Chapter 14, Problem 18PE

(a)

Interpretation Introduction

Interpretation:

Molarity of $0.50\text{ mol}$ of solute in $\text{125 mL}$ of solution has to be determined.

Concept Introduction:

Molarity is amount of solute per $1\text{ L}$ of solution. Formula for molarity is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$        (1)

Answer to Problem 18PE

Molarity of $0.50\text{ mol}$ of solute in $\text{125 mL}$ of solution is $4M$.

Explanation of Solution

Substitute $0.50\text{ mol}$ for moles of solute and $\text{125 mL}$ for volume of solution in equation (1).

  $\begin{array}{c}\text{Molarity of solution}=\left(\frac{0.50\text{ mol}}{\text{125 mL}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\\ =4M\end{array}$

Hence, molarity of given solution is $4M$.

(b)

Interpretation Introduction

Interpretation:

Molarity of $2.25\text{ mol}$ of ${\text{CaCl}}_2$ in $\text{1}\text{.50 L}$ of solution has to be determined.

Concept Introduction:

Molarity is amount of solute per $1\text{ L}$ of solution. Formula for molarity is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$        (1)

Answer to Problem 18PE

Molarity of $2.25\text{ mol}$ of ${\text{CaCl}}_2$ in $\text{1}\text{.50 L}$ of solution is $1.5M$.

Explanation of Solution

Formula for molarity of ${\text{CaCl}}_2$ solution is as follows:

  ${\text{Molarity of CaCl}}_2\text{ solution}=\frac{{\text{Moles of CaCl}}_2}{\text{Volume }\left(\text{L}\right){\text{ of CaCl}}_2\text{ solution}}$        (2)

Substitute $2.25\text{ mol}$ for moles and $\text{1}\text{.50 L}$ for volume of ${\text{CaCl}}_2$ solution in equation (2).

  $\begin{array}{c}{\text{Molarity of CaCl}}_2\text{ solution}=\frac{2.25\text{ mol}}{\text{1}\text{.50 L}}\\ =1.5M\end{array}$

Hence, molarity of given solution is $1.5M$.

(c)

Interpretation Introduction

Interpretation:

Molarity of $\text{275 g}$ of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ in $\text{775 mL}$ of solution has to be determined.

Concept Introduction:

Molarity is amount of solute per $1\text{ L}$ of solution. Formula for molarity is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$        (1)

Answer to Problem 18PE

Molarity of $\text{275 g}$ of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ in $\text{775 mL}$ of solution is $1.97M$.

Explanation of Solution

Formula for molarity of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ solution is as follows:

  ${\text{Molarity of C}}_6{\text{H}}_{12}{\text{O}}_6\text{ solution}=\frac{{\text{Moles of C}}_6{\text{H}}_{12}{\text{O}}_6}{\text{Volume }\left(\text{L}\right){\text{ of C}}_6{\text{H}}_{12}{\text{O}}_6\text{ solution}}$        (3)

Formula for moles of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ is as follows:

  ${\text{Moles of C}}_6{\text{H}}_{12}{\text{O}}_6=\frac{{\text{Mass of C}}_6{\text{H}}_{12}{\text{O}}_6}{{\text{Molar mass of C}}_6{\text{H}}_{12}{\text{O}}_6}$        (4)

Substitute $\text{275 g}$ for mass and $180.2\text{ g/mol}$ for molar mass of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ in equation (4).

  $\begin{array}{c}{\text{Moles of C}}_6{\text{H}}_{12}{\text{O}}_6=\frac{\text{275 g}}{180.2\text{ g/mol}}\\ =1.53\text{ mol}\end{array}$

Substitute $1.53\text{ mol}$ for moles and $\text{775 mL}$ for volume of ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$ solution in equation (3).

  $\begin{array}{c}{\text{Molarity of C}}_6{\text{H}}_{12}{\text{O}}_6\text{ solution}=\left(\frac{\text{1}\text{.53 mol}}{\text{775 mL}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\\ =1.97M\end{array}$

Hence, molarity of given solution is $1.97M$.

(d)

Interpretation Introduction

Interpretation:

Molarity of $\text{125 g}$ of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}$ in $\text{2}\text{.50 L}$ of solution has to be determined.

Concept Introduction:

Molarity is amount of solute per $1\text{ L}$ of solution. Formula for molarity is as follows:

  $\text{Molarity of solution}=\frac{\text{Moles of solute}}{\text{Volume }\left(\text{L}\right)\text{ of solution}}$        (1)

Answer to Problem 18PE

Molarity of $\text{125 g}$ of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}$ in $\text{2}\text{.50 L}$ of solution is $0.2028M$.

Explanation of Solution

Formula for molarity of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}$ solution is as follows:

  $\begin{array}{l}\text{Molarity of}\\ {\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O solution}=\frac{{\text{Moles of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}}{\text{Volume }\left(\text{L}\right){\text{ of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O solution}}\end{array}$        (5)

Formula for moles of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}$ is as follows:

  ${\text{Moles of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}=\frac{{\text{Mass of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}}{{\text{Molar mass of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}}$        (6)

Substitute $\text{125 g}$ for mass and $246.47\text{ g/mol}$ for molar mass of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}$ in equation (6).

  $\begin{array}{c}{\text{Moles of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}=\frac{\text{125 g}}{246.47\text{ g/mol}}\\ =0.507\text{ mol}\end{array}$

Substitute $0.507\text{ mol}$ for moles and $\text{2}\text{.50 L}$ for volume of ${\text{MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O}$ solution in equation (3).

  $\begin{array}{c}{\text{Molarity of MgSO}}_{\text{4}}\cdot 7{\text{H}}_2\text{O solution}=\frac{0.507\text{ mol}}{\text{2}\text{.50 L}}\\ =0.2028M\end{array}$

Hence, molarity of given solution is $0.2028M$.