1. No category

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(MnO4 ) V(MnO4 )
n(Fe2+)
n(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, 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 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 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 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.
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 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