Chapter 12: Solutions and Colloids

# [merged with Libre] 12.4 Solution Concentration

### Learning Outcomes

• Identify the concentration unit most appropriate for a particular application
• Calculate concentrations of solutions and convert between concentration units (molarity, molality, mole fraction, parts by mass (ppm, ppb), etc.)

## Mole Fraction and Molality

Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity ($M$) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:

$M=\dfrac{\text{mol solute}}{\text{L solution}}$

Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.
The mole fraction, X, of a component is the ratio of its molar amount to the total number of moles of all solution components:

${X}_{\text{A}}=\dfrac{\text{mol}\text{ A}}{\text{total mol of all components}}$

By this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one.
Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:

$m=\dfrac{\text{mol solute}}{\text{kg solvent}}$

Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this section.

### Example 12.4.1: Calculating Mole Fraction and Molality

The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, $\ce{C2H4(OH)2}$, in a solution prepared from 2.22 × 103 g of ethylene glycol and 2.00 × 103 g of water (approximately 2 L of glycol and 2 L of water)?

Show Solution

(a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.

$\begin{array}{l}\\ \text{mol}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}=2220\text{g}\times \dfrac{1\text{mol}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}}{62.07\text{g}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}}=35.8\text{mol}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}\\ \text{mol}{\ce{ H}}_{2}\ce{O}=2000\text{g}\times \dfrac{1\text{mol}{\ce{ H}}_{2}\ce{O}}{18.02\text{g}{\text{ H}}_{2}\ce{O}}=11.1\text{mol}{\ce{ H}}_{2}\ce{O}\\ {X}_{\text{ethylene }\text{glycol}}=\dfrac{35.8\text{mol}{\ce{ C}}_{2}{\text{H}}_{4}{\left(\ce{OH}\right)}_{2}}{\left(35.8+11.1\right)\text{mol total}}=0.763\end{array}$

Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:

$2220\text{g}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}\left(\dfrac{\text{mol}{\ce{ C}}_{2}{\ce{H}}_{2}{\left(\ce{OH}\right)}_{2}}{62.07\text{g}}\right)=35.8\text{mol}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}$

Then, convert the mass of the water from grams to kilograms:

$\text{2000 g}{\ce{ H}}_{2}\ce{O}\left(\dfrac{1\text{kg}}{1000\text{g}}\right)=\text{2 kg}{\ce{ H}}_{2}\ce{O}$

Finally, calculate molarity per its definition:

$\begin{array}{lll}\\ \hfill \text{molality}& =& \dfrac{\text{mol solute}}{\text{kg solvent}}\hfill \\ \hfill \text{molality}& =& \dfrac{35.8\text{mol}{\ce{ C}}_{2}{\ce{H}}_{4}{\left(\ce{OH}\right)}_{2}}{2\text{kg}{\ce{ H}}_{2}\ce{O}}\hfill \\ \hfill \text{molality}& =& 17.9m\hfill \end{array}$

### Example 12.4.2: Converting Mole Fraction and Molal Concentrations

Calculate the mole fraction of solute and solvent in a $3.0 m$ solution of sodium chloride.

Show Solution

Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:

$\dfrac{3.0\text{mol NaCl}}{1.0\text{kg}{\ce{ H}}_{2}\ce{O}}$

The numerator for this solution’s mole fraction is, therefore, $3.0\;mol\;\ce{NaCl}$. The denominator may be computed by deriving the molar amount of water corresponding to $1.0 kg$

$1.0\text{kg}{\ce{ H}}_{2}\ce{O}\left(\dfrac{1000\text{g}}{1\text{kg}}\right)\left(\dfrac{\text{mol}{\ce{ H}}_{2}\ce{O}}{18.02\text{g}}\right)=55\text{mol}{\ce{ H}}_{2}\ce{O}$

and then substituting these molar amounts into the definition for mole fraction.
Deleted:

$\begin{array}{ccc}\hfill {X}_{{\text{H}}_{2}\text{O}}& =& \dfrac{\text{mol}{\text{ H}}_{2}\text{O}}{\text{mol NaCl}+\text{mol}{\text{ H}}_{2}\text{O}}\hfill \\ \hfill {X}_{{\text{H}}_{2}\text{O}}& =& \dfrac{55\text{mol}{\text{ H}}_{2}\text{O}}{3.0\text{mol NaCl}+55\text{mol}{\text{ H}}_{2}\text{O}}\hfill \\ \hfill {X}_{{\text{H}}_{2}\text{O}}& =& 0.95\hfill \\ \hfill {X}_{\text{NaCl}}& =& \dfrac{\text{mol NaCl}}{\text{mol NaCl}+\text{mol}{\text{ H}}_{2}\text{O}}\hfill \\ \hfill {X}_{\text{NaCl}}& =& \dfrac{3.0\text{mol}{\text{ H}}_{2}\text{O}}{3.0\text{mol NaCl}+55\text{mol}{\text{ H}}_{2}\text{O}}\hfill \\ \hfill {X}_{\text{NaCl}}& =& 0.052\hfill \end{array}$

### Example 12.4.3: Molality and Molarity Conversions

Intravenous infusion of a $0.556 M$ aqueous solution of glucose (density of $1.04 g/mL$) is part of some post-operative recovery therapies. What is the molal concentration of glucose in this solution?

Show Solution

The provided molal concentration may be explicitly written as: $M = 0.556 \text{mol glucose/ 1 L solution}$

Consider the definition of molality:

$$m = \text{mol solute/ kg solvent}$$

The amount of glucose in $1 L$ of this solution is $0.556 mol$, so the mass of water in this volume of solution is needed.

First, compute the mass of $1.00 L$ of the solution:

$$(1.0 \text{L soln})(1.04 \text{g/mL})(1000 \text{mL/L})(1 \text{kg}/1000 \text{g}) = 1.04 \text{kg soln}$$

This is the mass of both the water and its solute, glucose, and so the mass of glucose must be subtracted. Compute the mass of glucose from its molar amount:

$$(0.556 \text{mol glucose})(180.2 \text{g/1 mol}) = 100.2 \text{g or } 0.1002 \text{kg}$$

Subtracting the mass of glucose yields the mass of water in the solution:

$$1.04 \text{kg solution} – 0.1002 \text{kg glucose} = 0.94 \text{kg water}$$

Finally, the molality of glucose in this solution is computed as:

$$m = 0.556 \text{mol glucose} / 0.94 \text{kg water} = 0.59 \text{m}$$

### Try It

1. Calculate the mole fraction of each solute and solvent:
1. $0.710 kg$ of sodium carbonate (washing soda), $\ce{Na_{2}CO_{3}}$, in $10.0 kg$ of water—a saturated solution at $0 °C$
2. $125 g$ of $\ce{NH_{4}NO_{3}}$ in $275 g$ of water—a mixture used to make an instant ice pack
3. $25 g$ of $\ce{Cl_{2}}$ in $125 g$ of dichloromethane, $\ce{CH_{2}Cl_2}$
4. $0.372 g$ of histamine, $\ce{C_{5}H_{9}N}$, in $125 g$ of chloroform, $\ce{CHCl_{3}}$
2. Calculate the molality of each of the following solutions:
1. $0.710 kg$ of sodium carbonate (washing soda), $\ce{Na_{2}CO_{3}}$, in $10.0 kg$ of water—a saturated solution at $0°C$
2. $125 g$ of $\ce{NH_{4}NO_{3}}$ in $275 g$ of water—a mixture used to make an instant ice pack
3. $25 g$ of $\ce{Cl_{2}}$ in $125 g$ of dichloromethane, $\ce{CH_{2}Cl_{2}}$
4. $0.372 g$ of histamine, $\ce{C_{5}H_{9}N}$, in $125 g$ of chloroform, $\ce{CHCl_{3}}$
Show Selected Solutions

1. The mole fractions are as follows:

1. $\begin{array}{l}\text{mol}{\ce{Na}}_{2}{\ce{CO}}_{3}=710\cancel{\text{g}{\ce{Na}}_{2}{\ce{CO}}_{3}}\times \frac{1\text{mol}}{105.9886\cancel{\text{g}{\ce{Na}}_{2}{\ce{CO}}_{3}}}=6.70\text{mol}\\ \text{mol}{\ce{H}}_{2}\ce{O}=\frac{10,000\text{g}}{18.0153\text{g/mol}}=555.08\text{mol}\end{array}$
• $\textbf{Total number of moles} = 555.08 mol + 6.70 mol = 561.78 mol$
$\begin{array}{l}{ }{X}_{{\text{Na}}_{2}{\text{CO}}_{3}}=\frac{6.70\text{mol}}{561.78\text{mol}}=0.0119\\ {X}_{{\text{H}}_{2}\text{O}}=\frac{555.08\text{mol}}{561.78\text{mol}}=0.988\end{array}$
2. $\begin{array}{l}\text{mol}{\ce{NH}}_{4}{\ce{NO}}_{3}=125\cancel{\text{g}{\ce{NH}}_{4}{\ce{NO}}_{3}}\times \frac{1\text{mol}}{80.0434\cancel{\text{g}{\ce{NH}}_{4}{\ce{NO}}_{3}}}=1.56\text{mol}\\ \text{mol}{\ce{H}}_{2}\ce{O}=\frac{275\text{g}}{18.0153\text{g/mol}}=15.26\text{mol}\end{array}$
• $\textbf{Total number of moles} = 15.26 mol + 1.56 mol = 16.82 mol$
$\begin{array}{l}{ }{X}_{{\ce{NH}}_{4}{\ce{NO}}_{3}}=\frac{1.56\text{mol}}{16.82\text{mol}}=0.9927\\ {X}_{{\ce{H}}_{2}\ce{O}}=\frac{15.26\text{mol}}{16.82\text{mol}}=0.907\end{array}$
3. $\begin{array}{l}\text{mol}{\ce{Cl}}_{2}=25\cancel{\text{g}{\ce{Cl}}_{\text{2}}}\times \frac{1\text{mol}}{70.9054\cancel{\text{g}{\ce{Cl}}_{2}}}=0.35\text{mol}\\ \text{mol}{\ce{CH}}_{\text{2}}{\ce{Cl}}_{\text{2}}=\frac{125\text{g}}{84.93\text{g/mol}}=1.47\text{mol}\end{array}$
• $\textbf{Total number of moles} = 1.47 mol + 0.35 mol = 1.82 mol$
$\begin{array}{l}{X}_{{\ce{Cl}}_{2}}=\frac{0.35\text{mol}}{1.82\text{mol}}=0.192\\ {X}_{{\ce{CH}}_{2}{\ce{Cl}}_{2}}=\frac{1.47\text{mol}}{1.82\text{mol}}=0.808\end{array}$
4. $\begin{array}{l}\text{mol}{\ce{C}}_{\text{5}}{\ce{H}}_{\text{9}}\text{N}=0.372\cancel{\text{g}{\ce{C}}_{5}{\ce{H}}_{9}\ce{N}}\times \frac{1\text{mol}}{83.1332\cancel{\text{g}{\ce{C}}_{5}{\ce{H}}_{9}\ce{N}}}=4.47\times {10}^{-3}\text{mol}\\ \text{mol}{\ce{CHCl}}_{3}=\frac{125\text{g}}{119.38\text{g/mol}}=1.047\text{mol}\end{array}$
• $\textbf{Total number of moles} = 1.047 mol + 0.00447 mol = 1.05 mol$
$\begin{array}{l}\\ {X}_{{\ce{C}}_{5}{\ce{H}}_{9}\ce{N}}=\frac{0.00447\text{mol}}{1.05\text{mol}}=0.00426\\ {X}_{{\ce{CHCl}}_{3}}=\frac{1.047\text{mol}}{1.05\text{mol}}=0.997\end{array}$

2. The molality of each is as follows:

1. $\begin{array}{l}\text{mol}{\ce{Na}}_{\text{2}}{\ce{CO}}_{\text{3}}=710\text{g}{\ce{Na}}_{2}{\ce{CO}}_{3}\times \frac{1\text{mol}}{105.9886\text{g}{\ce{Na}}_{2}{\ce{CO}}_{3}}\\ \text{molality of}{\ce{Na}}_{\text{2}}{\ce{CO}}_{\text{3}}=\frac{6.70\text{mol}}{10.0\text{kg}}=6.70\times {10}^{-1}m\end{array}$
2. $\begin{array}{l}\text{mol}{\ce{NH}}_{\text{4}}{\ce{NO}}_{\text{3}}=125\cancel{\text{g}{\ce{NH}}_{4}{\text{NO}}_{3}}\times \frac{1\text{mol}}{80.0434\cancel{\text{g}{\ce{NH}}_{4}{\ce{NO}}_{3}}}=1.56\text{mol}\\ \text{molality of}{\ce{NH}}_{\text{4}}{\ce{NO}}_{\text{3}}=\frac{1.56\text{mol}}{0.275\text{kg}}=5.67m\end{array}$
3. $\text{mol}{\ce{Cl}}_{\text{2}}=25\cancel{\text{g}{\ce{Cl}}_{\text{2}}}\times \frac{1\text{mol}}{70.9054\cancel{\text{g}{\ce{Cl}}_{\text{2}}}}=0.35\text{mol}$
4. $\begin{array}{l}\text{mol}{\ce{C}}_{\text{5}}{\text{H}}_{\text{9}}\text{N}=0.372\cancel{\text{g}{\ce{C}}_{\text{5}}{\ce{H}}_{\text{9}}\ce{N}}\times \frac{1\text{mol}}{83.1332\cancel{\text{g}{\ce{C}}_{\text{5}}{\ce{H}}_{\text{9}}\ce{N}}}=4.47\times {10}^{-3}\text{mol}\\ \text{molality of}{\ce{C}}_{\text{5}}{\ce{H}}_{\text{9}}\ce{N}=\frac{4.47\times {10}^{-3}\text{mol}}{0.125\text{kg}}=0.0358m\end{array}$

### Key Concepts and Summary

Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb); and molality (m), the number of moles of solute per kilogram of solvent.

## Glossary

molality (m): a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
mole fraction (X): the ratio of a solution component’s molar amount to the total number of moles of all solution components

definition