Chapter U4.13, Problem 6E

a.

Interpretation Introduction

Interpretation: The molarity of $\text{29}\text{.2 g NaCl per 0}\text{.50 L}$ needs to be determined.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 6E

The molarity of $\text{29}\text{.2 g NaCl per 0}\text{.50 L}$ is $\text{0}\text{.998 M}$ .

Explanation of Solution

The formula to calculate number of moles is:

  $\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$

Now, equation (1) an be rewritten as:

  $\text{Molarity = }\frac{\frac{\text{mass of the solute}}{\text{Molar mass of the solute}}}{\text{Volume of solution (L)}}$ - (2)

The molar mass of $\text{NaCl}$ is $\text{58}\text{.44 g/mol}$ .

Substituting the given values in equation (2) to determine the molarity of the solution as:

  $\begin{array}{l}\text{Molarity = }\frac{\frac{\text{29}\text{.2 g}}{\text{58}\text{.44 g/mol}}}{\text{0}\text{.50 L}}\\ \text{Molarity = }\frac{\text{0}\text{.499 mol}}{\text{0}\text{.50 L}}\\ \text{Molarity = 0}\text{.998 mol/L = 0}\text{.998 M}\end{array}$

Hence, the molarity of $\text{29}\text{.2 g NaCl per 0}\text{.50 L}$ is $\text{0}\text{.998 M}$ .

b.

Interpretation Introduction

Interpretation: The molarity of $\text{5}\text{.8 g NaCl per 50 mL}$ needs to be determined.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 6E

The molarity of $\text{5}\text{.8 g NaCl per 50 mL}$ is $\text{1}\text{.98 M}$ .

Explanation of Solution

The formula to calculate number of moles is:

  $\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$

Now, equation (1) an be rewritten as:

  $\text{Molarity = }\frac{\frac{\text{mass of the solute}}{\text{Molar mass of the solute}}}{\text{Volume of solution (L)}}$ - (2)

The molar mass of $\text{NaCl}$ is $\text{58}\text{.44 g/mol}$ .

Conversion of volume from mL to L as:

  $\text{50 mL}\times \frac{1\text{ L}}{1000\text{ mL}}=0.05\text{ L}$

Substituting the given values in equation (2) to determine the molarity of the solution as:

  $\begin{array}{l}\text{Molarity = }\frac{\frac{\text{5}\text{.8 g}}{\text{58}\text{.44 g/mol}}}{\text{0}\text{.05 L}}\\ \text{Molarity = }\frac{\text{0}\text{.099 mol}}{\text{0}\text{.05 L}}\\ \text{Molarity = 1}\text{.98 mol/L = 1}\text{.98 M}\end{array}$

Hence, the molarity of $\text{5}\text{.8 g NaCl per 50 mL}$ is $\text{1}\text{.98 M}$ .

c.

Interpretation Introduction

Interpretation: The molarity of $\text{2}\text{.9 g NaCl per 10}\text{.2 mL}$ needs to be determined and the solutions should be listed in increasing molarity.

Concept Introduction: The molar concentration of a solute in a solution can be expressed in terms of molarity as moles of solute present in per liter of solution. Mathematically,

  $\text{Molarity = }\frac{\text{number of moles of solute}}{\text{Volume of solution (L)}}$ - (1)

Answer to Problem 6E

The molarity of $\text{2}\text{.9 g NaCl per 10}\text{.2 mL}$ is $\text{4}\text{.863 M}$ .

The list of solutions in increasing order of molarity is:

  $\text{29}\text{.2 g NaCl per 0}\text{.50 L}<\text{5}\text{.8 g NaCl per 50 mL}<\text{2}\text{.9 g NaCl per 10}\text{.2 mL}$

Explanation of Solution

The formula to calculate number of moles is:

  $\text{number of moles = }\frac{\text{mass of the substance}}{\text{Molar mass of the substance}}$

Now, equation (1) an be rewritten as:

  $\text{Molarity = }\frac{\frac{\text{mass of the solute}}{\text{Molar mass of the solute}}}{\text{Volume of solution (L)}}$ - (2)

The molar mass of $\text{NaCl}$ is $\text{58}\text{.44 g/mol}$ .

Conversion of volume from mL to L as:

  $\text{10}\text{.2 mL}\times \frac{1\text{ L}}{1000\text{ mL}}=\text{0}\text{.0102 L}$

Substituting the given values in equation (2) to determine the molarity of the solution as:

  $\begin{array}{l}\text{Molarity = }\frac{\frac{\text{2}\text{.9 g}}{\text{58}\text{.44 g/mol}}}{\text{0}\text{.0102 L}}\\ \text{Molarity = }\frac{\text{0}\text{.0496 mol}}{\text{0}\text{.0102 L}}\\ \text{Molarity = 4}\text{.863 mol/L = 4}\text{.863 M}\end{array}$

Hence, the molarity of $\text{29}\text{.2 g NaCl per 0}\text{.50 L}$ is $\text{0}\text{.998 M}$ .

The molarity of all the three solutions is:

• The molarity of $\text{29}\text{.2 g NaCl per 0}\text{.50 L}$ is $\text{0}\text{.998 M}$ .
• The molarity of $\text{5}\text{.8 g NaCl per 50 mL}$ is $\text{1}\text{.98 M}$ .
• The molarity of $\text{2}\text{.9 g NaCl per 10}\text{.2 mL}$ is $\text{4}\text{.863 M}$ .

Hence, the list of solution in increasing order of molarity is:

  $\text{29}\text{.2 g NaCl per 0}\text{.50 L}<\text{5}\text{.8 g NaCl per 50 mL}<\text{2}\text{.9 g NaCl per 10}\text{.2 mL}$