Chapter 14, Problem 22PE

(a)

Interpretation Introduction

Interpretation:

Grams of solute in $\text{1}\text{.20 L}$ of $\text{18 }M{\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}$ have to be determined.

Concept Introduction:

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

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

Answer to Problem 22PE

$2118.5\text{ g}$ is present in $\text{1}\text{.20 L}$ of $\text{18 }M{\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

Formula for molarity of ${\text{H}}_2{\text{SO}}_4$ solution is as follows:

  ${\text{Molarity of H}}_2{\text{SO}}_4\text{ solution}=\frac{{\text{Moles of H}}_2{\text{SO}}_4}{\text{Volume }\left(\text{L}\right){\text{ of H}}_2{\text{SO}}_4\text{ solution}}$        (1)

Rearrange equation (1) for moles of ${\text{H}}_2{\text{SO}}_4$.

  ${\text{Moles of H}}_2{\text{SO}}_4=\left[\begin{array}{c}\left({\text{Molarity of H}}_2{\text{SO}}_4\text{ solution}\right)\\ \left(\text{Volume }\left(\text{L}\right){\text{ of H}}_2{\text{SO}}_4\text{ solution}\right)\end{array}\right]$        (2)

Substitute $\text{18 }M$ for molarity and $\text{1}\text{.20 L}$ for volume of ${\text{H}}_2{\text{SO}}_4$ solution in equation (2).

  $\begin{array}{c}{\text{Moles of H}}_2{\text{SO}}_4=\left(\text{18 }M\right)\left(\text{1}\text{.20 L}\right)\\ =21.6\text{ mol}\end{array}$

Formula for mass of ${\text{H}}_2{\text{SO}}_4$ is as follows:

  ${\text{Mass of H}}_2{\text{SO}}_4=\left({\text{Moles of H}}_2{\text{SO}}_4\right)\left({\text{Molar mass of H}}_2{\text{SO}}_4\right)$        (3)

Substitute $\text{21}\text{.6 mol}$ for moles and $\text{98}\text{.079 g/mol}$ for molar mass of ${\text{H}}_2{\text{SO}}_4$ in equation (3).

  $\begin{array}{c}{\text{Mass of H}}_2{\text{SO}}_4=\left(\text{21}\text{.6 mol}\right)\left(\text{98}\text{.079 g/mol}\right)\\ =2118.5\text{ g}\end{array}$

Hence, $2118.5\text{ g}$ is present in $\text{1}\text{.20 L}$ of $\text{18 }M{\text{ H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

(b)

Interpretation Introduction

Interpretation:

Grams of solute in $\text{27}\text{.5 mL}$ of $\text{1}\text{.50 }M{\text{ KMnO}}_4$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 22PE

$6.52\text{ g}$ is present in $\text{27}\text{.5 mL}$ of $\text{1}\text{.50 }M{\text{ KMnO}}_4$.

Explanation of Solution

Formula for molarity of ${\text{KMnO}}_4$ solution is as follows:

  ${\text{Molarity of KMnO}}_4\text{ solution}=\frac{{\text{Moles of KMnO}}_4}{\text{Volume }\left(\text{L}\right){\text{ of KMnO}}_4\text{ solution}}$        (4)

Rearrange equation (4) for moles of ${\text{KMnO}}_4$.

  ${\text{Moles of KMnO}}_4=\left[\begin{array}{c}\left({\text{Molarity of KMnO}}_4\text{ solution}\right)\\ \left(\text{Volume }\left(\text{L}\right){\text{ of KMnO}}_4\text{ solution}\right)\end{array}\right]$        (5)

Substitute $\text{1}\text{.50 }M$ for molarity and $\text{27}\text{.5 mL}$ for volume of ${\text{KMnO}}_4$ solution in equation (5).

  $\begin{array}{c}{\text{Moles of KMnO}}_4=\left(\text{1}\text{.50 }M\right)\left(\text{27}\text{.5 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.04125\text{ mol}\end{array}$

Formula for mass of ${\text{KMnO}}_4$ is as follows:

  ${\text{Mass of KMnO}}_4=\left[\begin{array}{c}\left({\text{Moles of KMnO}}_4\right)\\ \left({\text{Molar mass of KMnO}}_4\right)\end{array}\right]$        (6)

Substitute $0.04125\text{ mol}$ for moles and $\text{158}\text{.034 g/mol}$ for molar mass of ${\text{KMnO}}_4$ in equation (6).

  $\begin{array}{c}{\text{Mass of KMnO}}_4=\left(0.04125\text{ mol}\right)\left(\text{158}\text{.034 g/mol}\right)\\ =6.52\text{ g}\end{array}$

Hence, $6.52\text{ g}$ is present in $\text{27}\text{.5 mL}$ of $\text{1}\text{.50 }M{\text{ KMnO}}_4$.

(c)

Interpretation Introduction

Interpretation:

Grams of solute in $\text{120 mL}$ of $\text{0}\text{.025 }M{\text{ Fe}}_2{\left({\text{SO}}_4\right)}_3$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 22PE

$1.20\text{ g}$ is present in $\text{120 mL}$ of $\text{0}\text{.025 }M{\text{ Fe}}_2{\left({\text{SO}}_4\right)}_3$.

Explanation of Solution

Formula for molarity of ${\text{Fe}}_2{\left({\text{SO}}_4\right)}_3$ solution is as follows:

  ${\text{Molarity of Fe}}_2{\left({\text{SO}}_4\right)}_3\text{ solution}=\frac{{\text{Moles of Fe}}_2{\left({\text{SO}}_4\right)}_3}{\text{Volume }\left(\text{L}\right){\text{ of Fe}}_2{\left({\text{SO}}_4\right)}_3\text{ solution}}$        (7)

Rearrange equation (7) for moles of ${\text{Fe}}_2{\left({\text{SO}}_4\right)}_3$.

  ${\text{Moles of Fe}}_2{\left({\text{SO}}_4\right)}_3=\left[\begin{array}{c}\left({\text{Molarity of Fe}}_2{\left({\text{SO}}_4\right)}_3\text{ solution}\right)\\ \left(\text{Volume }\left(\text{L}\right){\text{ of Fe}}_2{\left({\text{SO}}_4\right)}_3\text{ solution}\right)\end{array}\right]$        (8)

Substitute $\text{0}\text{.025 }M$ for molarity and $\text{120 mL}$ for volume of ${\text{Fe}}_2{\left({\text{SO}}_4\right)}_3$ solution in equation (8).

  $\begin{array}{c}{\text{Moles of Fe}}_2{\left({\text{SO}}_4\right)}_3=\left(\text{0}\text{.025 }M\right)\left(\text{120 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.003\text{ mol}\end{array}$

Formula for mass of ${\text{Fe}}_2{\left({\text{SO}}_4\right)}_3$ is as follows:

  ${\text{Mass of Fe}}_2{\left({\text{SO}}_4\right)}_3=\left[\begin{array}{c}\left({\text{Moles of Fe}}_2{\left({\text{SO}}_4\right)}_3\right)\\ \left({\text{Molar mass of Fe}}_2{\left({\text{SO}}_4\right)}_3\right)\end{array}\right]$        (9)

Substitute $0.003\text{ mol}$ for moles and $\text{399}\text{.88 g/mol}$ for molar mass of ${\text{Fe}}_2{\left({\text{SO}}_4\right)}_3$ in equation (9).

  $\begin{array}{c}{\text{Mass of Fe}}_2{\left({\text{SO}}_4\right)}_3=\left(\text{0}\text{.003 mol}\right)\left(\text{399}\text{.88 g/mol}\right)\\ =1.20\text{ g}\end{array}$

Hence, $1.20\text{ g}$ is present in $\text{120 mL}$ of $\text{0}\text{.025 }M{\text{ Fe}}_2{\left({\text{SO}}_4\right)}_3$.