Chapter 13, Problem 24QP

The density of an aqueous solution containing 15.0 percent of ethanol $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)$ by mass is 0.984 g/mL. (a) Calculate the molality of this solution. (b) Calculate its molarity. (c) What volume of the solution would contain 0.250 mole of ethanol?

Interpretation Introduction

Interpretation: The molarity, molality, and volume of the given solutions are to be calculated.

Concept Introduction:

Molarity $\left(M\right)$ is the ratio of moles of solute to the volume of solution (in liters).

Its expression is

$M=\frac{\text{moles of solute}}{\text{liters of solution}}$

Molality $\left(m\right)$ is the ratio of moles of solute to the mass of solvent $\left(\text{in kg}\right)$. Its formula is

$m=\frac{\text{moles of solute}}{\text{mass of solvent}\left(\text{in kg}\right)}$

The expression to calculate moles is

$\text{moles}=\frac{\text{mass}}{\text{molar mass}}$

Answer to Problem 24QP

Solution:

(a) $3.83\text{ m}$

(b) $3.19\text{ M}$

(c) $0.0784\text{ L or 78}\text{.4 mL}$

Explanation of Solution

Given information:

Density of the solution is $0.984\text{ g/mL}$.

Percent by mass of ethanol is $15.0\text{ percent}$.

Moles of ethanol is $0.250\text{ mol}$.

Explanation:

a) The molality of the solution

The mass of ethanol is $15.0\%$, that is 15 g.

Mass of solution = mass of solvent + mass of solute.

$\text{Mass of solvent }=\text{ mass of solution }-\text{ mass of solute}$.

Substitute $100.0\text{ g}$ for mass of solution and $15.0\text{ g}$ for mass of solute in the above expression as:

$\begin{array}{c}\text{Mass of solvent }=100.0\text{ g}-15.0\text{ g}\\ =85.0\text{ g}\text{.}\end{array}$

The expression to calculate moles is

$\text{moles}=\frac{\text{mass}}{\text{molar mass}}.$

Ethanol has a molar mass of $46.07\text{ g/mol}$.

Substitute $15.0\text{ g}$ for mass and $46.07\text{ g/mol}$ for molar mass in the above expression:

$\begin{array}{c}\text{moles}=\frac{15.0\text{ g}}{46.07\text{ g/mol}}\\ =0.3256\text{ mol}\text{.}\end{array}$

The expression to calculate molality is

$m=\frac{\text{moles of solute}}{\text{mass of solvent}\left(\text{in kg}\right)}.$ …… (1)

The mass of solvent is in grams and can be converted into kilograms by the following relation:

$1g={10}^{-3}kg.$

Thus, the conversion factor used to convert volume from milliliters into liters is $\left(\frac{{10}^{-3}kg}{1\text{ g}}\right)$.

Convert $85.0\text{ g}$ to $\text{kg}$ with the help of the conversion factor as

$\begin{array}{c}\mathrm{Mass}\text{ of solvent}=\left(\frac{{10}^{-3}\text{ kg}}{1\cancel{\text{g}}}\right)\times 85.0\cancel{\text{ g}}\\ =0.085\text{ kg}\text{.}\end{array}$

Substitute $0.3256\text{ mol}$ for moles of solute and $0.085\text{ kg}$ for mass of solvent in equation (1):

$\begin{array}{c}m=\frac{0.3256\text{ mol}}{0.085\text{ kg}}\\ =3.83\text{ m}\text{.}\end{array}$

Therefore, the molality of the solution is $3.83\text{ m}$.

Explanation:

b) Molarity of solutions

Density of the solution is $0.984\text{ g/mL,}$ which means that $0.984\text{ g}$ in present in $1\text{ mL}$ of solution.

Therefore, the conversion factor becomes $\left(\frac{1\text{ mL}}{0.984\text{ g}}\right)$.

Multiply the conversion factor with $100.0\text{ g}$ in order to calculate the volume of solution as

$\begin{array}{c}\text{Volume}=100.0\cancel{\text{g}}\times \left(\frac{1\text{ mL}}{0.984\cancel{\text{g}}}\right)\\ =102\text{ mL}\text{.}\end{array}$

The volume of solvent is in $\text{mL}$ and can be converted into $\text{L}$ by the following relation:

$1mL={10}^{-3}L$

Thus, the conversion factor used to convert volume from milliliters into liters is $\left(\frac{{10}^{-3}\text{ L}}{1\text{ mL}}\right)$.

Convert $\text{102 mL}$ to $\text{L}$ with the help of the conversion factor as:

$\begin{array}{c}\text{Volume}=\left(\frac{{10}^{-3}\text{ L}}{1\cancel{\text{mL}}}\right)\times \text{102 }\cancel{\text{mL}}\\ =0.102\text{ L}\text{.}\end{array}$

The expression to calculate molarity is

$M=\frac{\text{moles of solute}}{\text{liters of solution}}.$

Substitute $0.3256\text{ mol}$ for moles of solute and $0.102\text{ L}$ for liters of solution in the above expression as

$\begin{array}{c}M=\frac{0.3256\text{ mol}}{0.102\text{ L}}\\ =3.19\text{ M}\text{.}\end{array}$

Therefore, the molarity of the solution is $3.19\text{ M}$.

Explanation:

c) The volume of the solution that would contain 0.250 mol of ethanol.

Since, molarity is the number of moles of solute dissolved in $1\text{}L$ of solution. Thus, the calculated molarity, that is, $3.19\text{ M}$ suggests that $3.19\text{ mol}$ is present in $1L$ of solution. Thus, the given molarity can be expressed mathematically as $\frac{3.19\text{ mol}}{1\text{ L}}$.

Invert the conversion factor and multiply it with moles of ethanol in order to calculate the volume of solution as:

$\begin{array}{c}\text{Volume}=0.250\cancel{\text{mol}}\times \frac{1\text{ L}}{3.19\cancel{\text{mol}}}\\ =0.0784\text{ L or 78}\text{.4 mL}\text{.}\end{array}$

Therefore, the volume of the solution is $0.0784\text{ L or 78}\text{.4 mL}$.