13.4 Ways of Expressing Concentration

The concentration of a solution can be expressed either qualitatively or quantitatively. The terms dilute and concentrated are used to describe a solution qualitatively. A solution with a relatively small concentration of solute is said to be dilute; one with a large concentration is said to be concentrated. We use several different ways to express concentration in quantitative terms, and we examine four of these in this section: mass percentage, mole fraction, molarity, and molality.

Mass Percentage, ppm, and ppb

One of the simplest expressions of concentration is the mass percentage of a component in a solution, given by

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[13.5]

where we have abbreviated solution as "soln." Thus, a solution of hydrochloric acid that is 36 percent HCl by mass contains 36 g of HCl for each 100 g of solution.

We often express the concentrations of very dilute solution in parts per million (ppm), defined as

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[13.6]

A solution whose solute concentration is 1 ppm contains 1 g of solute for each million (106) grams of solution or, equivalently, 1 mg of solute per kilogram of solution. Because the density of water is 1 g/mL, 1 kg of a dilute aqueous solution will have a volume very close to 1 L. Thus, 1 ppm also corresponds to 1 mg of solute per liter of solution. The acceptable maximum concentrations of toxic or carcinogenic substances are often expressed in ppm. For example, the maximum allowable concentration of arsenic in drinking water in the United States is 0.05 ppm; that is, 0.05 mg of arsenic per liter of water.

For solutions that are even more dilute, parts per billion (ppb) is used. A concentration of 1 ppb represents 1 g of solute per billion (109) grams of solution, or 1 microgram ( g) of solute per liter of solution. Thus, the allowable concentration of arsenic in water can be expressed as 50 ppb.

SAMPLE EXERCISE 13.3

(a) A solution is made by dissolving 13.5 g of glucose, C6H12O6, in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn2+. What is the concentration of Zn2+ in parts per million?

SOLUTION (a) We calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Because 0.100 kg equals 100 g,

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The mass percentage of water in this solution is (100 - 11.9)% = 88.1%.

(b) Because 1 g is 1 10–6 g, 5.4 g = 5.4 10–6 g. Thus,

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PRACTICE EXERCISE

(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? Answer: (a) 2.91 percent; (b) 90.5 g of NaOCl

Mole Fraction, Molarity, and Molality

We often use concentration expressions based on the number of moles of one or more components of the solution. The three we use most commonly are mole fraction, molarity, and molality.

Recall from Section 10.6 that the mole fraction of a component of a solution is given by

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[13.7]

The symbol X is commonly used for mole fraction, with a subscript to indicate the component of interest. For example, the mole fraction of HCl in a hydrochloric acid solution is represented as XHCl. Thus a solution containing 1.00 mol of HCl (36.5 g) and 8.00 mol of water (144 g) has a mole fraction of HCl of XHCl = (1.00 mol)/(1.00 mol + 8.00 mol) = 0.111. Note that mole fractions have no units because the units in the numerator and the denominator cancel. The sum of the mole fractions of all components of a solution must equal 1. Thus, in the aqueous HCl solution, XH2O = 1.000 - 0.111 = 0.889. Mole fractions are very useful when dealing with gases as we saw in Section 10.6 but have limited use when dealing with liquid solutions.

We introduced molarity (M) in Section 4.5. Recall that the molarity of a solute in a solution is defined as

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[13.8]

For example, if you dissolve 0.500 mol of Na2CO3 in enough water to form 0.250 L of solution, then the solution has a concentration of (0.500 mol)/(0.250 L) = 2.00 M in Na2CO3. Molarity is especially useful for relating the volume of a solution to the quantity of solute it contains, as we saw in our earlier discussions of titrations.

The molality of a solution, denoted m, is a unit that we haven't encountered in previous chapters. This concentration unit equals the number of moles of solute per kilogram of solvent:

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[13.9]

Thus, if you form a solution by mixing 0.200 mol of NaOH (40.0 g) and 0.500 kg of water (500 g), the concentration of the solution is (0.200 mol)/(0.500 kg) = 0.400 m (that is, 0.400 molal) in NaOH.

Notice the difference between the definitions of molarity and molality. Because these two ways of expressing concentration are so similar, they can be easily confused. Molarity is defined in terms of the volume of solution, whereas molality is defined in terms of the mass of solvent. When water is the solvent, the molality and molarity of dilute solutions are numerically about the same, because 1 kg of solvent is nearly the same as 1 kg of solution, and 1 kg of the solution has a volume of about 1 L.

The molality of a given solution does not vary with temperature because masses do not vary with temperature. Molarity, however, changes with temperature because the expansion or contraction of the solution changes its volume. Thus molality is often the concentration unit of choice when a solution is to be used over a range of temperatures.

SAMPLE EXERCISE 13.4

A solution is made by dissolving 4.35 g glucose, C6H12O6, in 25.0 mL of water. Calculate the molality of glucose in the solution.

SOLUTION In order to calculate the molality, we must know the number of moles of solute (glucose) and the number of kilograms of the solvent (water). The molar mass of glucose is 180.2 g/mol. Hence, there are 0.0241 mol of glucose:

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Because water has a density of 1.00 g/mL, the mass of the solvent is 25.0 g = 0.0250 kg. Using Equation 13.9, then gives

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PRACTICE EXERCISE

What is the molality of a solution made by dissolving 36.5 g of naphthalene, C10H8, in 425 g of toluene, C7H8? Answer: 0.670 m

Conversion of Concentration Units

Sometimes the concentration of a given solution needs to be known in several different concentration units. It is possible to interconvert concentration units as the following examples show.

SAMPLE EXERCISE 13.5

A solution of hydrochloric acid contains 36 percent HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.

SOLUTION (a) In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let's assume that there is exactly 100 g of solution. The solution therefore contains 36 g of HCl and (100 - 36) g = 64 g of H2O. To calculate the mole fraction of HCl, we first convert mass to moles and then use Equation 13.7:

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(b) To calculate the molality of HCl in the solution, we use Equation 13.9, where water is the solvent. We calculated the number of moles of HCl in part (a), and the mass of solvent is 64 g = 0.064 kg.

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PRACTICE EXERCISE

A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution. Answers: (a) 0.505 m; (b) 9.00 10–3

In order to interconvert molality and molarity, we need to know the density of the solution. Figure 13.19 outlines the calculation of the molarity and molality of a solution from the mass of solute and the mass of solvent. The mass of the solution is the sum of masses of the solvent and solute. The volume of the solution can be calculated from its mass and density.

Figure 13.19 Diagram summarizing the calculation of molality and molarity from the mass of the solute, the mass of the solvent, and the density of the solution.

SAMPLE EXERCISE 13.6

Given that the density of a solution of 5.0 g of toluene and 225 g of benzene is 0.876 g/mL, calculate the molarity of the solution.

SOLUTION The total mass of the solution is equal to the mass of the solvent plus the mass of the solute:

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The density of the solution is used to convert the mass of the solution to its volume:

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The number of moles of solute is:

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Molarity is moles of solute per liter of solution:

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By comparison, the molality of this solution is (0.054 mol C7H8)/(0.225 kg solvent) = 0.24 m.

PRACTICE EXERCISE

A solution containing equal masses of glycerol, C3H8O3, and water has a density of 1.10 g/mL. Calculate (a) the molality of glycerol; (b) the mole fraction of glycerol; (c) the molarity of glycerol in the solution. Answers: (a) 10.9 m; (b) XC3H8O3 = 0.163; (c) 5.97 M