Chapter 14, Problem 17PE

(a)

Interpretation Introduction

Interpretation:

Molarity of $0.25\text{ mol}$ of solute in $75.0\text{ mL}$ 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 17PE

Molarity of $0.25\text{ mol}$ of solute in $75.0\text{ mL}$ of solution is $\text{3}\text{.33 }M$.

Explanation of Solution

Substitute $0.25\text{ mol}$ for moles of solute and $75.0\text{ mL}$ for volume of solution in equation (1).

  $\begin{array}{c}\text{Molarity of solution}=\left(\frac{0.25\text{ mol}}{\text{75}\text{.0 mL}}\right)\left(\frac{1\text{ mL}}{{10}^{-3}\text{ L}}\right)\\ =3.33M\end{array}$

Hence, molarity of given solution is $\text{3}\text{.33 }M$.

(b)

Interpretation Introduction

Interpretation:

Molarity of $1.75\text{ mol}$ of $\text{KBr}$ in $\text{0}\text{.75 L}$ 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 17PE

Molarity of $1.75\text{ mol}$ of $\text{KBr}$ in $\text{0}\text{.75 L}$ of solution is $2.33M$.

Explanation of Solution

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

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

Substitute $1.75\text{ mol}$ for moles of $\text{KBr}$ and $\text{0}\text{.75 L}$ for volume of solution in equation (2).

  $\begin{array}{c}\text{Molarity of KBr solution}=\frac{1.75\text{ mol}}{\text{0}\text{.75 L}}\\ =2.33M\end{array}$

Hence, molarity of given solution is $2.33M$.

(c)

Interpretation Introduction

Interpretation:

Molarity of $\text{35}\text{.0 g}$ of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ in $\text{1}\text{.25 L}$ 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 17PE

Molarity of $\text{35}\text{.0 g}$ of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ in $\text{1}\text{.25 L}$ of solution is $0.3416M$.

Explanation of Solution

Formula for molarity of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ solution is as follows:

  ${\text{Molarity of NaC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}=\frac{{\text{Moles of NaC}}_2{\text{H}}_3{\text{O}}_2}{\text{Volume }\left(\text{L}\right){\text{ of NaC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}}$        (3)

Formula for moles of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{Moles of NaC}}_2{\text{H}}_3{\text{O}}_2=\frac{{\text{Mass of NaC}}_2{\text{H}}_3{\text{O}}_2}{{\text{Molar mass of NaC}}_2{\text{H}}_3{\text{O}}_2}$        (4)

Substitute $\text{35}\text{.0 g}$ for mass of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ and $82.03\text{ g/mol}$ for molar mass of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ in equation (4).

  $\begin{array}{c}{\text{Moles of NaC}}_2{\text{H}}_3{\text{O}}_2=\frac{\text{35}\text{.0 g}}{82.03\text{ g/mol}}\\ =0.427\text{ mol}\end{array}$

Substitute $0.427\text{ mol}$ for moles and $\text{1}\text{.25 L}$ for volume of ${\text{NaC}}_2{\text{H}}_3{\text{O}}_2$ solution in equation (3).

  $\begin{array}{c}{\text{Molarity of NaC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}=\frac{0.427\text{ mol}}{\text{1}\text{.25 L}}\\ =0.3416M\end{array}$

Hence, molarity of given solution is $0.3416M$.

(d)

Interpretation Introduction

Interpretation:

Molarity of $\text{75 g}$ of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ in $\text{1}\text{.0 L}$ 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 17PE

Molarity of $\text{75 g}$ of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ in $\text{1}\text{.0 L}$ of solution is $0.3004M$.

Explanation of Solution

Formula for molarity of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ solution is as follows:

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

Formula for moles of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ is as follows:

  ${\text{Moles of CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}=\frac{{\text{Mass of CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}}{{\text{Molar mass of CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}}$        (6)

Substitute $\text{75 g}$ for mass of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ and $249.659\text{ g/mol}$ for molar mass of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ in equation (6).

  $\begin{array}{c}{\text{Moles of CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}=\frac{\text{75 g}}{249.659\text{ g/mol}}\\ =0.3004\text{ mol}\end{array}$

Substitute $0.3004\text{ mol}$ for moles and $\text{1}\text{.0 L}$ for volume of ${\text{CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O}$ solution in equation (3).

  $\begin{array}{c}{\text{Molarity of CuSO}}_{\text{4}}\cdot 5{\text{H}}_2\text{O solution}=\frac{0.3004\text{ mol}}{\text{1}\text{.0 L}}\\ =0.3004M\end{array}$

Hence, molarity of given solution is $0.3004M$.