemical reaction follows the Law of conservation of mass that states before and after the reaction the mass of the reactant and product remain the same. It means mass is neither created nor destroyed.

Answer to Problem 9SP

  $\text{ 13}\text{.5 g}$of $N{a}_{2}C{O}_3.10H_{2}O$ is needed.

Explanation of Solution

  $N{a}_{2}C{O}_3.10H_{2}O\to N{a}_{2}C{O}_3+10H_{2}O$

Hence, 1 mole of $N{a}_{2}C{O}_3.10H_{2}O$ prepare 1 mole of $N{a}_{2}C{O}_3$ .

Mass of $N{a}_{2}C{O}_3$ = $\text{5}\text{.00 g}$

Molar mass of $N{a}_{2}C{O}_3$ = $105.98\text{g/mol}$

Moles of $N{a}_{2}C{O}_3$ = $\frac{5.00g}{105.98\text{g/mol}}=\text{0}\text{.047}moles$

Hence, $\text{0}\text{.047}moles$ of $N{a}_{2}C{O}_3.10H_{2}O$ is needed.

Molar mass of $N{a}_{2}C{O}_3.10H_{2}O$ = $286\text{g/mol}$

Mass of $N{a}_{2}C{O}_3.10H_{2}O$ = $\text{0}\text{.047}moles\times 286g/mol=\text{13}\text{.5}g$

Conclusion

Moles of substances can be determined with the help of mass and molar mass of it.