Chapter 11, Problem 112QP

Interpretation Introduction

Interpretation:

The molality of acetic acid solution is to be determined.

Concept Introduction:

Molarity $\left(M\right)$ is defined as the number of moles $\left(n\right)$ of solute present in the volume of solution $\left(V\right)$ in litres. Molarity describes the concentration in terms of the volume of the solution.

$M=\frac{n}{V}$ …… (1)

Molality $\left(m\right)$ is defined as the number of moles $\left(n\right)$ of solute present in the mass(kilogram) of solvent $\left(w\right)$ in kilograms unit. Molality describes the concentration in terms of the amount of solvent but not the solution.

$m=\frac{n}{w}$ …… (2)

Answer to Problem 112QP

Solution:

Molality of given acetic acid solution is 1.058 m.

Explanation of Solution

Given Information: The molarity of Acetic acid solution is $1\text{ M}$ and density of the solution is $1.005{\text{ g mL}}^{-1}$ .

Substitute $M$ as $1\text{ M}$ and $V$ as $1\text{ L}$ in equation (1).

$\begin{array}{c}1{\text{ mol L}}^{-1}=\frac{n}{\text{1 L }}\\ n=\text{1 mol}\end{array}$

The number of moles $\left(n\right)$ of solute is equal to the ratio of mass $\left(m\right)$ of solute in the solution to the molar mass $\left(MM\right)$ of solute.

$n=\frac{m}{MM}$ …… (3)

The molar mass of acetic acid is as follows:

$\begin{array}{c}\text{CH3COOH }=\left(\text{2}\times {\text{12 g mol}}^{-1}\right)+\left(1\times 4{\text{ g mol}}^{-1}\right)+\left(2\times 16{\text{ g mol}}^{-1}\right)\\ ={\text{ 24 g mol}}^{-1}+{\text{4 g mol}}^{-1}+{\text{32 g mol}}^{-1}\\ ={\text{ 60 g mol}}^{-1}\end{array}$

Substitute $MM$ as ${\text{60 g mol}}^{-1}$ and $n$ as $\text{1 mol}$ in equation (3) to determine $m$ as follows:

$\begin{array}{c}\text{1 mol}=\frac{m}{{\text{60 g mol}}^{-1}}\\ m=\text{60 g}\end{array}$

The density $\left(\rho \right)$ of solution is the ratio of mass $\left(m\right)$ of the solution to the volume $\left(V\right)$ of the solution.

$\rho =\frac{w}{V}$ …… (4)

The density of acetic acid solution is $1.005{\text{ g mL}}^{-1}$ . Substitute $\rho$ as $1.005{\text{ g mL}}^{-1}$ and $V$ as $1\text{ mL}$ in equation (4).

$\begin{array}{c}1.005{\text{ g mL}}^{-1}=\frac{m}{1\text{ mL}}\\ m=1.005\text{ g}\end{array}$

$1\text{ mL}$ of acetic acid solution contains $1.005\text{ g}$ solute. So, $1000\text{mL}$ of solution contains mass as follows:

$\begin{array}{c}1\text{ mL}=\frac{1.005\text{ g}}{1\text{ mL}}\\ \frac{1\text{ L}}{1000\text{ mL}}=\frac{1.005\text{ g}}{1\text{ mL}}\\ 1\text{L}=\frac{1.005\text{ g}}{1\text{ mL}}\times 1000\text{ mL}\\ =1005\text{ g}\end{array}$

The mass of solvent is equal to the mass of solution minus the mass of solute.

$\begin{array}{c}\left(\begin{array}{l}\text{mass of }\\ \text{solvent}\end{array}\right)=\left(\begin{array}{l}\text{mass of }\\ \text{solution}\end{array}\right)-\left(\begin{array}{l}\text{mass of }\\ \text{solute}\end{array}\right)\\ =\text{ 1005 g}-\text{60 g}\\ =\text{945 g }\end{array}$

$60{\text{ g CH}}_3\text{COOH}$ is present in $945\text{ g solvent}$ as follows:

$\begin{array}{c}60{\text{ g CH}}_3\text{COOH}=945\text{ g solvent}\\ 1\text{ g solvent}=\frac{60{\text{ g CH}}_3\text{COOH}}{945\text{ g solvent}}\end{array}$

$1000\text{ g}$ of solvent contains ${\text{CH}}_3\text{COOH}$ as follows:

$\begin{array}{c}1000\text{ g solvent}=\frac{60{\text{ g CH}}_3\text{COOH}}{945\text{ g solvent}}\times 1000\text{ g solvent}\\ =63.49{\text{g CH}}_3\text{COOH}\end{array}$

Substitute $w$ as $63.49\text{g}$ and $MM$ as $60{\text{ g mol}}^{-1}$ in equation (3) to determine number of moles of acetic acid as follows:

$\begin{array}{c}n=\frac{63.49\text{ g}}{60{\text{ g mol}}^{-1}}\\ =1.058\text{ mol}\end{array}$

Substitute $n$ as $1.058\text{ mol}$ and $w$ as $1\text{ kg}$ in equation (2) as follows:

$\begin{array}{c}m=\frac{1.058\text{ mol}}{1\text{ kg}}\\ =1.058{\text{ mol kg}}^{-1}\end{array}$

Conclusion

The molality of the acetic acid solution is 1.058 m.