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13.3: Units of Concentration

The Learning Objective of this module is to describe the concentration of a solution in the way that is most appropriate for a particular problem or application.

There are several different ways to quantitatively describe the concentration of a solution. For example, molarity was introduced in Chapter 4 as a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions, introduced in Chapter 10, are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example 4 reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known.

Example 4

Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL.

  1. What is its molarity?
  2. What is its mole fraction?

Given: mass of substance and mass and density of solution

Asked for: molarity and mole fraction

Strategy:

  1. Calculate the number of moles of acetic acid in the sample. Then calculate the number of liters of solution from its mass and density. Use these results to determine the molarity of the solution.
  2. Determine the mass of the water in the sample and calculate the number of moles of water. Then determine the mole fraction of acetic acid by dividing the number of moles of acetic acid by the total number of moles of substances in the sample.

Solution:

A The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. The volume of the solution equals its mass divided by its density. The calculations follow:

\[moles\; CH_3CO_2H=\dfrac{3.78\; \cancel{g}\; CH_3CO_2H}{60.05\; \cancel{g}/mol}=0.0629 \;mol\]

\[volume=\dfrac{mass}{density}=\dfrac{100.0\; \cancel{g}\; solution}{1.00\; \cancel{g}/mL}=100\; mL\]

\[  molarity\; of\; CH_3CO_2H=\dfrac{moles\; CH_3CO_2H }{\text{liter solution}}=\dfrac{0.0629\; mol\; CH_3CO_2H}{(100\; \cancel{mL})(1\; L/1000\; \cancel{mL})}=0.629\; M \;CH_3CO_2H \]

This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than \(\frac{1}{2}\) mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL.

B To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have

\[moles\; H_2O=\dfrac{96.2\; \cancel{g}\; H_2O}{18.02\; \cancel{g}/mol}=5.34\; mol\; H_2O\]

The mole fraction \(X\) of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present:

\[X_{CH_3CO_2H}=\dfrac{moles\; CH_3CO_2H}{moles \;CH_3CO_2H + moles\; H_2O}=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol}=0.0116=1.16 \times 10^{−2}\]

This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01.

Exercise 5

A solution of \(HCl\) gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of \(HCl\) per 100.0 g of solution, and its density is 1.10 g/mL.

  1. What is its molarity?
  2. What is its mole fraction?

Answer:

  1. 6.10 M HCl
  2. XHCl = 0.111

The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent:

\[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \tag{13.5}\]

Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of \(H_2O\) under these conditions is very close to 1.0 L, and a 0.50 M solution of \(KBr\) in water, for example, has approximately the same concentration as a 0.50 m solution.

Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb):

\[\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \tag{13.6}\]

\[\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \tag{13.7}\]

\[\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \tag{13.8}\]

In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution.

The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of \(H_2SO_4\) per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL).

 

Example 5

Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL.

  1. What is the molarity of the solution?
  2. What is the mass of benzene in the sample?

Given: volume of sample, solute concentration, and density of solution

Asked for: molarity of solute and mass of solute in 250 mL

Strategy:

  1. Use the concentration of the solute in parts per million to calculate the molarity.
  2. Use the concentration of the solute in parts per million to calculate the mass of the solute in the specified volume of solution.

Solution:

a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore

\[ molarity=\dfrac{moles}{liter solution}=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L}=1.63 \times 10^{-4} M\]

b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is

\[ mass of benzene=\dfrac{(12.7\; mg\; benzene)(250\; \cancel{mL})}{1000\; mL}=3.18\; mg =3.18 \times 10^{-3}\; g\; benzene\]
Exercise 6

The maximum allowable concentration of lead in drinking water is 9.0 ppb. What is the molarity of \(Pb^{2+}\) in a 9.0 ppb aqueous solution? Use your calculated concentration to determine how many grams of \(Pb^{2+}\) are in an 8 oz glass of water.

Answer: 4.3 × 10−8 M; 2 × 10−6 g

How do chemists decide which units of concentration to use for a particular application? 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 we saw in Chapter 10. As you will learn in Section 13.5, 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 13.5 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 6.

Table 13.5 Different Units for Expressing the Concentrations of Solutions*

Unit Definition Application
*The molarity of a solution is temperature dependent, but the other units shown in this table are independent of temperature.
molarity (M) moles of solute/liter of solution (mol/L) Used for quantitative reactions in solution and titrations; mass and molecular mass of solute and volume of solution are known.
mole fraction (X) moles of solute/total moles present (mol/mol) Used for partial pressures of gases and vapor pressures of some solutions; mass and molecular mass of each component are known.
molality (m) moles of solute/kg of solvent (mol/kg) Used in determining how colligative properties vary with solute concentration; masses and molecular mass of solute are known.
mass percentage (%) [mass of solute (g)/mass of solution (g)] × 100 Useful when masses are known but molecular masses are unknown.
parts per thousand (ppt) [mass of solute/mass of solution] × 103 (g solute/kg solution) Used in the health sciences, ratio solutions are typically expressed as a proportion, such as 1:1000.
parts per million (ppm) [mass of solute/mass of solution] × 106 (mg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
parts per billion (ppb) [mass of solute/mass of solution] × 109 (µg solute/kg solution) Used for trace quantities; masses are known but molecular masses may be unknown.
Example 6

Vodka is essentially a solution of pure 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.

  1. the mass percentage
  2. the mole fraction
  3. the molarity
  4. the molality

Given: volume percent and density

Asked for: mass percentage, mole fraction, molarity, and molality

Strategy:

  1. Use the density of the solute to calculate the mass of the solute in 100.0 mL of solution. Calculate the mass of water in 100.0 mL of solution.
  2. Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.
  3. Convert grams of solute and solvent to moles of solute and solvent. Calculate the mole fraction of solute by dividing the moles of solute by the total number of moles of substances present in solution.
  4. Calculate the molarity of the solution: moles of solute per liter of solution. Determine the molality of the solution by dividing the number of moles of solute by the kilograms of solvent.

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:

\[mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\]

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:

\[\%EtOH=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100)=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) = 34.5\%\]

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:

\[ moles\; EtOH=(31.6\; \cancel{g\; EtOH}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; EtOH}}\right)=0.686 \;mol\; CH_3CH_2OH\]

Similarly, the number of moles of water is

\[ moles \;H_2O=(60.0\; \cancel{g \; H_2O}) \left(\dfrac{1 \;mol\; H_2O}{18.02\; \cancel{g\; H_2O}}\right)=3.33\; mol\; H_2O\]

The mole fraction of ethanol is thus

\[ X_{EtOH}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\]

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

\[ M_{EtOH}=\left(\dfrac{0.686\; mol}{100\; \cancel{mL}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=6.86 \;M\]

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:

\[ m_{EtOH}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m \]

Exercise 7

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.

  1. mass percentage
  2. mole fraction
  3. molarity
  4. molality

Answer:

  1. mass percentage toluene = 24.8%
  2. \(X_{toluene} = 0.219\)
  3. 2.35 M toluene
  4. 3.59 m toluene

Summary

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.

Key Takeaway

  • Different units are used to express the concentrations of a solution depending on the application.

Key Equations

molality

\[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \tag{13.5}\]

mass percentage

\[\text{mass percentage}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \tag{13.6}\]

parts per million

\[\text{parts per million (ppm)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \tag{13.7}\]

parts per billion

\[\text{parts per billion (ppb)}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \tag{13.8}\]

Conceptual Problems

  1. Does the molality have the same numerical value as the molarity for a highly concentrated aqueous solution of fructose (\(C_6H_{12}O_6\)) (approximately 3.2 M)? Why or why not?
  2. Explain why the molality and molarity of an aqueous solution are not always numerically identical. Will the difference between the two be greater for a dilute or a concentrated solution? Explain your answer.
  3. Under what conditions are molality and molarity likely to be equal? Is the difference between the two greater when water is the solvent or when the solvent is not water? Why?
  4. What is the key difference between using mole fraction or molality versus molarity to describe the concentration of a solution? Which unit(s) of concentration is most appropriate for experiments that must be carried out at several different temperatures?
  5. An experiment that relies on very strict control of the reaction stoichiometry calls for adding 50.0 mL of a 0.95 M solution of A to 225 mL of a 1.01 M solution of B, followed by heating for 1 h at 60°C. To save time, a student decided to heat solution B to 60°C before measuring out 225 mL of solution B, transferring it to the flask containing solution A, and proceeding normally. This change in procedure caused the yield of product to be less than usual. How could such an apparently minor change in procedure have resulted in a decrease in the yield?

Numerical Problems

  1. Complete the following table for aqueous solutions of the compounds given.
Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X)
\(H_2SO_4\) 18.0 1.84  
\(CH_3COOH\)   1.00 \(7.21 \times 10^{−3}\)
\(KOH\) 3.60 1.16  
  1. Complete the following table for each compound given.
Compound Mass (g) Volume of Solution (mL) Molarity (M)
\(Na_2SO_4\) 7.80 225  
\(KNO_3\)   125 1.27
\(NaO_2CCH_3\) 18.64   0.95
  1. How would you prepare 100.0 mL of an aqueous solution with 0.40 M KI? a solution with 0.65 M \(NaCN\)?
  2. Calculate the molality of a solution with 775 mg of \(NaCl\) in 500.0 g of water. Do you expect the molarity to be the same as the molality? Why or why not?
  3. What is the molarity of each solution?
  1. 12.8 g of glucose (\(C_6H_{12}O_{6}\)) in water, total volume 150.0 mL
  2. 9.2 g of \(Na_3PO_4\) in water, total volume 200.0 mL
  3. 843 mg of I2 in \(EtOH\), total volume 150.0 mL
  1. A medication used to treat abnormal heart rhythms is labeled “Procainamide 0.5 g/250 cc.” Express this concentration in parts per thousand.
  2. Meperidine is a medication used for pain relief. A bottle of meperidine is labeled as 50 mg/mL. Express this concentration in parts per thousand.
  3. An aqueous solution that is 4.61% NaOH by mass has a density of 1.06 g/mL. Calculate the molarity of the solution, the mole fraction of \(NaOH\), and the molality of the solution.
  4. A solution of concentrated phosphoric acid contains 85.0% \(H_3PO_4\) by mass and has a density of 1.684 g/mL. Calculate the following.
  1. the molarity of the solution
  2. the mole fraction of \(H_3PO_4\)
  3. the molality of the solution
  1. A solution of commercial concentrated nitric acid is 16 M HNO3 and has a density of 1.42 g/mL. What is the percentage of HNO3 in the solution by mass? What is the molality?
  2. A commercial aqueous ammonia solution contains 28.0% NH3 by massand has a density of 0.899 g/mL. Calculate the following.
    1. the molarity
    2. the mole fraction
  3. Concentrated, or glacial, acetic acid is pure acetic acid and has a density of 1.053 g/mL. It is widely used in organic syntheses, in the manufacture of rayon and plastics, as a preservative in foods, and occasionally to treat warts. What volume of glacial acetic acid is required to prepare 5.00 L of a 1.75 M solution of acetic acid in ethanol?
  4. Solutions of sodium carbonate decahydrate, also known as washing soda, are used as skin cleansers. The solubility of this compound in cold water is 21.52 g/100 mL. If a saturated solution has a density of 1.20 g/mL, what is its molarity? What is the mole fraction of sodium carbonate decahydrate in this solution?
  5. Hydrogen peroxide (\(H_2O_2\)) is usually sold over the counter as an aqueous solution that is 3% by mass. Assuming a solution density of 1.01 g/mL, what is the molarity of hydrogen peroxide? What is the molar concentration of a solution that is 30% hydrogen peroxide by mass (density = 1.112 g/mL)? How would you prepare 100.0 mL of a 3% solution from the 30% solution?
  6. Determine the concentration of a solution with 825 mg of \(Na_2HPO_4\) dissolved in 450.0 mL of \(H_2O\) at 20°C in molarity, molality, mole fraction, and parts per million. Assume that the density of the solution is the same as that of water. Which unit of concentration is most convenient for calculating vapor pressure changes? Why?
  7. How many moles of \(Cl^−\) are there in 25.0 mL of a 0.15 M \(CaCl_2\) solution?
  8. How many moles of Na+ are there in 25.0 g of a 1.33 × 10−3 m Na2HPO4 solution? What is the sodium concentration of this solution in ppb?
  9. How many grams of copper are there in 30.0 mL of a 0.100 M \(CuSO_4\) solution?
  10. How many grams of nitrate ion are there in 75.0 g of a \(1.75 \times 10^{−4} m\) \(Pb(NO_3)_2\) solution? What is the nitrate concentration of the solution in ppb?
  11. How many milliliters of a 0.750 M solution of \(K_2CrO_4\) are required to deliver 250 mg of chromate ion?
  12. How many milliliters of a \(1.95 \times 10^{−6} M\) solution of \(Ag_3PO_4\) are required to deliver 14.0 mg of \(Ag^+\)?
  13. Iron reacts with bromine according to the following equation:\[2Fe_{(s)} + 3Br_{2(aq)} \rightarrow 2FeBr_{3(aq)}\]
  14. How many milliliters of a \(5.0 \times 10^{−2} M\) solution of bromine in water are required to react completely with 750.0 mg of iron metal?
  15. Aluminum reacts with HCl according to the following equation:\[2Al_{(s)} + 6HCl_{(aq)} \rightarrow 2AlCl_{3(aq)} + 3H_{2(g)}\]
  16. If 25.0 mL of a solution of HCl in water is required to react completely with 1.05 g of aluminum metal, what is the molarity of the HCl solution?
  17. The precipitation of silver chloride is a diagnostic test for the presence of chloride ion. If 25.0 mL of 0.175 M \(AgNO_3\) are required to completely precipitate the chloride ions from 10.0 mL of an \(NaCl\) solution, what was the original concentration of \(NaCl\)?
  18. Barium sulfate is virtually insoluble. If a 10.0 mL solution of 0.333 M \(Ba(NO_3)_2\) is stirred with 40.0 mL of a 0.100 M \(Na_2SO_4\), how many grams of barium sulfate will precipitate? Which reactant is present in excess? What is its final concentration?

Numerical Answers

1.

Compound Molarity (M) Solution Density (g/mL) Mole Fraction (X)
\(H_2SO_4\) 18.0 1.84 0.82
\(CH_3COOH\) 0.393 1.00 7.21 × 10−3
\(KOH\) 3.60 1.16 6.33 × 10 −2

3. 100.0 ml of 0.40 M KI: dissolve 6.64 g of KI in enough water to make 100.0 mL of solution; 100.0 ml of 0.65 M NaCN: dissolve 3.18 g of NaCN in enough water to make 100.0 mL of solution.

5.

  1. 0.474 M glucose
  2. 0.28 M Na3PO4
  3. 0.0221 M I2

9.

  1. 14.6 M
  2. X = 0.510
  3. 57.7 m

11.

  1. 14.8 M
  2. X = 0.292

13. The molarity is 0.745 M, and the mole fraction is 0.0134.

15. The molarity is 0.0129 M, the molality is 0.0129 m, the mole fraction is \(2.33 \times 10^{-4}, and the solution contains 1830 ppm \(Na_2HPO_4\). Mole fraction is most useful for calculating vapor pressure, because Raoult’s law states that the vapor pressure of a solution containing a non-volatile solute is equal to the mole fraction of solvent times the vapor pressure of the pure solvent. The mole fraction of the solvent is just one minus the mole fraction of solute.

17. \(6.65 \times 10^{-5} mol sodium; \(6.14 \times 10^4\; ppb\)

19. \(1.63 \times 10^{-3} g; \(2.17 \times 10^4\) ppb

21. \(2.22 \times 10^4\; mL\) or 22.2 L

23. 4.68 M HCl

25 0.777 g \(BaSO_4\); \(Na_2SO_4\); 0.0134 M \(Na_2SO_4\)

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