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:

A 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.

B 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:

1. 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:

   $${\text{moles CH}}_3{\text{CO}}_2\text{H}=\frac{3.78\cancel{\text{g}}{\text{ CH}}_3{\text{CO}}_2\text{H}}{60.05\cancel{\text{g}}\text{/mol}}=0.0629\text{ mol}$$ $$\text{volume}=\frac{\text{mass}}{\text{density}}=\frac{100.0\cancel{\text{g}}\text{ solution}}{1.00\cancel{\text{g}}\text{/mL}}=100\text{ mL}$$ $${\text{molarity of CH}}_3{\text{CO}}_2\text{H}=\frac{{\text{moles CH}}_3{\text{CO}}_2\text{H}}{\text{liter solution}}=\frac{0.0629{\text{ mol CH}}_3{\text{CO}}_2\text{H}}{(100\cancel{\text{mL}})(1\text{ L/1000 }\cancel{\text{mL}}\text{)}}=0.629{\text{ M CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}$$

   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.

2. 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

   $${\text{moles H}}_2\text{O}=\frac{96.2\cancel{\text{g}}{\text{ H}}_2\text{O}}{18.02\cancel{\text{g}}\text{/mol}}=5.34{\text{ mol H}}_2\text{O}$$

   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:

   $$X_{{\text{CH}}_3{\text{CO}}_2\text{H}}=\frac{\text{moles }{\text{CH}}_3{\text{CO}}_2\text{H}}{\text{moles }{\text{CH}}_3{\text{CO}}_2\text{H }+\text{ moles }{\text{H}}_2\text{O}}=\frac{0.0629\text{ mol}}{0.0629\text{ mol}+5.\text{34 mol}}=0.0116=1.16\times {10}^{-2}$$

   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

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. X_HCl = 0.111

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:

A Use the concentration of the solute in parts per million to calculate the molarity.

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

Solution:

1. 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

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

2. 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

   $$\text{mass of benzene}=\frac{(12.7\text{ mg benzene})(250\cancel{\text{mL}})}{1000\cancel{\text{mL}}}=3.18\text{ mg}=3.18\times {10}^{-3}\text{ g benzene}$$

Exercise

The maximum allowable concentration of lead in drinking water is 9.0 ppb. What is the molarity of Pb²⁺ in a 9.0 ppb aqueous solution? Use your calculated concentration to determine how many grams of Pb²⁺ are in an 8 oz glass of water.

Answer: 4.3 × 10⁻⁸ M; 2 × 10⁻⁶ g

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:

A 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.

B Determine the mass percentage of solute by dividing the mass of ethanol by the mass of the solution and multiplying by 100.

C 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.

D 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 (CH₃CH₂OH), 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:

$$\text{mass of EtOH}=\left(40.0\cancel{\text{mL}}\right)\left(\frac{0.789\text{ g}}{\cancel{\text{mL}}}\right)=31.6\text{ g EtOH}$$

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.

1. 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:

   $$\%\text{EtOH}=\left(\frac{\text{mass of EtOH}}{\text{mass of solution}}\right)(100)=\left(\frac{31.6\cancel{\text{g}}\text{ EtOH}}{31.6\cancel{\text{g}}\text{ EtOH}+60.0\cancel{\text{g}}{{\rm H}}_2\text{O}}\right)(100)=34.5\%$$

2. 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:

   $$\text{moles EtOH}=(31.6\cancel{\text{g EtOH}})\left(\frac{1\text{ mol}}{46.07\cancel{\text{g EtOH}}}\right)=0.686{\text{ mol CH}}_3{\text{CH}}_2\text{OH}$$

   Similarly, the number of moles of water is

   $${\text{moles H}}_2{\rm O}=(60.0\cancel{{\text{g H}}_2\text{O}})\left(\frac{1{\text{ mol H}}_2\text{O}}{18.02\cancel{{\text{g H}}_2\text{O}}}\right)=3.33{\text{ mol H}}_2\text{O}$$

   The mole fraction of ethanol is thus

   $$X_{\text{EtOH}}=\frac{0.686\cancel{\text{mol}}}{0.686\cancel{\text{mol}}+3.33\cancel{\text{mol}}}=0.171$$

3. 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

   $${\text{M}}_{\text{EtOH}}=\left(\frac{0.686\text{ mol}}{100\cancel{\text{mL}}}\right)\left(\frac{1000\cancel{\text{mL}}}{\text{L}}\right)=6.86\text{ M}$$

4. 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:

   $$m_{\text{EtOH}}=\left(\frac{0.686\text{ mol EtOH}}{60.0\cancel{\text{g}}{\text{ H}}_2\text{O}}\right)\left(\frac{1000\cancel{\text{g}}}{\text{kg}}\right)=\frac{11.4\text{ mol EtOH}}{{\text{kg H}}_2\text{O}}=11.4m$$

Exercise

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