Chapter U4, Problem SIII7RE

(a)

Interpretation Introduction

Interpretation: The molarity of the solution that is prepared by dissolving 7.30 g of NaCl to make 500 mL of the 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 SIII7RE

  $\text{Molarity = 0}\text{.250M}$

Explanation of Solution

Mass of NaCl = 7.30 g

Volume of solution = 500 mL = 0.500 L

Molar mass of NaCl = 58.5 g/mol

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

Substitute the values to calculate the molarity:

  $\begin{array}{l}\text{Molarity = }\frac{\text{mass }}{\text{molar mass ×volume }}\\ \text{Molarity = }\frac{\text{7}\text{.30 g}}{\text{58}\text{.5 g/mol ×0}\text{.500L }}\\ \text{Molarity = 0}\text{.250M}\end{array}$

(b)

Interpretation Introduction

Interpretation: The molarity of the solution that is prepared by dissolving 43.84 g of NaCl to make 1 L of the 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 SIII7RE

  $\text{Molarity = 0}\text{.750 M}$

Explanation of Solution

Mass of NaCl = 43.84 g

Volume of solution = 1.0 L

Molar mass of NaCl = 58.5 g/mol

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

Substitute the values to calculate the molarity:

  $\begin{array}{l}\text{Molarity = }\frac{\text{mass }}{\text{molar mass ×volume }}\\ \text{Molarity = }\frac{\text{43}\text{.84 g}}{\text{58}\text{.5 g/mol × 1 L }}\\ \text{Molarity = 0}\text{.750 M}\end{array}$

(c)

Interpretation Introduction

Interpretation: The molarity of the solution that is prepared by dissolving 25.00 g of NaCl to make 1 L of the 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 SIII7RE

  $\text{Molarity = 0}\text{.428 M}$

Explanation of Solution

Mass of NaCl = 25.00 g

Volume of solution = 1.0 L

Molar mass of NaCl = 58.5 g/mol

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

Substitute the values to calculate the molarity:

  $\begin{array}{l}\text{Molarity = }\frac{\text{mass }}{\text{molar mass ×volume }}\\ \text{Molarity = }\frac{\text{25}\text{.00 g}}{\text{58}\text{.5 g/mol × 1 L }}\\ \text{Molarity = 0}\text{.428 M}\end{array}$

(d)

Interpretation Introduction

Interpretation: The molarity of the solution that is prepared by dissolving 25.00 g of NaCl to make 2 L of the 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 SIII7RE

  $\text{Molarity = 0}\text{.214 M}$

Explanation of Solution

Mass of NaCl = 25.00 g

Volume of solution = 2.0 L

Molar mass of NaCl = 58.5 g/mol

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

Substitute the values to calculate the molarity:

  $\begin{array}{l}\text{Molarity = }\frac{\text{mass }}{\text{molar mass ×volume }}\\ \text{Molarity = }\frac{\text{25}\text{.00 g}}{\text{58}\text{.5 g/mol × 2 L }}\\ \text{Molarity = 0}\text{.214 M}\end{array}$