General
Final test 2009.12.10.
Chemistry
Point
Name:
Max
32p
In 50 g H2O dissolve 0.1235 g of AlCl3. The density of the solution is 1.010 g/cm3.
Ar(Al): 27.0; Ar(Cl): 35.5. R=8.314 J/mol K
1. The osmotic pressure at 10 °C:
(4p)
-4
M(AlCl3)=27+35.5·3=135.5 g/mol  n=9.25 10 mol dissociates to 4 ions!
V=50.1235 g/(1.010 g/cm3)=49.63 cm3, c=9.25 10-4mol/0.04963 dm3=0.01864 mol/ dm3
=4·c·R·T=175.4 kPa
The molar freezing point depression of the water is Kf-= -1.86 °C· kg H2O/mol.
cR=9.25 10-4 mol/0.050 kg H2O. T = 4 · cR · TM = -0.1376 °C
2. The freezing point of the solution: -0.1373 °C
(6p)
3. 3 P + 5
(3p)
HNO3 + 2 H2O = 3 H3PO4 + 5 NO
4. 2 MnO4− + 5 H2O2- + 3 H2SO4 → 5 O2 + 3 SO42- + 8 H2O+ 2 Mn2+ (3p)
O2 + 2e- +2H+ = H2O2
5. What is the volume of the O2 gas at T=100 °C and p=151.4 kPa that develops from
equation 4?:
n(O2) = 5, T=373 K, V 
6. 1 Bi + 6 HNO3
(4p)
nRT
p
V = 102.41 dm3
= 1 Bi(NO3)3 + 3 NO2 + 3 H2O
(3p)
7. 2 ICl3 + 4 H2O= 1 HIO3 + 1 HIO + 6 HCl
(3p)
8. 1 Na2S2O3 + 4 Br2 + 5 H2O= 1 H2SO4 + 1 Na2SO4 + 8 HBr
(3p)
9. 2 MnO4− + 10 I− + 16 H+ = 2 Mn2+ + 5 I2 + 8 H2O
(3p)
Point
General
Final test 2009.12.10.
Chemistry
Name:
Max
32p
Name the following compounds (1p for each, total 10 points):
10. MnO4− permanganate anion
11. Na2S2O3 sodium thiosulphate
12. Bi(NO3)3 bismuth nitrate
13. HIO hypoiodous acic
14. HIO3 iodic acic
15. ICl3 iodine trichloride
16. H3PO4 phosphoric acide
17. H2O2 hydrogen peroxide (dihydrogen dioxide)
18. K2S4O6 potassium tetrathionate
19. CaSO3 calcium sulfite
Reaction: N2O4 (g)  2 NO2 (g).
Ar(N): 14; Ar(O): 16.
Volume = 0.5 dm3, T= 80 °C, the initial mass of the N2O4 is 5.6 g. We wait for the
dissociation equilibrium and in the equilibrium we measure the total pressure of 400 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)=0.06087 mol, (5.6 g/92 g/mol)
final (in equilibrium): ntotal=pV/(RT)= 0.06815 mol
N2O4
2 NO2

Initial
0.06087 mol
0 mol
Reaction
−x mol
+ 2x mol
Equilibrium
0.06087 −x mol
2x mol
ntotal=0.06087 −x+2x = 0.06815 mol thus x=0.00728 mol.
mol%(N2O4)= (0.06087 −0.00728 ) /0.06815 *100=78.64% = volume %(N2O4),
# N2O4
NO2
K#
n 0.05359 mol
0.01456 mol
0.01456 2
Kn 
c
p
0.05359/0.5 M
0.7864·400=314.5 kPa
0.01456/0.5 M
0.2136 400=85.5 kPa
0.05359
 3.956  10 3
2*Kn = Kc = 7.912 10-3
(85.5/101.325)2/
(314.5/101.325)=0.23
20. Kn =
3.956 10-3
(4p)
21. Kc =
7.912 10-3
(4p)
22. Kp = 0.23
(4p)
23.  = x/0.06087 * 100% =11.96 %
(4p)
24. volume % =
78.64 N2O4 (g)
25. average molar mass: 82.17 g/mol
21.36 NO2
(g)
(4p)
(2p)
Point
General
Final test 2009.12.10.
Chemistry
Max
Name:
36p
2+
Calculate the following non standard electrode potentials if °(Cu/Cu )= +0,340 V:
26. Metal Cu in 0.005 M CuSO4 solution:
(Cu/Cu2+)= °(Cu/Cu2+)-
a
0.0591
log  red
2
 a ox
(4p)

0.0591V
 =+0.340 V+
log(0.005)= 0.272 V
2

27. Metal Cu in saturated Cu3(PO4)2 solution:
(6p)
Ksp[Cu3(PO4)2]= 1.08 x 10−13
Ksp=(3S)3·(2S)2=108·S5=1.08·10−13  S=1.00·10−3 M  [Cu2+]=3S=3.0·10−3 M
0.059V
log(3.0·10−3) =
+0.266 V
2
28. Metal Fe in saturated Fe(OH)2 solution at pH=11.0, °(Fe/Fe2+)= -0,440 V: (6p)
(Cu/Cu2+)=+0.340 V+
Ksp[Fe(OH)2]= 5.0·x 10−15
pH = 11.00  pOH = 3.00  [OH−] = 10−3 Ksp = [Fe2+]·[OH−]2  [Fe2+] = 5·10−15/(10−3)2
= 5.0·10−9 M (3p)
0.059V
log(5.0·10−9) =
2
29. Hydrogen electrode in pH 7 solution:
-0.686 V
(H2/2H+)=°(H2/2H+)+0.059 V·lg[H+]=0.000 V + 0.059 V·(−pH) =
−0.414 V
(Fe/Fe2+)=-0.440 V+
(6p)
30. Hydrogen electrode in 0.02 M weak acid solution, the electrode potential is
 = -0.168 V. What is the value of Ks(weak acid):
(8p)
Ks = [H+]2/(c-[H+]) where -0.168 = -0.0592*pH  0.0592 log [H+] = -0.168
 
0.168
[H+] = 10 01 H   10 0.0592 =1.453 10-3
Ks = 1.138 10-4
31. Connect the electrodes in 29. and 30. What is the cell potential? Which electrode
is the cathode and which electrode is the anode? How the cell potential will change
in time as we use the cell?
(6p)
Ecell = E30-E29 = 0.2464 V
Cathode E30 (more positive) reduction is going on
Anode E29 (less positive) oxidation
The cell potential will decrease to zero as the concentrations (activities) equalize in
time (the salt bridge