Chapter U4.13, Problem 7E

a.

Interpretation Introduction

Interpretation: The molarity of ${\text{25 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ 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 7E

The molarity of ${\text{25 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ is $\text{0}\text{.56 M}$ .

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 glucose, ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $\text{180}\text{.156 g/mol}$ .

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

  $\begin{array}{l}\text{Molarity = }\frac{\frac{\text{25 g}}{\text{180}\text{.156 g/mol}}}{\text{0}\text{.25 L}}\\ \text{Molarity = }\frac{\text{0}\text{.1388 mol}}{\text{0}\text{.25 L}}\\ \text{Molarity = 0}\text{.56 mol/L = 0}\text{.56 M}\end{array}$

Hence, the molarity of ${\text{25 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ is $\text{0}\text{.56 M}$ .

b.

Interpretation Introduction

Interpretation: The molarity of ${\text{50 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ 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 7E

The molarity of ${\text{50 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ is $\text{1}\text{.11 M}$ .

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 glucose, ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ is $\text{180}\text{.156 g/mol}$ .

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

  $\begin{array}{l}\text{Molarity = }\frac{\frac{\text{50 g}}{\text{180}\text{.156 g/mol}}}{\text{0}\text{.25 L}}\\ \text{Molarity = }\frac{\text{0}\text{.278 mol}}{\text{0}\text{.25 L}}\\ \text{Molarity = 1}\text{.11 mol/L = 1}\text{.11 M}\end{array}$

Hence, the molarity of ${\text{50 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ is $\text{1}\text{.11 M}$ .

c.

Interpretation Introduction

Interpretation: The molarity of ${\text{50 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}\text{ per 0}\text{.25 L}$ 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 7E

The molarity of ${\text{50 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}\text{ per 0}\text{.25 L}$ is $\text{0}\text{.58 M}$ .

The list of solution in increasing order of molarity is:

  ${\text{25 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}{\text{.25 L < 50 g C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{ per 0}{\text{.25 L < 50 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$

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 sucrose, ${\text{C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}$ is $\text{342}\text{.3 g/mol}$ .

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

  $\begin{array}{l}\text{Molarity = }\frac{\frac{\text{50 g}}{\text{342}\text{.3 g/mol}}}{\text{0}\text{.25 L}}\\ \text{Molarity = }\frac{\text{0}\text{.146 mol}}{\text{0}\text{.25 L}}\\ \text{Molarity = 0}\text{.58 mol/L = 0}\text{.58 M}\end{array}$

Hence, the molarity of ${\text{50 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}\text{ per 0}\text{.25 L}$ is $\text{0}\text{.58 M}$ .

The molarity of all the three solutions is:

• The molarity of ${\text{25 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ is $\text{0}\text{.56 M}$ .
• The molarity of ${\text{50 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$ is $\text{1}\text{.11 M}$ .
• The molarity of ${\text{50 g C}}_{12}{\text{H}}_{\text{22}}{\text{O}}_{11}\text{ per 0}\text{.25 L}$ is $\text{0}\text{.58 M}$ .

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

  ${\text{25 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}{\text{.25 L < 50 g C}}_{\text{12}}{\text{H}}_{\text{22}}{\text{O}}_{\text{11}}\text{ per 0}{\text{.25 L < 50 g C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\text{ per 0}\text{.25 L}$