Chapter 11 Worked Example 1 (a) Calculate the

Chapter 11 Worked Example 1 (a) Calculate the

Chapter 11 Worked Example 1 (a) Calculate the mass of FeSO4.7H2O 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 KMnO4 solution. If the balanced ionic equation is 5Fe2+ + MnO4- + 8H+ 5Fe3+ + Mn2+ + 4H2O (all aq), Calculate the molarity of the KMnO4 solution. [molar masses (g mol-1): Fe = 55.845, S = 32.065, O = 15.999, H = 1.008] Solutions (a) Molar mass of FeSO4.7H2O = 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 (b) Using the reaction stoichiometry, M(Fe2+)V(Fe2+) = M(MnO 4 ) V(MnO4 ) n(Fe2+) n(MnO4-) M(MnO 4-) = 0.020 M x 28.14 mL x 1 5 x 25.00 mL = 4.502 x 10-3 M Chapter 11 Worked Example 2 A solution of hydrochloric acid in water is 38.00% HCl by mass.

Its density is 1.1886 g/cm3 at 20 oC. Compute its molarity, mole fraction, and molality at this temperature. Molar masses (g/mol): H = 1.00794; Cl = 35.45; O = 15.9993 Solution Exactly 100 g of solution contain 38.00 g HCl and 62.00 g water 1 mL soln Its volume is100 g x = 84.133 mL 1.1886 g soln 1 mol 1 mol No. Moles HCl = 38.00 g HCl No. moles water = 62.00 g water 36.461 g 18.015 = 1.0422 = 3.4416 Molarity of HCl =

1.0422 mol 84.133 mL 1000 mL 1L Molality of HCl = 1.0422 mol 1000 g 62.00 g 1kg 1.0422 mol Mole fraction of HCl 1.0422 + 3.4415 mol = 12.39 mol/L = 16.81 mol/kg = 0.2324

Chapter 11 Worked Example 3 At 27oC, 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 nhexane 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. Solution Firstly, XB = XH = 0.5 Hence from Raoults Law, PB = 0.5 x 0.1355 = 0.068 atm PH = 0.5 x 0.2128 = 0.106 atm and PTOTAL = 0.174 atm For vapor, pB = XB x PTOTAL, or 0.068 = XB x 0.174 Hence, XB = 0.389 Since the mole fraction of benzene is lower in the vapor, hexane must be the more volatile component.

Chapter 11 Worked Example 4 1. Complete and balance the equation for reaction in acidic solution: VO2+(aq) + SO2(aq) VO2+(aq) + SO42- (aq) 2. Complete and balance the equation for reaction in basic solution: ZrO(OH)2(s) + SO32-(aq) Zr(s) + SO42- (aq) Solutions 1. Oxidation half equation: SO2(g) +2H2O(l) SO42-(aq) + 4H+(aq) + 2eReduction half reaction: VO2+(aq) + 2H+(aq) + eVO2+(aq) + H2O(l) Multiply the 2nd half equation by 2 and add, gives 2VO2+(aq) + SO2(g)

2 VO2+(aq) + SO42-(aq) 2. Oxidation half reaction: SO32-(aq) + 2OH-(aq) SO42-(aq) + H2O(l) + 2e- Reduction half reaction: ZrO(OH)2(s) + H2O(l) + 4e- Zr(s) + 4OH-(aq) Multiply the top half equation by 2 and add, gives 2SO32-(aq) + ZrO(OH)2(s) 2SO42-(aq) + H2O(l) 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. Solution A is of pure water B is of pure HClC 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. Chapter 11 Worked Example 6 Henrys Law constant for CO2 dissolved in water is 1.65 x 103 Atm. If a carbonated drink is bottled under a CO2 pressure of 5.0 atm:

(a) Calculate the molar concentration of CO2 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 (a) pCO2 = 5.0 atm in the closed bottled = k(CO2)X(CO2) (Henry's Law) Hence 5.0 atm = 1.65 x 103 atm x X(CO2) X(CO 2) = 0.0030 (and X(H2O) = 0.9970) 0.0030 mol of CO2 per 0.9970 mol H2O Since density of drink = 1.00 g cm -3, 1.00 L of solution contains 55.5 mol H2O (1000 g/18.00 g/mol) Hence molar concentration of CO 2 = = 0.17 mol L-1 (0.17 M) 0.0030 mol CO2

x 55.5 mol H2O 0.9970 mol H2O 1.00 L soln (b) In the closed bottle the CO2 at 5.0 atm pressure in the small space above the liquid is in dynamic equilibrium with the dissolved CO2. When the bottle is opened, the pressure becomes 1 atm, CO2 escapes because the partial pressure of CO2 in the atmosphere is far lower than 1 atm, thus equilibrium is eventually established with a much lower concentration of CO2 in solution. 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

Solution 1 c 2 a 3 d 4 b 5 e

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