In chapter 4, we introduced molarity, a very useful measurement unit for evaluating the concentration of solutions. However, molarity is only one measure of concentration. In this section, we will introduce some other units of concentration that are commonly used in various applications, either for convenience or by convention.
Earlier in this chapter, we introduced percent composition as a measure of the relative amount of a given element in a compound. Percentages are also commonly used to express the composition of mixtures, including solutions. The mass percentage of a solution component is defined as the ratio of the component’s mass to the solution’s mass, expressed as a percentage: \[ \text{mass percentage} = \dfrac{\text{mass of component}}{\text{mass of solution}} \times100\% \label{3.5.1}\] We are generally most interested in the mass percentages of solutes, but it is also possible to compute the mass percentage of solvent. Mass percentage is also referred to by similar names such as percent mass, percent weight, weight/weight percent, and other variations on this theme. The most common symbol for mass percentage is simply the percent sign, %, although more detailed symbols are often used including %mass, %weight, and (w/w)%. Use of these more detailed symbols can prevent confusion of mass percentages with other types of percentages, such as volume percentages (to be discussed later in this section). Mass percentages are popular concentration units for consumer products. The label of a typical liquid bleach bottle (Figure \(\PageIndex{1}\)) cites the concentration of its active ingredient, sodium hypochlorite (\(\ce{NaOCl}\)), as being 7.4%. A 100.0-g sample of bleach would therefore contain 7.4 g of \(\ce{NaOCl}\). 
A 5.0-g sample of spinal fluid contains 3.75 mg (0.00375 g) of glucose. What is the percent by mass of glucose in spinal fluid? Solution The spinal fluid sample contains roughly 4 mg of glucose in 5000 mg of fluid, so the mass fraction of glucose should be a bit less than one part in 1000, or about 0.1%. Substituting the given masses into the equation defining mass percentage yields: \[\mathrm{\%\,glucose=\dfrac{3.75\;mg \;glucose \times \frac{1\;g}{1000\; mg}}{5.0\;g \;spinal\; fluid}=0.075\%} \nonumber\] The computed mass percentage agrees with our rough estimate (it’s a bit less than 0.1%). Note that while any mass unit may be used to compute a mass percentage (mg, g, kg, oz, and so on), the same unit must be used for both the solute and the solution so that the mass units cancel, yielding a dimensionless ratio. In this case, we converted the units of solute in the numerator from mg to g to match the units in the denominator. We could just as easily have converted the denominator from g to mg instead. As long as identical mass units are used for both solute and solution, the computed mass percentage will be correct.
A bottle of a tile cleanser contains 135 g of \(\ce{HCl}\) and 775 g of water. What is the percent by mass of \(\ce{HCl}\) in this cleanser? Answer14.8%
“Concentrated” hydrochloric acid is an aqueous solution of 37.2% \(\ce{HCl}\) that is commonly used as a laboratory reagent. The density of this solution is 1.19 g/mL. What mass of \(\ce{HCl}\) is contained in 0.500 L of this solution? Solution The HCl concentration is near 40%, so a 100-g portion of this solution would contain about 40 g of HCl. Since the solution density isn’t greatly different from that of water (1 g/mL), a reasonable estimate of the HCl mass in 500 g (0.5 L) of the solution is about five times greater than that in a 100 g portion, or \(\mathrm{5 \times 40 = 200\: g}\). To derive the mass of solute in a solution from its mass percentage, we need to know the corresponding mass of the solution. Using the solution density given, we can convert the solution’s volume to mass, and then use the given mass percentage to calculate the solute mass. This mathematical approach is outlined in this flowchart: For proper unit cancellation, the 0.500-L volume is converted into 500 mL, and the mass percentage is expressed as a ratio, 37.2 g HCl/g solution: \[ \mathrm{500\; mL\; solution \left(\dfrac{1.19\;g \;solution}{mL \;solution}\right) \left(\dfrac{37.2\;g\; HCl}{100\;g \;solution}\right)=221\;g\; HCl} \nonumber\] This mass of HCl is consistent with our rough estimate of approximately 200 g.
What volume of concentrated HCl solution contains 125 g of HCl? Answer282 mL
Liquid volumes over a wide range of magnitudes are conveniently measured using common and relatively inexpensive laboratory equipment. The concentration of a solution formed by dissolving a liquid solute in a liquid solvent is therefore often expressed as a volume percentage, %vol or (v/v)%: \[ \text{volume percentage} = \dfrac{\text{volume solute}}{\text{volume solution}} \times100\% \label{3.5.2}\]
Rubbing alcohol (isopropanol) is usually sold as a 70%vol aqueous solution. If the density of isopropyl alcohol is 0.785 g/mL, how many grams of isopropyl alcohol are present in a 355 mL bottle of rubbing alcohol? Solution Per the definition of volume percentage, the isopropanol volume is 70% of the total solution volume. Multiplying the isopropanol volume by its density yields the requested mass: \[ \text {355 mL solution}(\frac{\text{70 mL isopropyl alcohol}}{\text{100 mL solution}})(\frac{\text{0.785 g isopropyl alcohol}}{\text{1 mL isopropyl alcohol}})=\text{195 g isopropyl alcohol} \nonumber\]
Wine is approximately 12% ethanol (\(\ce{CH_3CH_2OH}\)) by volume. Ethanol has a molar mass of 46.06 g/mol and a density 0.789 g/mL. How many moles of ethanol are present in a 750-mL bottle of wine? Answer1.5 mol ethanol
“Mixed” percentage units, derived from the mass of solute and the volume of solution, are popular for certain biochemical and medical applications. A mass-volume percent is a ratio of a solute’s mass to the solution’s volume expressed as a percentage. The specific units used for solute mass and solution volume may vary, depending on the solution. For example, physiological saline solution, used to prepare intravenous fluids, has a concentration of 0.9% mass/volume (m/v), indicating that the composition is 0.9 g of solute per 100 mL of solution. The concentration of glucose in blood (commonly referred to as “blood sugar”) is also typically expressed in terms of a mass-volume ratio. Though not expressed explicitly as a percentage, its concentration is usually given in milligrams of glucose per deciliter (100 mL) of blood (Figure \(\PageIndex{2}\)).
Very low solute concentrations are often expressed using appropriately small units such as parts per million (ppm) or parts per billion (ppb). Like percentage (“part per hundred”) units, ppm and ppb may be defined in terms of masses, volumes, or mixed mass-volume units. There are also ppm and ppb units defined with respect to numbers of atoms and molecules. The mass-based definitions of ppm and ppb are given here: \[\text{ppm}=\dfrac{\text{mass solute}}{\text{mass solution}} \times 10^6\; \text{ppm} \label{3.5.3A}\] \[\text{ppb}=\dfrac{\text{mass solute}}{\text{mass solution}} \times 10^9\; \text{ppb} \label{3.5.3B}\] Both ppm and ppb are convenient units for reporting the concentrations of pollutants and other trace contaminants in water. Concentrations of these contaminants are typically very low in treated and natural waters, and their levels cannot exceed relatively low concentration thresholds without causing adverse effects on health and wildlife. For example, the EPA has identified the maximum safe level of fluoride ion in tap water to be 4 ppm. Inline water filters are designed to reduce the concentration of fluoride and several other trace-level contaminants in tap water (Figure \(\PageIndex{3}\)).
According to the EPA, when the concentration of lead in tap water reaches 15 ppb, certain remedial actions must be taken. What is this concentration in ppm? At this concentration, what mass of lead (μg) would be contained in a typical glass of water (300 mL)? Solution The definitions of the ppm and ppb units may be used to convert the given concentration from ppb to ppm. Comparing these two unit definitions shows that ppm is 1000 times greater than ppb (1 ppm = 103 ppb). Thus: \[ \mathrm{15\; \cancel{ppb} \times \dfrac{1\; ppm}{10^3\;\cancel{ppb}} =0.015\; ppm} \nonumber\] The definition of the ppb unit may be used to calculate the requested mass if the mass of the solution is provided. However, only the volume of solution (300 mL) is given, so we must use the density to derive the corresponding mass. We can assume the density of tap water to be roughly the same as that of pure water (~1.00 g/mL), since the concentrations of any dissolved substances should not be very large. Rearranging the equation defining the ppb unit and substituting the given quantities yields: \[\text{ppb}=\dfrac{\text{mass solute}}{\text{mass solution}} ×10^9\; \text{ppb} \nonumber\] \[\text{mass solute} = \dfrac{\text{ppb} \times \text{mass solution}}{10^9\;\text{ppb}} \nonumber\] \[\text{mass solute}=\mathrm{\dfrac{15\:ppb×300\:mL×\dfrac{1.00\:g}{mL}}{10^9\:ppb}=4.5 \times 10^{-6}\;g} \nonumber\] Finally, convert this mass to the requested unit of micrograms: \[\mathrm{4.5 \times 10^{−6}\;g \times \dfrac{1\; \mu g}{10^{−6}\;g} =4.5\; \mu g} \nonumber\]
A 50.0-g sample of industrial wastewater was determined to contain 0.48 mg of mercury. Express the mercury concentration of the wastewater in ppm and ppb units. Answer9.6 ppm, 9600 ppb Measures of Concentration: https://youtu.be/RjMGaUpkg8g
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: \[M=\dfrac{\text{mol solute}}{\text{L solution}} \label{11.5.1}\] 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, \(\chi\), of a component is the ratio of its molar amount to the total number of moles of all solution components: \[\chi_\ce{A}=\dfrac{\text{mol A}}{\text{total mol of all components}} \label{11.5.2}\] Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms: \[m=\dfrac{\text{mol solute}}{\text{kg solvent}} \label{11.5.3}\] 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.
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 \(\mathrm{2.22 \times 10^3 \;g}\) of ethylene glycol and \(\mathrm{2.00 \times 10^3\; g}\) 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 unit definition. \(\mathrm{mol\:C_2H_4(OH)_2=2220\:g×\dfrac{1\:mol\:C_2H_4(OH)_2}{62.07\:g\:C_2H_4(OH)_2}=35.8\:mol\:C_2H_4(OH)_2}\) \(\mathrm{mol\:H_2O=2000\:g×\dfrac{1\:mol\:H_2O}{18.02\:g\:H_2O}=111\:mol\:H_2O}\) \(\chi_\mathrm{ethylene\:glycol}=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{(35.8+111)\:mol\: total}=0.245}\) 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: \[\mathrm{2220\:g\:C_2H_4(OH)_2\left(\dfrac{mol\:C_2H_2(OH)_2}{62.07\:g}\right)=35.8\:mol\:C_2H_4(OH)_2} \]
Then, convert the mass of the water from grams to kilograms: \[\mathrm{2000\: g\:H_2O\left(\dfrac{1\:kg}{1000\:g}\right)=2\: kg\:H_2O} \] Finally, calculate molarity per its definition: \[\begin{align*} \ce{molality}&=\mathrm{\dfrac{mol\: solute}{kg\: solvent}}\\ \ce{molality}&=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{2\:kg\:H_2O}}\\ \ce{molality}&=17.9\:m \end{align*}\]
What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH3, dissolved in 125 g of water? Answer7.14 × 10−3; 0.399 m
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: \[\mathrm{\dfrac{3.0\;mol\; NaCl}{1.0\; kg\; H_2O}} \] 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 \[\mathrm{1.0\:kg\:H_2O\left(\dfrac{1000\:g}{1\:kg}\right)\left(\dfrac{mol\:H_2O}{18.02\:g}\right)=55\:mol\:H_2O} \] and then substituting these molar amounts into the definition for mole fraction. \[\begin{align*} X_\mathrm{H_2O}&=\mathrm{\dfrac{mol\:H_2O}{mol\: NaCl + mol\:H_2O}}\\ X_\mathrm{H_2O}&=\mathrm{\dfrac{55\:mol\:H_2O}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\\ X_\mathrm{H_2O}&=0.95\\ X_\mathrm{NaCl}&=\mathrm{\dfrac{mol\: NaCl}{mol\: NaCl+mol\:H_2O}}\\ X_\mathrm{NaCl}&=\mathrm{\dfrac{3.0\:mol\:NaCl}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\\ X_\mathrm{NaCl}&=0.052 \end{align*}\]
The mole fraction of iodine, \(\ce{I_2}\), dissolved in dichloromethane, \(\ce{CH_2Cl_2}\), is 0.115. What is the molal concentration, m, of iodine in this solution? Answer1.50 m A Video Discussing how to Convert Measures of Concentration. Video Link: https://youtu.be/MG5CZOrazRA
Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka.
Given: volume percent and density Asked for: mass percentage, mole fraction, molarity, and molality Strategy:
Solution: The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (\(CH_3CH_2OH\)), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density: \[mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber\] If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution. B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage: \[ \begin{align*} \%EtOH &=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100) \\[4pt] &=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) \\[4pt]&= 34.5\%\end{align*}\] C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL: \[ \begin{align*} moles\; \ce{CH_3CH_2OH}&=(31.6\; \cancel{g\; \ce{CH_3CH_2OH}}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; \ce{CH_3CH_2OH}}}\right) \\[4pt] &=0.686 \;mol\; \ce{CH_3CH_2OH} \end{align*}\] Similarly, the number of moles of water is \[ moles \;\ce{H_2O}=(60.0\; \cancel{g \; \ce{H_2O}}) \left(\dfrac{1 \;mol\; \ce{H_2O}}{18.02\; \cancel{g\; \ce{H_2O}}}\right)=3.33\; mol\; \ce{H_2O}\nonumber\] The mole fraction of ethanol is thus \[ \chi_{\ce{CH_3CH_2OH}}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber\] D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is \[ M_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M\nonumber\]The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward: \[ m_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m\nonumber\]
A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene.
mass percentage toluene = 24.8% Answer b\(\chi_{toluene} = 0.219\) Answer c2.35 M toluene Answer d3.59 m toluene
Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature. Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies. Table \(\PageIndex{1}\) summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example \(\PageIndex{3}\).
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, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent. mass percentage ratio of solute-to-solution mass expressed as a percentage mass-volume percent ratio of solute mass to solution volume, expressed as a percentage parts per billion (ppb) ratio of solute-to-solution mass multiplied by 109 molality (m) a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms parts per million (ppm) ratio of solute-to-solution mass multiplied by 106 volume percentage ratio of solute-to-solution volume expressed as a percentage |
What is the percent by volume of isopropyl alcohol in a solution that contains 25 mL isopropanol?
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