(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] 

(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 sulfateHence, 250 mL requires 5.560/4 = 1.390 g

Solutions

Chapter 11 Worked Example 1

(b) Using the reaction stoichiometry,

M(Fe2+)V(Fe2+)

n(Fe2+) n(MnO4-)

M(MnO4-) V(MnO4

-)=

M(MnO4-) =

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, molefraction, and molality at this temperature.Molar masses (g/mol): H = 1.00794; Cl = 35.45; O = 15.9993

Exactly 100 g of solution contain 38.00 g HCl and 62.00 g water

Its volume is100 g x1 mL soln

1.1886 g soln= 84.133 mL

No. Moles HCl = 38.00 g HCl1 mol

36.461 gNo. moles water = 62.00 g water

=

1 mol

18.0151.0422 = 3.4416

Molarity of HCl =1.0422 mol

84.133 mL

1000 mL

1 L= 12.39 mol/L

Molality of HCl = 1.0422 mol

62.00 g

1000 g

1kg= 16.81 mol/kg

Mole fraction of HCl 1.0422 mol1.0422 + 3.4415 mol

= 0.2324

Solution

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 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, XB = XH = 0.5

Hence from Raoult’s 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, p’B = X’B x PTOTAL, 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

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 inbasic solution: ZrO(OH)2(s) + SO3

2-(aq) Zr(s) + SO42- (aq)

Solutions

1. Oxidation half equation: SO2(g) +2H2O(l) SO4

2-(aq) + 4H+(aq) + 2e-

Reduction half reaction:VO2

+(aq) + 2H+(aq) + e- VO2+(aq) + H2O(l)

Multiply the 2nd half equation by 2 and add, gives

2VO2+(aq) + SO2(g) 2 VO2+(aq) + SO4

2-(aq)

2. Oxidation half reaction:

SO32-(aq) + 2OH-(aq) SO4

2-(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) 2SO4

2-(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.

A is B.pt. of pure water B is B.pt. of pure HCl…C is azeotropic mixture……If solution of composition X is distilledHCl 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

Chapter 11 Worked Example 6

Henry’s 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 theseconditions, using 1.00 g cm-3 as the density of the drink.(b) Explain what happens on a microscopic level when the bottleis opened.

SolutionpCO2 = 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(CO2) = 0.0030 (and X(H2O) = 0.9970)

0.0030 mol of CO2 per 0.9970 mol H2OSince 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 CO2 = 0.0030 mol CO2

0.9970 mol H2Ox 55.5 mol H2O

1.00 L soln

= 0.17 mol L-1 (0.17 M)

(a)

(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 foam1. milk 2. sodium chloride in cell fluid 3. hemoglobin in blood 4. smoke 5. meringue 

1 2 3 4 5c a d b e  

 

Solution