Exercise practice: Solutions

6.1 How the exercise set is arranged

The chapter has 41 exercises numbered 1.1 to 1.41, along with eleven intext questions that are answered inside the chapter. The exercises fall into four groups. Concentration and units covers 1.1 to 1.9, with 1.26 and 1.28 asking for molarity and mass percentage in other contexts. Solubility and Henry's law covers 1.10 to 1.13, with 1.35 and 1.39 giving the gas calculations. Raoult's law and non-ideal mixtures covers 1.14 to 1.19 and 1.34 to 1.38. Colligative properties cover the rest, mostly 1.20 to 1.25 and 1.29 to 1.33, 1.40 and 1.41. The worked items below are the ones whose method transfers to the other questions in the same group.

6.2 Group 1: concentration and units

Exercise 1.4. Concentrated nitric acid is 68 percent by mass with density 1.504 g mL^-1. One litre has mass 1504 g, acid mass 1022.7 g and acid amount 16.2 mol, so the molarity is 16.2 M. The molality follows from the 481.3 g of water, giving 33.7 mol kg^-1.

Exercise 1.5. For 10 percent w/w glucose, molality is 0.617 m, mole fractions are 0.0110 for glucose and 0.9890 for water, and the molarity with density 1.2 g mL^-1 is 0.667 M. The three answers come from the same 100 g sample, which is the efficient way to do this type.

Exercise 1.7. Mixing 300 g of 25 percent solution with 400 g of 40 percent solution gives 75 plus 160, or 235 g of solute in 700 g of solution. The mass percentage of the mixture is 33.6 percent. Mass percentages of mixed solutions are weighted averages, so they cannot be averaged without weights.

Exercise 1.9. A chloroform level of 15 ppm by mass is 15 g of chloroform per 10^6 g of solution. That is 1.5 times 10^-3 percent. The solvent mass is close to 1000 kg, so the molality is about 1.26 times 10^-4 mol kg^-1. Notice how small the molality is: ppm solutions are extremely dilute.

6.3 Group 2: solubility and Henry's law

Exercise 1.35. The Henry constant for methane in benzene is 4.27 times 10^5 mm Hg at 298 K. At 760 mm Hg, the mole fraction of methane is 760 divided by 4.27 times 10^5, which is 1.78 times 10^-3. In the mixture questions such as 1.39, the partial pressure of each gas is found from its mole fraction in air before Henry's law is applied.

Exercise 1.12 asks for the statement of Henry's law and its applications, which should be answered with the equation, the definition of K_H, and two uses such as scuba air and carbonated drinks. Exercise 1.13 asks why the solubility of a gas decreases with temperature; the answer follows from the exothermic nature of dissolution.

6.4 Group 3: Raoult's law and non-ideal mixtures

Exercise 1.17. The vapour pressure of water at 300 K is 12.3 kPa. A one molal solution contains 1 mol of solute in 55.5 mol of water, so the water mole fraction is 0.9823. The vapour pressure of the solution is 12.3 times 0.9823, which is 12.08 kPa. The lowering is small because the solution is dilute.

Exercise 1.18 follows the same route but works backwards from a 20 percent lowering to the mass of solute. In 1.19 the vapour pressure after adding more water is used to find the molar mass of the solute from the two states. Exercise 1.38 on benzene and toluene is a two volatile liquid problem: find the mole fraction of each, multiply by its pure vapour pressure and add the partial pressures.

6.5 Group 4: colligative properties

Exercise 1.20. A 5 percent cane sugar solution freezes at 271 K, which is a depression of 2.15 K from 273.15 K. Glucose has a smaller molar mass, 180 against 342, so the same mass percentage gives a molality larger by the ratio 342 divided by 180, which is 1.9. The depression for the glucose solution is 2.15 times 1.9, which is 4.09 K, and its freezing point is therefore 269.1 K.

Exercise 1.22. A litre of solution with 36 g of glucose has an osmotic pressure of 4.98 bar. For 1.52 bar at the same temperature the concentration must be 0.2 times 1.52 divided by 4.98, which is 0.061 mol L^-1, or about 11 g of glucose per litre.

Exercise 1.32. For 10 g of CH3CH2CHClCOOH of molar mass 122.5 in 250 g of water, the molality is 0.327. The acid is weak, so the degree of dissociation is the square root of Ka divided by concentration, which is 0.066, and i is 1.066. The depression is 1.066 times 1.86 times 0.327, which is 0.65 K.

Exercise 1.40. Calcium chloride gives three ions per formula unit and has i equal to 2.47 at the stated concentration. From the osmotic pressure, the concentration is 0.75 divided by the product of 2.47, 0.0821 and 300, which is 0.0123 mol L^-1 in 2.5 L. That is 0.0308 mol, and with a molar mass of 111 g mol^-1 the mass is about 3.4 g.

Exercise 1.41. Potassium sulphate dissociates completely into three ions. The 25 mg in 2 L is 7.18 times 10^-5 mol L^-1, so the osmotic pressure is 3 times 7.18 times 10^-5 times 0.0821 times 298, which is 5.3 times 10^-3 bar. The answer is small because the amount of solute is small.