Chapter U4, Problem SIII8RE

Interpretation Introduction

Interpretation: The mass in grams for magnesium sulfate needed to make 100 mL of 0.1 M solution needs to be determined.

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:

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

The relation between concentration and moles is as given below;

  $\text{Molarity = }\frac{\text{moles }}{\text{volume }}$

Answer to Problem SIII8RE

  $\text{mass =1}\text{.2g}$

Explanation of Solution

Given information:

Volume of solution = 100 mL = 0.100 L

Molar mass of MgSO₄ = 120 g/mol

Concentration = 0.1 M

  $\begin{array}{l}\text{Molarity = }\frac{\text{moles }}{\text{volume }}\\ \text{Molarity = }\frac{\text{mass }}{\text{molar mass }\times \text{volume }}\end{array}$

Substitute the values to calculate the molarity:

  $\begin{array}{l}\text{Molarity = }\frac{\text{mass }}{\text{molar mass ×volume }}\\ \text{0}\text{.10 mol/L = }\frac{\text{mass }}{\text{120 g/mol ×0}\text{.100 L }}\\ \text{mass =0}\text{.10 mol/L×120 g/mol ×0}\text{.100 L}\\ \text{mass =1}\text{.2g}\end{array}$

Conclusion

Thus,

  $\text{mass =1}\text{.2g}$