Chapter 14, Problem 7PE

(a)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $15.0\text{ g KCl}+100.0{\text{ g H}}_2\text{O}$

Concept Introduction:

Mass percent is concentration term used to determine concentration of solution in terms of solute percent in particular mass of solution. The 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 7PE

Mass percent of $\text{KCl}$ is $13.04\text{ \%}$.

Explanation of Solution

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

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

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

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

Substitute $15.0\text{ g}$ for mass of $\text{KCl}$ and $100.0\text{ g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (2).

  $\begin{array}{c}\text{Mass of solution}=15.0\text{ g}+100.0\text{ g}\\ =115.0\text{ g}\end{array}$

Substitute $15.0\text{ g}$ for mass of $\text{KCl}$ and $115.0\text{ g}$ for mass of solution in equation (1).

  $\begin{array}{c}\text{Mass percent of KCl}=\left(\frac{15.0\text{ g}}{115.0\text{ g}}\right)\left(100\right)\\ =13.04\text{ \%}\end{array}$

Hence, mass percent of $\text{KCl}$ is $13.04\text{ \%}$.

(b)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $2.50{\text{ g Na}}_3{\text{PO}}_4+10.0{\text{ g H}}_2\text{O}$

Concept Introduction:

Refer to part (a).

Answer to Problem 7PE

Mass percent of ${\text{Na}}_3{\text{PO}}_4$ is $\text{20 \%}$.

Explanation of Solution

Expression for mass percent of ${\text{Na}}_3{\text{PO}}_4$ is as follows:

  ${\text{Mass percent of Na}}_3{\text{PO}}_4=\left(\frac{{\text{Mass of Na}}_3{\text{PO}}_4}{\text{Mass of solution}}\right)\left(100\right)$        (3)

Expression to calculate mass of ${\text{Na}}_3{\text{PO}}_4$ solution is as follows:

  $\text{Mass of solution}={\text{Mass of Na}}_3{\text{PO}}_4+{\text{Mass of H}}_2\text{O}$        (4)

Substitute $2.50\text{ g}$ for mass of ${\text{Na}}_3{\text{PO}}_4$ and $10.0\text{ g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (4).

  $\begin{array}{c}\text{Mass of solution}=2.50\text{ g}+10.0\text{ g}\\ =12.5\text{ g}\end{array}$

Substitute $2.50\text{ g}$ for mass of ${\text{Na}}_3{\text{PO}}_4$ and $12.5\text{ g}$ for mass of solution in equation (3).

  $\begin{array}{c}{\text{Mass percent of Na}}_3{\text{PO}}_4=\left(\frac{\text{2}\text{.50 g}}{12.5\text{ g}}\right)\left(100\right)\\ =20\text{ \%}\end{array}$

Hence, mass percent of ${\text{Na}}_3{\text{PO}}_4$ is $\text{20 \%}$.

(c)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $0.20{\text{ mol NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2\text{ in }125{\text{ g H}}_2\text{O}$

Concept Introduction:

Refer to part (a).

Answer to Problem 7PE

Mass percent of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ is $10.98\text{ \%}$.

Explanation of Solution

Expression to calculate mass of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{Mass of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2=\left[\begin{array}{c}\left({\text{Moles of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2\right)\\ \left({\text{Molar mass of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2\right)\end{array}\right]$        (5)

Substitute $0.20\text{ mol}$ for moles of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ and $77.083\text{ g/mol}$ for molar mass of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ in equation (5).

  $\begin{array}{c}{\text{Mass of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2=\left(0.20\text{ mol}\right)\left(77.083\text{ g/mol}\right)\\ =15.42\text{ g}\end{array}$

Expression for mass percent of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ is as follows:

  ${\text{Mass percent of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2=\left(\frac{{\text{Mass of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2}{\text{Mass of solution}}\right)\left(100\right)$        (6)

Expression to calculate mass of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ solution is as follows:

  $\text{Mass of solution}={\text{Mass of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2+{\text{Mass of H}}_2\text{O}$        (7)

Substitute $15.42\text{ g}$ for mass of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ and $125\text{ g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (7).

  $\begin{array}{c}\text{Mass of solution}=15.42\text{ g}+125\text{ g}\\ =140.42\text{ g}\end{array}$

Substitute $15.42\text{ g}$ for mass of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ and $140.42\text{ g}$ for mass of solution in equation (6).

  $\begin{array}{c}{\text{Mass percent of NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2=\left(\frac{\text{15}\text{.42 g}}{140.42\text{ g}}\right)\left(100\right)\\ =10.98\text{ \%}\end{array}$

Hence, mass percent of ${\text{NH}}_4{\text{C}}_2{\text{H}}_3{\text{O}}_2$ is $10.98\text{ \%}$.

(d)

Interpretation Introduction

Interpretation:

Mass percent of below solution has to be determined.

  $1.50\text{ mol NaOH in }33.0{\text{ mol H}}_2\text{O}$

Concept Introduction:

Refer to part (a).

Answer to Problem 7PE

Mass percent of $\text{NaOH}$ is $9.15\text{ \%}$.

Explanation of Solution

Formula to calculate mass of $\text{NaOH}$ is as follows:

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

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

  $\begin{array}{c}\text{Mass of NaOH}=\left(1.50\text{ mol}\right)\left(\text{39}\text{.9 g/mol}\right)\\ =59.85\text{ g}\end{array}$

Formula 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 $33.0\text{ mol}$ 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).

  $\begin{array}{c}{\text{Mass of H}}_2\text{O}=\left(33.0\text{ mol}\right)\left(\text{18}\text{.01 g/mol}\right)\\ =594.33\text{ g}\end{array}$

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

  $\text{Mass percent of NaOH}=\left(\frac{\text{Mass of NaOH}}{\text{Mass of solution}}\right)\left(100\right)$        (10)

Formula to calculate mass of $\text{NaOH}$ solution is as follows:

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

Substitute $59.85\text{ g}$ for mass of $\text{NaOH}$ and $\text{594}\text{.33 g}$ for mass of ${\text{H}}_{\text{2}}\text{O}$ in equation (11).

  $\begin{array}{c}\text{Mass of solution}=59.85\text{ g}+594.33\text{ g}\\ =654.18\text{ g}\end{array}$

Substitute $59.85\text{ g}$ for mass of $\text{NaOH}$ and $654.18\text{ g}$ for mass of solution in equation (10).

  $\begin{array}{c}\text{Mass percent of NaOH}=\left(\frac{\text{59}\text{.85 g}}{\text{654}\text{.18 g}}\right)\left(100\right)\\ =9.15\text{ \%}\end{array}$

Hence, mass percent of $\text{NaOH}$ is $9.15\text{ \%}$.