Point
General
Final test 2009.12.03.
Chemistry
Name:
Corrected
Max
32p
In 100 g H2O dissolve 0.635 g of FeCl2. The density of the solution is 1.012 g/cm3.
1. The osmotic pressure at 23 °C:
371.4 kPa
n
  RT , M(FeCl2)=56+35.5·2=127 g/mol  n=0.005 mol dissociates to 3 ions!
V
(4p)
V=100.635 g/(1.012 g/cm3)=99.442 cm3, c=0.005 mol/0.099442 dm3=0.05028 mol/ dm3
=3·c·R·T=3·0.05028·8.314·296.15 kPa=371.4 kPa
The molar boiling point increase of the water is 0.52 °C * kg H2O/mol.
Ar(Fe): 56.0; Ar(Cl): 35.5.
T = TM CR, CR is the Roault concentration (also noted by m)
cR=0.005 mol/0.1 kg=0.05 mol/kg H2O. T = 3 · cR · TM = 3 · 0.05 · 0.52 °C = 0.078 °C.
2. The boiling point of the solution: 100.078 °C
(6p)
Next we add H2SO4 and oxidize the FeCl2 completely by KMnO4:
3. 1 MnO4− + 5 Fe2+ + 8 H+
= 1 Mn2+ + 5 Fe3+ + 4 H2O
MnO4− + 5e- + 8 H+= Mn2+ + 4 H2O and Fe3+ + e- = Fe2+ (5 e- change)
4. 2 MnO4− + 10 Cl− + 16 H+
= 2 Mn2+ + 5 Cl2 + 8 H2O
2 MnO4− + 10e- + 16 H+= 2 Mn2+ + 8 H2O and Cl2 + 2e- = 2Cl5. What is the volume of the Cl2 gas at T=25 °C and p=101.41 kPa?:
nRT
, note: if you use p in Pa the V is in m3, if you use p in kPa the V is in dm3
V 
p
n(FeCl2)= 0.005 mol = n(Cl2), R=8.314 J/mol K, T=298 K,
(3p)
(3p)
(4p)
V = 0.1222 dm3
The Cl2 gas reacts with NaI solution (bubbling):
6. 1 I− + 3 Cl2 + 3 H2O
= 1 IO3− + 6 Cl− + 6 H+
(3p)
Cl2 + 2e- = 2Cl- and IO3- + 6e- + 6H+ = I- + 3H2O (6e- change)
7. 1 IO3− + 5 I− + 6 H+
= 3 I2 + 3 H2O
(3p)
IO3- + 5e- + 6H+ = 0.5 I2 + 3H2O and I2 + 2e- = 2I- (5e- change)
8. 2 Na2S2O3 + 1 I2
= 1 Na2S4O6 + 2 NaI
I2 + 2e- = 2I- and S4O62-+2e- =2 S2O32-
(3p)
(2e- change)
9. 6 Br- + 1 Cr2O72- + 7 H2SO4 → 3 Br2 + 2 Cr3+ + 7 SO42- + 7 H2O (3p)
Cr2O72- + 6e- 14 H+= 2 Cr3+ + 7 H2O and Br2 + 2e- = 2Br- (6e- change)
Point
General
Final test 2009.12.03.
Chemistry
Name:
Max
32p
Name the following compounds (1p for each, total 10 points):
10. NO
nitrogen monoxide
11. NO2
nitrogen dioxide
12. N2O
dinitrogen monoxide
13. N2O3
dinitrogen trioxide
14. N2O4
dinitrogen tetroxide
15.NaClO
sodium hypochlorite
16. KClO2
potassium chlorite
17. HClO3
chloric acid
18. HClO4
perchloric acid
19. CaSO3
calcium sulfite
Reaction: N2O4 (g)  2 NO2 (g).
Ar(N): 14; Ar(O): 16.
Volume = 1 dm3, T= 70 °C, the initial mass of the N2O4 is 9.2 g. We wait for the dissociation
equilibrium and in the equilibrium we measure the total pressure of 399.24 kPa
Calculate equilibrium constant with mols (Kn), with concentrations (Kc), and with
pressure (Kp), the degree of dissociation (), the volume %, and the average molar
mass. M(N2O4)=92, M(NO2)=46.
initial: ni(N2O4)=0.100 mol, (9.2 g/92 g/mol)
final (in equilibrium): ntotal=pV/(RT)=0.140 mol
N2O4
Initial
0.1 mol
Reaction
−x mol
Equilibrium
0.1−x mol
ntotal=0.1−x+2x=0.1+x=0.14 mol thus x=0.04 mol.
mol%(N2O4)= 0.06/0.14 *100=42.9% = volume %(N2O4)
# N2O4
NO2
c 0.06 M
0.08 M
n
p
0.06 mol
0.429·399.24=171.1 kPa
2 NO2
0 mol
+ 2x mol
2x mol

K#
0.08 2
Kc 
 0.1067
0.06
0.08 mol
0.571·399.24=228.1 kPa
0.082/0.06=0.1067
(228.1/101.325)2/
(171.1/101.325)=3.003
20. Kn =
0.1067
mol
(4p)
21. Kc =
0.1067
mol/dm3
(4p)
22. Kp =
3.003
(with activities dimensionless)
(4p)
23.  =
40%
24. volume % =
25. average molar mass:
(4p)
42.9 N2O4 (g)
65.73 g/mol
51.7 NO2(g)
(4p)
(2p)
Point
General
Final test 2009.12.03.
Chemistry
Max
Name:
36p
2+
Calculate the following non standard electrode potentials if °(Cu/Cu )= +0.340 V:
26. Metal Cu in 0.05 M CuSO4 solution:
(Cu/Cu2+)= °(Cu/Cu2+)note that Q 
a
0.0591
log  red
2
 a ox
ared
a
1
 Cu2 
aox
Cu
Cu 2

 
(4p)

0.0591V
 = +0.340 V+
log(0.05)= 0.302 V
2


27. Metal Cu in saturated Cu3(PO4)2 solution:
(6p)
Ksp[Cu3(PO4)2]= 1.08 x 10
Ksp=(3S)3·(2S)2=108·S5=1.08·10−13  S=1.00·10−3 M  [Cu2+]=3S=3.0·10−3 M
−13
0.059V
log(3.0·10−3) =
2
28. Metal Cu in saturated Cu(OH)2 solution at pH=9:
(Cu/Cu2+)=+0.340 V+
+0.266 V
(6p)
Ksp[Cu(OH)2]= 5.0·x 10
pH = 9.00  pOH = 5.00  [OH−] = 10−5 Ksp = [Cu2+]·[OH−]2  [Cu2+] = 5·10−18/(10−5)2
= 5.0·10−8 M (3p)
−18
0.059V
log(5.0·10−8) =
2
29. Hydrogen electrode in pH 12.5 NaOH solution:
+0.125 V
(H2/2H+)=°(H2/2H+)+0.059 V·lg[H+]=0.000 V + 0.059 V·(−pH) =
−0.738 V
(Cu/Cu2+)=+0.340 V+
30. Hydrogen electrode in 0.02 M weak acid solution:
Ks(weak acid)=1.0 x 10−4
[H+]=(Ksc0)1/2=(1.00·10−4·0.02)1/2=1.414·10−3 M (4p)  pH=2.85
(H2/2H+)=°(H2/2H+)+0.059 V·lg[H+]=0.000 V + 0.059 V·(−pH)=
(6p)
(8p)
−0.168 V
31. Connect the electrodes in 29. and 30. What is the cell potential? Which electrode
is the cathode and which electrode is the anode?
(6p)
Ecell = -0.168 V –(-0.738 V) =
+0.57 V
Cathode (reduction): 30 (smaller pH, larger H+ concentration) 2H+ 2e-  H2
Anode (oxidation): 29