11.4  Concentration Units

Nicole Bouvier-Brown; Saori Shiraki; J. Ryan Hunt; Emily Jarvis

Chapter 11 Solutions and Colloids

11.4 Concentration Units

Learning Objectives

By the end of this section, you will be able to:

Express concentrations of solution components using mole fraction and molality
Convert between different concentration units

The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.

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:

$M = \frac{n}{V} = \frac{m o l s o l u t e}{L s o l u t i o n}$

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:

$X_{A} = \frac{n_{A}}{n_{T}} = \frac{m o l A}{t o t a l m o l s o f a l l c o m p o n e n t s}$

By this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one.

Molality (m) is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:

$m = \frac{n}{m a s s} = \frac{m o l s o l u t e}{k g s o l v e n t}$

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 chapter module.

Example 11.3 – 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, 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)?

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 definition of mole fraction.

$m o l C_{2} H_{4} (O H)_{2} = 2.22 \times 10^{3} g (\frac{m o l C_{2} H_{4} (O H)_{2}}{62.07 g}) = 35.8 m o l C_{2} H_{4} (O H)_{2}$

$m o l H_{2} O = 2.00 \times 10^{3} g \frac{m o l H_{2} O}{18.02 g} = 111 m o l H_{2} O$

$X_{e t h y l e n e g l y c o l} = \frac{n_{e t h y l e n e g l y c o l}}{n_{T}} = \frac{35.8 m o l}{35.8 m o l + 111 m o l} = 0.244$

Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).

(b) Derive moles of solute and mass of solvent (in kg).

First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:

$2.22 \times 10^{3} g C_{2} H_{4} (O H)_{2} \frac{m o l C_{2} H_{4} (O H)_{2}}{62.07 g C_{2} H_{4} (O H)_{2}} = 35.8 m o l C_{2} H_{4} (O H)_{2}$

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

$2.00 \times 10^{3} g H_{2} O \frac{1 k g}{1, 000 g} = 2.00 k g H_{2} O$

Finally, calculate molality per its definition:

$m = \frac{n}{m a s s} = \frac{m o l C_{2} H_{4} (O H)_{2}}{k g H_{2} O} = \frac{35.8 m o l}{2.00 k g} = 17.9 m$

Check Your Learning

Click here to see this problem worked through!

First, we calculate the mols of ammonia and water:

$mol N H_{3} = 0.850 g (\frac{1 mol}{17.031 g}) = 0.0499 mol$

$mol H_{2} O = 125 g (\frac{1 mol}{18.0153 g}) = 6.94 mol$

Then, we can calculate the mol fraction of ammonia:

$X_{N H_{3}} = \frac{n_{N H_{3}}}{n_{T}} = \frac{0.0499 mol}{(0.0499 mol + 6.94 mol)} = 0.00714 = 7.14 \times 10^{- 3}$

Click here to see this problem worked through!

First, we calculate the mols of ammonia:

$mol N H_{3} = 0.850 g (\frac{1 mol}{17.031 g}) = 0.0499 mol$

Next, we convert the mass of the water from grams to kilograms:

$125 g (\frac{1 k g}{1000 g}) = 0.125 k g$

Finally, we calculate molality using its definition:

$m_{N H_{3}} = \frac{n_{N H_{3}}}{m a s s_{H_{2} O}} = \frac{0.0499 mol}{0.125 k g} = 0.399 m$

Example 11.4 – Converting Mole Fraction and Molal Concentrations

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

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:

\frac{3.0 m o l N a C l}{1 k g H_{2} O}

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

1.0 k g H_{2} O (\frac{1, 000 g}{1 k g}) (\frac{m o l H_{2} O}{18.02 g}) = 55 m o l H_{2} O

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

$X_{H_{2} O} = \frac{m o l H_{2} O}{m o l N a C l + m o l H_{2} O} = \frac{55 m o l H_{2} O}{3.0 m o l N a C l + 55 m o l H_{2} O} = 0.95$

$X_{N a C l} = \frac{m o l N a C l}{m o l N a C l + m o l H_{2} O} = \frac{3.0 m o l N a C l}{3.0 m o l N a C l + 55 m o l H_{2} O} = 0.052$

Check Your Learning

Click here to see this problem worked through!

If we assume that the total number of mols is 1, we can write an I₂ mole fraction of 0.115 as

$\frac{0.115 mol I_{2}}{1 mol total} = \frac{0.115 mol I_{2}}{0.115 mol I_{2} + 0.885 mol C H_{2} C l_{2}}$

implying that we have 0.115 mol I₂ and 0.885 mol CH₂Cl₂. We can convert 0.885 mol CH₂Cl₂ to kg using its molar mass:

$0.885 mol C H_{2} C l_{2} (\frac{84.93 g}{mol}) (\frac{1 k g}{1000 g}) = 0.0752 k g$

Finally, we can calculate the molality using its definition:

$m_{I_{2}} = \frac{n_{I_{2}}}{m a s s_{C H_{2} C l_{2}}} = \frac{0.115 mol}{0.0752 k g} = 1.53 m$

Example 11.5 – 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?

Solution

The provided molal concentration may be explicitly written as:

M = \frac{0.556 m o l g l u c o s e}{1 L s o l u t i o n}

Consider the definition of molality:

m = \frac{m o l s o l u t e}{k g s o l v e n t}

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 L s o l n (\frac{1.04 g}{m L}) (\frac{1, 000 m L}{1 L}) (\frac{1 k g}{1, 000 g}) = 1.04 k g s o l n

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 m o l g l u c o s e (\frac{180.2 g}{m o l g l u c o s e}) = 100.2 g o r 0.1002 k g

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

1.04 k g s o l u t i o n - 0.1002 k g g l u c o s e = 0.94 k g w a t e r

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

m = \frac{n}{m} = \frac{0.556 m o l g l u o c s e}{0.94 k g w a t e r} = 0.59 m

Check Your Learning

Click here to see a walkthrough for this problem!

Using the definition of molality, we can write

$molality = \frac{33.7 mol H N O_{3}}{1 k g H_{2} O}$

Let’s assume we have 33.7 mols of HNO₃ and 1 kg of water.

Consider the definition of molarity:

molarity = \frac{mol solute}{volume of solution (L)}

We can find the total mass of the solution and then use density to calculate the volume of the solution. The mass of HNO₃is

33.7 mol (\frac{63.01 g}{1 mol}) = 2123 g = 2.123 k g

so the total mass of the solution is

1 k g H_{2} O + 2.123 k g H N O_{3} = 3.123 k g = 3123 g

Using the density of the solution, the total volume is

3123 g (\frac{1 m L}{1.35 g}) (\frac{1 L}{1000 m L}) = 2.31 L

Finally, we can calculate the molarity:

Molarity = \frac{33.7 mol}{2.31 L} = 14.6 M

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

11.4 Concentration Units Copyright © by Nicole Bouvier-Brown; Saori Shiraki; J. Ryan Hunt; and Emily Jarvis is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.