Chapter 6, Problem 20P

Interpretation Introduction

(a)

Interpretation:

The formation of following solutions should be explained:

500.0 mL of a 5.32 % w/w ${\text{H}}_{\text{2}}\text{S}$ solution in water.

Concept Introduction:

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. In mass of solute per unit mass of solvent, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/w is as follows:

${\text{C}}_{w/w}=\frac{m_{\text{solute}}}{M_{\text{solvent}}}\times 100\%$.

Here, $m_{\text{solute}}$ is mass of solute and $m_{\text{solvent}}$ is mass of solvent.

When solving for the required amount of solute, the following formula is used:

${\text{C}}_{w/w}=\frac{m_{\text{solute}}}{m_{\text{solute}}+M_{\text{solvent}}}\times 100\%$.

Answer to Problem 20P

$28.0\text{9 g}$ of the solute must be added.

Explanation of Solution

Given Information:

$\begin{array}{l}V\text{}=500.0\text{ mL}\\ \text{C}\text{}=\text{}5.32\%w/w\end{array}$.

The concentration is shown as w/w meaning this is mass of solute per mass of solvent.

Since, density of water is 1 g/mL thus, 500 mL of water contains 500 g of water. Substitute known data and solve for mass of solute.

$\begin{array}{c}5.32\%=\frac{m_{\text{solute}}}{m_{\text{solute}}+500.\text{0 g}}\times 100\%\\ \frac{5.32\%}{100\%}=\frac{m_{\text{solute}}}{m_{\text{solute}}+500.\text{0 g}}\\ 0.0532\left(m_{\text{solute}}+500.0\mathrm{g}\right)=m_{\text{solute}}\\ 0.0532\left(m_{\text{solute}}\right)+26.6\text{ g}=m_{\text{solute}}\\ \left(1-0.0532\right)\left(m_{\text{solute}}\right)=26.6\text{ g}\\ m_{\text{solute}}=\frac{26.\text{6 g}}{0.9468}\\ =28.0\text{9 g}\end{array}$.

Thus, $28.0\text{9 g}$ of solute is added to the total $500.\text{0 mL}$ of solvent in order to get a $5.32\%$ of w/w solution.

(b)

Interpretation Introduction

Interpretation:

The formation of following solutions should be explained:

342.0 mL of a 0.443 % w/w benzene solution in toluene.

Concept Introduction:

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. In mass of solute per unit mass of solvent, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/w is as follows:

${\text{C}}_{w/w}=\frac{m_{\text{solute}}}{M_{\text{solvent}}}\times 100\%$.

Here, $m_{\text{solute}}$ is mass of solute and $m_{\text{solvent}}$ is mass of solvent.

When solving for the required amount of solute, the following formula is used:

${\text{C}}_{w/w}=\frac{m_{\text{solute}}}{m_{\text{solute}}+M_{\text{solvent}}}\times 100\%$.

Answer to Problem 20P

$1.51\text{5 mL}$ of the solute must be added.

Explanation of Solution

Given Information:

$\begin{array}{l}V\text{}=34\text{2 mL}\\ \text{C}\text{}=0.443\%w/w\\ D_{\text{toluene}}=0.867\text{ g/mL}\\ D_{\text{benzene}}=0.876 \text{g/mL}\end{array}$.

The concentration is shown as w/w meaning this is mass of solute per mass of solvent.

In this specific case, we do not know which volume or mass each material is, but we do know which one the solvent is, toluene.

$\begin{array}{l}{\text{Mass}}_{\text{total}}={\text{Mass}}_{\text{benzene}}+{\text{Mass}}_{\text{toluene}}\\ {\text{Mass}}_{\text{total}}={\text{D}}_{\text{benzene}}{\text{V}}_{\text{benzene}}+{\text{D}}_{\text{toluene}}{\text{V}}_{\text{toluene}}\\ {\text{C}}_{w/w}=\frac{m_{\text{solute}}}{m_{\text{solute}}+m_{\text{solvent}}}\times 100\%\\ 0.443\%=\frac{D_{\text{benzene}}{V}_{\text{benzene}}}{D_{\text{benzene}}{V}_{\mathrm{benzene}}+D_{\text{toluene}}{V}_{\text{toluene}}}\times 100\%\\ 0.443\%=\frac{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})}{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})+0.867\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.\text{X}(V_{\text{toluene}})}\times 100\%\end{array}$.

Now, get the total volume equation.

$\begin{array}{l}{V}_{\text{total}}=V_{\text{benzene}}+V_{\text{toluene}}\\ 342.00\text{ mL}=V_{\text{benzene}}+V_{\text{toluene}}\\ V_{\text{toluene}}=342.00\text{ mL}-V_{\text{benzene}}\end{array}$.

Solve simultaneously.

$\begin{array}{l}0.443\%=\frac{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})}{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})+0.867\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.\text{X}(V_{\text{toluene}})}\times 100\%\\ V_{\text{toluene}}=342.00\text{ mL}-V_{\text{benzene}}\\ 0.443\%=\frac{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})}{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})+0.867\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(342.00\text{ mL}-V_{\text{benzene}})}\times 100\%\\ \frac{0.443\%}{100\%}=\frac{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{benzene}})}{0.876\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m\text{l}$}\right.(V_{\text{benzene}})+0.867\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(342.00\text{ mL}-V_{\text{benzene}})}\\ \frac{0.876\times V_{\text{benzene}}}{0.00443}=0.876\times V_{\text{benzene}}+296.5-0.867\times V_{\text{benzene}}\\ 195.7V_{\text{benzene}}=0.876\times V_{\text{benzene}}+296.5-0.867\times V_{\text{benzene}}\\ V_{\text{benzene}}(195.7-0.876+0.867)=296.5\\ V_{\text{benzene}}=\frac{296.5}{195.691}\\ V_{\text{benzene}}=1.515\text{ mL}\end{array}$.

Then, add $1.51\text{5 mL}$ to the mixture. Add all remaining volume of toluene:

$\begin{array}{l}{V}_{\text{toluene}}=342.00m\text{l}-V_{\text{benzene}}\\ V_{\text{toluene}}=342-1.515=340.48\text{5 mL}\end{array}$.

Interpretation Introduction

(c)

Interpretation:

The formation of following solutions should be explained:

125.5 mL of a 34.2 % w/w dimethyl sulfoxide solution in acetone.

Concept Introduction:

Concentration has several ways to be calculated. It typically relates an amount of solute and the solution. In mass of solute per unit mass of solvent, or w/w, the total mass of the solute, solvent and solution must be known.

The formula for w/w is as follows:

${\text{C}}_{w/w}=\frac{m_{\text{solute}}}{M_{\text{solvent}}}\times 100\%$.

Here, $m_{\text{solute}}$ is mass of solute and $m_{\text{solvent}}$ is mass of solvent.

When solving for the required amount of solute, the following formula is used:

${\text{C}}_{w/w}=\frac{m_{\text{solute}}}{m_{\text{solute}}+M_{\text{solvent}}}\times 100\%$.

Answer to Problem 20P

$V_{DMS}=3.37\text{9 mL}$ of the solute must be added.

Explanation of Solution

Given:

$\begin{array}{l}V\text{}=12.5m\text{l}\\ \text{C}\text{}=34.2\%w/w\\ D_{\text{dimethylsulfoxide}}=1.1004 g/m\text{l}\\ D_{\text{a}\text{C}\text{e}t\text{O}n\text{e}}=0.784 g/m\text{l}\end{array}$.

In this specific case, we do not know which volume or mass each material is, but we do know which one is the solvent that is acetone.

$\begin{array}{l}{\text{Mass}}_{\text{total}}{\text{=Mass}}_{\text{benzene}}{\text{+Mass}}_{\text{toluene}}\\ {\text{Mass}}_{\text{total}}\text{=}{D}_{\text{benzene}}{V}_{\text{benzene}}\text{+}{D}_{\text{toluene}}{V}_{\text{toluene}}\\ 34.2\%=\frac{m_{\text{solute}}}{m_{\text{solute}}+m_{\text{solvent}}}\times 100\%\\ 34.2\%=\frac{D_{\text{MDS}}{V}_{\text{MDS}}}{D_{\text{MDS}}{V}_{\text{MDS}}+D_{\text{Acetone}}{V}_{\text{Acetone}}}\times 100\%\\ 34.2\%=\frac{1.1004\times V_{\text{MDS}}}{1.1004\times V_{MDS}+0.784\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.\times (V_{\text{Acetone}})}\times 100\%\end{array}$.

Now, get the total volume equation.

$\begin{array}{l}{V}_{\text{total}}=V_{\text{DMS}}+V_{\text{Acetone}}\\ 12.\text{5 mL}=V_{\text{DMS}}+V_{\text{Acetone}}\\ V_{\text{acetone}}=12.5m\text{l}-V_{\text{DMS}}\end{array}$.

Solve simultaneously.

$\begin{array}{l}34.2\%=\frac{0.876\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mL}$}\right.(V_{\text{DMS}})}{0.876\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m\text{l}$}\right.(V_{DMS})+0.867\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m\text{l}$}\right.\text{X}(342.00m\text{l}-V_{DMS})}\times 100\%\\ \frac{34.2}{100\%}=\frac{1.1\times V_{DMS}}{1.1\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m\text{l}$}\right.(V_{DMS})+0.784\raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$m\text{l}$}\right.\times (12.5m\text{l}-V_{DMS})}\\ \frac{1.1\times V_{DMS}}{0.342}=1.1V_{DMS}+9.8-0.784V_{DMS}\\ (3.216-1.1+0.784)V_{DMS}=9.\text{8 mL}\\ V_{DMS}=\frac{9.8}{2.9}\\ V_{DMS}=3.37\text{9 mL}\end{array}$.

Then, add $V_{DMS}=3.379m\text{l}$ to the mixture. Add all remaining volume of acetone:

$\begin{array}{l}V=12.\text{5 mL}-3.37\text{9 mL}\\ V_{\text{a}\text{C}\text{e}t\text{O}n\text{e}}=9.12\text{1 mL}\end{array}$.