1. (a) Calculate the mass of FeSO 4.7H 2 O that is needed to make 250.0 mL of a 0.02000 molar solution in water. Show all working. (b) By titration, 28.14 mL of the solution above is equivalent to 25.00 mL of an acidified KMnO 4 solution. If the balanced ionic equation is 5Fe 2+ + MnO 4 - + 8H + 5Fe 3+ + Mn 2+ + 4H 2 O (all aq), Calculate the molarity of the KMnO 4 solution. [molar masses (g mol -1 ): Fe = 55.845, S = 32.065, O = 15.999, H = 1.008] (a) Molar mass of FeSO 4.7H 2 O = 278.011 g/mol 1L of 0.020 M solution requires 278.011 x 0.020 = 5.560 g of ferrous sulfate Hence, 250 mL requires 5.560/4 = 1.390 g Solutions Chapter 11 Worked Example 1

3. Chapter 11 Worked Example 2 A solution of hydrochloric acid in water is 38.00% HCl by mass. Its density is 1.1886 g/cm 3 at 20 o C. Compute its molarity, mole fraction, and molality at this temperature. Molar masses (g/mol): H = 1.00794; Cl = 35.45; O = 15.9993 Solution

4. At 27 o C, the vapor pressure of pure benzene is 0.1355 atm and the vapor pressure of pure n-hexane is 0.2128 atm. If equal amounts (equal number of moles) of benzene and n- hexane are mixed to form an ideal solution, calculate the mole fraction of benzene in the vapor at equilibrium with the solution. Show working. Explain which (benzene or n-hexane) Is the more volatile component. Firstly, X B = X H = 0.5 Hence from Raoult’s Law, P B = 0.5 x 0.1355 = 0.068 atm P H = 0.5 x 0.2128 = 0.106 atm and P TOTAL = 0.174 atm For vapor, p’ B = X’ B x P TOTAL, or 0.068 = X’ B x 0.174 Hence, X’ B = 0.389 Since the mole fraction of benzene is lower in the vapor, hexane must be the more volatile component. Chapter 11 Worked Example 3 Solution

5. Chapter 11 Worked Example 4 1.Complete and balance the equation for reaction in acidic solution: VO 2 + (aq) + SO 2 (aq) VO 2+ (aq) + SO 4 2- (aq) 2. Complete and balance the equation for reaction in basic solution: ZrO(OH) 2 (s) + SO 3 2- (aq) Zr(s) + SO 4 2- (aq) Solutions 1.Oxidation half equation: SO 2 (g) +2H 2 O(l) SO 4 2- (aq) + 4H + (aq) + 2e - Reduction half reaction: VO 2 + (aq) + 2H + (aq) + e - VO 2+ (aq) + H 2 O(l) Multiply the 2 nd half equation by 2 and add, gives 2VO 2 + (aq) + SO 2 (g) 2 VO 2+ (aq) + SO 4 2- (aq)

6. 2. Oxidation half reaction: SO 3 2- (aq) + 2OH - (aq) SO 4 2- (aq) + H 2 O(l) + 2e - Reduction half reaction: ZrO(OH) 2 (s) + H 2 O(l) + 4e - Zr(s) + 4OH - (aq) Multiply the top half equation by 2 and add, gives 2SO 3 2- (aq) + ZrO(OH) 2 (s) 2SO 4 2- (aq) + H 2 O(l)

7. Chapter 11 Worked Example 5 The following is temperature-composition diagram for the distillation of a hydrogen chloride/water solution. Identify points A, B and C on the diagram. Explain what would happen if a solution of composition X is distilled.

8. A is B.pt. of pure water B is B.pt. of pure HCl…C is azeotropic mixture…… If solution of composition X is distilled HCl will distil first, until composition of liquid in flask reaches that of the azeotropic mixture. Then, the azeotrope distils until the flask is empty. Solution

9. Chapter 11 Worked Example 6 Henry’s Law constant for CO 2 dissolved in water is 1.65 x 10 3 Atm. If a carbonated drink is bottled under a CO 2 pressure of 5.0 atm: (a)Calculate the molar concentration of CO 2 in water under these conditions, using 1.00 g cm -3 as the density of the drink. (b) Explain what happens on a microscopic level when the bottle is opened. Solution

10. (b)In the closed bottle the CO 2 at 5.0 atm pressure in the small space above the liquid is in dynamic equilibrium with the dissolved CO 2. When the bottle is opened, the pressure becomes 1 atm, CO 2 escapes because the partial pressure of CO 2 in the atmosphere is far lower than 1 atm, thus equilibrium is eventually established with a much lower concentration of CO 2 in solution.

11. Chapter 11 Worked Example 7 15.Classify each of 1 – 5 as (a) a true solution (b) an aerosol (c) an emulsion (d) a sol (e) a foam 1. milk 2. sodium chloride in cell fluid 3. hemoglobin in blood 4. smoke 5. meringue 1 2 3 4 5 c a d b e Solution