Chapter 14, Problem 21PE

(a)

Interpretation Introduction

Interpretation:

Grams of solute in $2.5\text{ L}$ of $\text{0}\text{.75 }M{\text{ K}}_2{\text{CrO}}_4$ 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 21PE

$364.11\text{ g}$ is present in $2.5\text{ L}$ of $\text{0}\text{.75 }M{\text{ K}}_2{\text{CrO}}_4$.

Explanation of Solution

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

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

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

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

Substitute $\text{0}\text{.75 }M$ for molarity and $2.5\text{ L}$ for volume of ${\text{K}}_2{\text{CrO}}_4$ solution in equation (2).

  $\begin{array}{c}{\text{Moles of K}}_2{\text{CrO}}_4=\left(\text{0}\text{.75 }M\right)\left(\text{2}\text{.5 L}\right)\\ =1.875\text{ mol}\end{array}$

Formula for mass of ${\text{K}}_2{\text{CrO}}_4$ is as follows:

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

Substitute $1.875\text{ mol}$ for moles of ${\text{K}}_2{\text{CrO}}_4$ and $\text{194}\text{.19 g/mol}$ for molar mass of ${\text{K}}_2{\text{CrO}}_4$ in equation (3).

  $\begin{array}{c}{\text{Mass of K}}_2{\text{CrO}}_4=\left(1.875\text{ mol}\right)\left(\text{194}\text{.19 g/mol}\right)\\ =364.11\text{ g}\end{array}$

Hence, $364.11\text{ g}$ is present in $2.5\text{ L}$ of $\text{0}\text{.75 }M{\text{ K}}_2{\text{CrO}}_4$.

(b)

Interpretation Introduction

Interpretation:

Grams of solute in $\text{75}\text{.2 mL}$ of $\text{0}\text{.050 }M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 21PE

$0.226\text{ g}$ is present in $\text{75}\text{.2 mL}$ of $\text{0}\text{.050 }M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$.

Explanation of Solution

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

  ${\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}=\frac{{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2}{\text{Volume }\left(\text{L}\right){\text{ of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}}$        (4)

Rearrange equation (4) for moles of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$.

  ${\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left[\begin{array}{c}\left({\text{Molarity of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}\right)\\ \left(\text{Volume }\left(\text{L}\right){\text{ of HC}}_2{\text{H}}_3{\text{O}}_2\text{ solution}\right)\end{array}\right]$        (5)

Substitute $\text{0}\text{.050 }M$ for molarity and $\text{75}\text{.2 mL}$ for volume of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ solution in equation (5).

  $\begin{array}{c}{\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2=\left(\text{0}\text{.050 }M\right)\left(\text{75}\text{.2 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =0.00376\text{ mol}\end{array}$

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

  ${\text{Mass of HC}}_2{\text{H}}_3{\text{O}}_2=\left[\begin{array}{c}\left({\text{Moles of HC}}_2{\text{H}}_3{\text{O}}_2\right)\\ \left({\text{Molar mass of HC}}_2{\text{H}}_3{\text{O}}_2\right)\end{array}\right]$        (6)

Substitute $0.00376\text{ mol}$ for moles and $\text{60}\text{.052 g/mol}$ for molar mass of ${\text{HC}}_2{\text{H}}_3{\text{O}}_2$ in equation (6).

  $\begin{array}{c}{\text{Mass of HC}}_2{\text{H}}_3{\text{O}}_2=\left(0.00376\text{ mol}\right)\left(\text{60}\text{.052 g/mol}\right)\\ =0.226\text{ g}\end{array}$

Hence, $0.226\text{ g}$ is present in $\text{75}\text{.2 mL}$ of $\text{0}\text{.050 }M{\text{ HC}}_2{\text{H}}_3{\text{O}}_2$.

(c)

Interpretation Introduction

Interpretation:

Grams of solute in $\text{250 mL}$ of $\text{16 }M{\text{ HNO}}_3$ have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 21PE

$252.04\text{ g}$ is present in $\text{250 mL}$ of $\text{16 }M{\text{ HNO}}_3$.

Explanation of Solution

Formula for molarity of ${\text{HNO}}_3$ solution is as follows:

  ${\text{Molarity of HNO}}_3\text{ solution}=\frac{{\text{Moles of HNO}}_3}{\text{Volume }\left(\text{L}\right){\text{ of HNO}}_3\text{ solution}}$        (7)

Rearrange equation (7) for moles of ${\text{HNO}}_3$.

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

Substitute $\text{16 }M$ for molarity and $\text{250 mL}$ for volume of ${\text{HNO}}_3$ solution in equation (8).

  $\begin{array}{c}{\text{Moles of HNO}}_3=\left(\text{16 }M\right)\left(\text{250 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)\\ =4\text{ mol}\end{array}$

Formula for mass of ${\text{HNO}}_3$ is as follows:

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

Substitute $\text{4 mol}$ for moles and $\text{63}\text{.01 g/mol}$ for molar mass of ${\text{HNO}}_3$ in equation (9).

  $\begin{array}{c}{\text{Mass of HNO}}_3=\left(\text{4 mol}\right)\left(\text{63}\text{.01 g/mol}\right)\\ =252.04\text{ g}\end{array}$

Hence, $252.04\text{ g}$ is present in $\text{250 mL}$ of $\text{16 }M{\text{ HNO}}_3$.