Chapter 14, Problem 8PE

(a)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $25.0{\text{ g NaNO}}_3+125.0{\text{ g H}}_2\text{O}$

Concept Introduction:

Mass percent is concentration term used to determine concentration of solution in terms of solute percent in particular mass of solution. Expression for mass percent is as follows:

  $\text{Mass percent}=\left(\frac{\text{Mass of solute}}{\text{Mass of solution}}\right)\left(100\right)$

Answer to Problem 8PE

Mass percent of ${\text{NaNO}}_3$ is $16.7\text{ \%}$.

Explanation of Solution

Expression for mass percent of ${\text{NaNO}}_3$ is as follows:

  ${\text{Mass percent of NaNO}}_3=\left(\frac{{\text{Mass of NaNO}}_3}{\text{Mass of solution}}\right)\left(100\right)$        (1)

Expression to calculate mass of ${\text{NaNO}}_3$ solution is as follows:

  $\text{Mass of solution}={\text{Mass of NaNO}}_3+{\text{Mass of H}}_2\text{O}$        (2)

Substitute $25.0\text{ g}$ for mass of ${\text{NaNO}}_3$ and $125.0\text{ g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (2).

  $\begin{array}{c}\text{Mass of solution}=25.0\text{ g}+125.0\text{ g}\\ =150\text{ g}\end{array}$

Substitute $25.0\text{ g}$ for mass of ${\text{NaNO}}_3$ and $150\text{ g}$ for mass of solution in equation (1).

  $\begin{array}{c}{\text{Mass percent of NaNO}}_3=\left(\frac{25.0\text{ g}}{150\text{ g}}\right)\left(100\right)\\ =16.7\text{ \%}\end{array}$

Hence, mass percent of ${\text{NaNO}}_3$ is $16.7\text{ \%}$.

(b)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $\text{1}{\text{.25 g CaCl}}_2+35.0{\text{ g H}}_2\text{O}$

Concept Introduction:

Refer to part (a).

Answer to Problem 8PE

Mass percent of ${\text{CaCl}}_2$ is $3.45\text{ \%}$.

Explanation of Solution

Expression for mass percent of ${\text{CaCl}}_2$ is as follows:

  ${\text{Mass percent of CaCl}}_2=\left(\frac{{\text{Mass of CaCl}}_2}{\text{Mass of solution}}\right)\left(100\right)$        (3)

Expression to calculate mass of ${\text{CaCl}}_2$ solution is as follows:

  $\text{Mass of solution}={\text{Mass of CaCl}}_2+{\text{Mass of H}}_2\text{O}$        (4)

Substitute $\text{1}\text{.25 g}$ for mass of ${\text{CaCl}}_2$ and $35.0\text{ g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (4).

  $\begin{array}{c}\text{Mass of solution}=1.25\text{ g}+35.0\text{ g}\\ =36.25\text{ g}\end{array}$

Substitute $\text{1}\text{.25 g}$ for mass of ${\text{CaCl}}_2$ and $36.25\text{ g}$ for mass of solution in equation (3).

  $\begin{array}{c}{\text{Mass percent of CaCl}}_2=\left(\frac{\text{1}\text{.25 g}}{36.25\text{ g}}\right)\left(100\right)\\ =3.45\text{ \%}\end{array}$

Hence, mass percent of ${\text{CaCl}}_2$ is $3.45\text{ \%}$.

(c)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $0.75{\text{ mol K}}_{\text{2}}{\text{CrO}}_4\text{ in }225{\text{ g H}}_2\text{O}$

Concept Introduction:

Refer to part (a).

Answer to Problem 8PE

Mass percent of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ is $39.29\text{ \%}$.

Explanation of Solution

Expression to calculate mass of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ is as follows:

  ${\text{Mass of K}}_{\text{2}}{\text{CrO}}_4=\left[\begin{array}{c}\left({\text{Moles of K}}_{\text{2}}{\text{CrO}}_4\right)\\ \left({\text{Molar mass of K}}_{\text{2}}{\text{CrO}}_4\right)\end{array}\right]$        (5)

Substitute $0.75\text{ mol}$ for moles of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ and $\text{194}\text{.19 g/mol}$ for molar mass of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ in equation (5).

  $\begin{array}{c}{\text{Mass of K}}_{\text{2}}{\text{CrO}}_4=\left(0.75\text{ mol}\right)\left(\text{194}\text{.19 g/mol}\right)\\ =145.64\text{ g}\end{array}$

Expression for mass percent of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ is as follows:

  ${\text{Mass percent of K}}_{\text{2}}{\text{CrO}}_4=\left(\frac{{\text{Mass of K}}_{\text{2}}{\text{CrO}}_4}{\text{Mass of solution}}\right)\left(100\right)$        (6)

Formula to calculate mass of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ solution is as follows:

  $\text{Mass of solution}={\text{Mass of K}}_{\text{2}}{\text{CrO}}_4+{\text{Mass of H}}_2\text{O}$        (7)

Substitute $145.64\text{ g}$ for mass of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ and $225\text{ g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (7).

  $\begin{array}{c}\text{Mass of solution}=145.64\text{ g}+225\text{ g}\\ =370.64\text{ g}\end{array}$

Substitute $145.64\text{ g}$ for mass of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ and $370.64\text{ g}$ for mass of solution in equation (6).

  $\begin{array}{c}{\text{Mass percent of K}}_{\text{2}}{\text{CrO}}_4=\left(\frac{145.64\text{ g}}{370.64\text{ g}}\right)\left(100\right)\\ =39.29\text{ \%}\end{array}$

Hence, mass percent of ${\text{K}}_{\text{2}}{\text{CrO}}_4$ is $39.29\text{ \%}$.

(d)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $1.20{\text{ mol H}}_2{\text{SO}}_4\text{ in }72.5{\text{ mol H}}_2\text{O}$

Concept Introduction:

Refer to part (a).

Answer to Problem 8PE

Mass percent of ${\text{H}}_2{\text{SO}}_4$ is $\text{8}\text{.27 \%}$.

Explanation of Solution

Expression to calculate mass of ${\text{H}}_2{\text{SO}}_4$ is as follows:

  ${\text{Mass of H}}_2{\text{SO}}_4=\left[\begin{array}{c}\left({\text{Moles of H}}_2{\text{SO}}_4\right)\\ \left({\text{Molar mass of H}}_2{\text{SO}}_4\right)\end{array}\right]$        (8)

Substitute $1.50\text{ mol}$ for moles of ${\text{H}}_2{\text{SO}}_4$ and $\text{39}\text{.9 g/mol}$ for molar mass of ${\text{H}}_2{\text{SO}}_4$ in equation (8).

  $\begin{array}{c}{\text{Mass of H}}_2{\text{SO}}_4=\left(1.20\text{ mol}\right)\left(\text{98}\text{.08 g/mol}\right)\\ =117.696\text{ g}\end{array}$

Expression to calculate mass of ${\text{H}}_2\text{O}$ is as follows:

  ${\text{Mass of H}}_2\text{O}=\left[\begin{array}{c}\left({\text{Moles of H}}_2\text{O}\right)\\ \left({\text{Molar mass of H}}_2\text{O}\right)\end{array}\right]$        (9)

Substitute $\text{72}\text{.5 mol}$ for moles of ${\text{H}}_2\text{O}$ and $\text{18}\text{.01 g/mol}$ for molar mass of ${\text{H}}_2\text{O}$ in equation (9).

  $\begin{array}{c}{\text{Mass of H}}_2\text{O}=\left(\text{72}\text{.5 mol}\right)\left(\text{18}\text{.01 g/mol}\right)\\ =1305.7\text{ g}\end{array}$

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

  ${\text{Mass percent of H}}_2{\text{SO}}_4=\left(\frac{{\text{Mass of H}}_2{\text{SO}}_4}{\text{Mass of solution}}\right)\left(100\right)$        (10)

Expression to calculate mass of ${\text{H}}_2{\text{SO}}_4$ solution is as follows:

  $\text{Mass of solution}={\text{Mass of H}}_2{\text{SO}}_4+{\text{Mass of H}}_2\text{O}$        (11)

Substitute $117.696\text{ g}$ for mass of ${\text{H}}_2{\text{SO}}_4$ and $\text{1305}\text{.7 g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (11).

  $\begin{array}{c}\text{Mass of solution}=117.696\text{ g}+1305.7\text{ g}\\ =1423.396\text{ g}\end{array}$

Substitute $117.696\text{ g}$ for mass of ${\text{H}}_2{\text{SO}}_4$ and $1423.396\text{ g}$ for mass of solution in equation (10).

  $\begin{array}{c}{\text{Mass percent of H}}_2{\text{SO}}_4=\left(\frac{117.696\text{ g}}{1423.396\text{ g}}\right)\left(100\right)\\ =8.27\text{ \%}\end{array}$

Hence, mass percent of ${\text{H}}_2{\text{SO}}_4$ is $8.27\text{ \%}$.