Chapter 12, Problem 12.89QE

(a)

Interpretation Introduction

Interpretation:

Alcohol concentration has to be expressed in molality if the dinner wines contain $9.00\%$ by volume and $7.23\%$ by mass and has a density of $\text{0}\text{.9877}\text{g/mL}$.

Answer to Problem 12.89QE

Molality of alcohol is $1.69m$.

Explanation of Solution

Mass of alcohol that is present in the wine is given as $\text{7}\text{.23}\text{g}$.  The molar mass of alcohol is $46.0\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of alcohol is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{alcohol}=7.23\text{g}\times \frac{1\text{mol}}{46.0\text{g}}\\ =0.1572\text{mol}\end{array}$

Mass of solvent that is present in the wine is calculated as shown below;

    $\begin{array}{c}\text{Mass of solvent}=100\text{g}-7.23\text{g}\\ \text{=92}\text{.8}\text{g}\\ \text{=0}\text{.093}\text{kg}\end{array}$

Molality of alcohol is calculated as shown below;

    $\begin{array}{c}\text{Molality}=\frac{0.157\text{mol}}{0.093\text{kg}}\\ =1.69m\end{array}$

Therefore, the molality of alcohol is $1.69m$.

(b)

Interpretation Introduction

Interpretation:

Alcohol concentration has to be expressed in mole fraction if the dinner wines contain $9.00\%$ by volume and $7.23\%$ by mass and has a density of $\text{0}\text{.9877}\text{g/mL}$.

Answer to Problem 12.89QE

Mole fraction of alcohol is $0.0296$.

Explanation of Solution

Mass of alcohol that is present in the wine is given as $\text{7}\text{.23}\text{g}$.  The molar mass of alcohol is $46.0\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of alcohol is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{alcohol}=7.23\text{g}\times \frac{1\text{mol}}{46.0\text{g}}\\ =0.1572\text{mol}\end{array}$

Mass of solvent that is present in the wine is calculated as shown below;

    $\begin{array}{c}\text{Mass of solvent}=100\text{g}-7.23\text{g}\\ \text{=92}\text{.8}\text{g}\\ \text{=0}\text{.093}\text{kg}\end{array}$

Moles of solvent is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{ofsolvent}=92.8\text{g}\times \frac{1\text{mol}}{18.0\text{g}}\\ =5.15\text{mol}\end{array}$

Mole fraction of alcohol is calculated as follows;

    $\begin{array}{c}\text{Mole}\text{fraction}\text{of}\text{alcohol}({\chi }_{\text{alcohol}})=\frac{\text{0}\text{.1572}\text{mol}}{\text{0}\text{.1572}\text{mol}\text{+}\text{5}\text{.15}\text{mol}}\\ =0.0296\end{array}$

Therefore, the mole fraction of alcohol is $0.0296$.

(c)

Interpretation Introduction

Interpretation:

Alcohol concentration has to be expressed in molarity if the dinner wines contain $9.00\%$ by volume and $7.23\%$ by mass and has a density of $\text{0}\text{.9877}\text{g/mL}$.

Answer to Problem 12.89QE

Molarity of alcohol is $1.55M$.

Explanation of Solution

Volume of solution is calculated as shown below;

    $\begin{array}{c}\text{Volume}=\frac{\text{Mass}}{\text{density}}\\ =\frac{\text{100}\text{g}}{\text{0}\text{.9877}\text{g/mL}}\\ =101.2\text{mL}\\ \text{=0}\text{.1012}\text{L}\end{array}$

Mass of alcohol that is present in the wine is given as $\text{7}\text{.23}\text{g}$.  The molar mass of alcohol is $46.0\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of alcohol is calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{alcohol}=7.23\text{g}\times \frac{1\text{mol}}{46.0\text{g}}\\ =0.1572\text{mol}\end{array}$

Molarity of alcohol is calculated as shown below;

    $\begin{array}{c}\text{Molarity}=\frac{0.157\text{mol}}{0.1012\text{L}}\\ =1.55M\end{array}$

Therefore, the molarity of alcohol is $1.55M$.

(d)

Interpretation Introduction

Interpretation:

Alcohol concentration has to be expressed in grams of alcohol per $\text{100}\text{mL}$ of solution if the dinner wines contain $9.00\%$ by volume and $7.23\%$ by mass and has a density of $\text{0}\text{.9877}\text{g/mL}$.

Answer to Problem 12.89QE

Mass of alcohol in $\text{100}\text{mL}$ is $\text{7}\text{.13}\text{g}$.

Explanation of Solution

Molarity of alcohol is calculated as shown below;

    $\begin{array}{c}\text{Molarity}=\frac{\text{Moles}\text{of}\text{alcohol}}{\text{Volume}\text{of}\text{solution}}\\ 1.55M=\frac{\text{Moles}\text{of}\text{alcohol}}{\text{0}\text{.10}\text{L}}\\ \text{Moles}\text{of}\text{alcohol}=0.155\text{mol}\end{array}$

Mass of alcohol is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}\text{alcohol}=\frac{\text{Weight}}{\text{Molar}\text{mass}}\\ 0.155\text{mol}=\frac{\text{Weight}}{\text{46}\text{.0}\text{g/mol}}\\ \text{Weight}=7.13\text{g}\end{array}$

Therefore, the mass of alcohol is $\text{7}\text{.13}\text{g}$.