THE UNIVERSITY OF THE FREE STATE
QWAQWA CAMPUSES
CEM 214
DEPARTMENT: CHEMISTRY
CONTACT NUMBER: 058 718 5317 / 084 704 4565
ASSESSOR(S):
M A Malimabe
MODERATOR(S):
DURATION : 70 min
MARKS : 50
ANSWER ALL QUESTIONS AND SHOW ALL CALCULATIONS (UNLESS SPECIFIED
OTHERWISE) WITH AN INK PEN -- NO PENCILS OR "TIPPEX" ALLOWED / PAPER
CONSISTS OF 2 PAGES AND 3 QUESTIONS / POCKET CALCULATORS ALLOWED.
Question 1(10)
1.1 Exactly one dm-3 of a 0.100 M solutin of a substance A is added to 3.00 liters of a 0.05
M solution of a substance B. Assume ideal behaviour and culculate the entropy of mixing.(5)
1.2 Calculate the entropy change when 1 mol of ice is heated from 250 K to 300 K. Using
the heat capacities of ice, and water to be respectively 37.7 anf 75.3 J K-1 mol-1 and the
latent heat of fusion of ice is 6.02 kJ mol-1.
(5)
Question 2(20)
2.1 Derive the equation that can be used to determine Λ, Ka and α, for weak electrolytes,
utilizing equivalent conductivity measurements.
(6)
2.2 Draw and explain a typical graph of conductivity vs. base added for the conductometric
titration of hydrochloric acid HCl with sodium hydroxide (NaOH).
(4)
2.3 Calculate the equivalent and molar conductance of aqueous BaSO 4 solution at infinite
dilution. Given
o1 / 2 Ba( NO3 ) 2  135.04 x104  1m 2equiv 1
o1 / 2 H 2 SO4  429.60 x10 4  1m 2equiv 1
oHNO3  421.21x10 4  1m 2equiv 1
(3)
2.4 The molar conductivities, Ls/C, of KCl in H2O (25 oC) at different concentrations are
104 / mol dm-3
Ls 106/ S cm-1
1.566
1.788
2.600
2.418
6.219
4.009
10.441
5.336
Determine 0 and m for KCl.
(7)
Question 3 (20)
1.1 Six-tenths, 0.6 mol of Mg and 0.40 mol of Zn are heated to 650 oC, Mg and Zn melt at
640 and 420 oC
respectively. They form one compound MgZn2 which congruently melts
at 590 oC to give a solid and melt containing 38 mol % Mg. The eutectic point of this
compound with Mg is at 380 oC and 15 mol % Mg. Finally, at 320 oC, when the liquid is 75
mol % in Mg and 25 mol % in Zn, the whole solution freezes, and solid MgZn 2 and solid Mg
come out together. From this temperature down there is no further change in the phases. At
all temperatures below 320 oC there are pure solids Mg and MgZn2.
1.1.1 Sketch the phase diagram and label it.
(10)
1.1.2 Draw cooling curves for a mixture containing 60 mol % of Mg.
(4)
1.2
Phase diagram of sulphur system is shown in the diagram below:
1.2.1Discuss the phase diagram of Sulpher system
(4)
1.3
(2)
Explain the following concepts
1.3.1 Allotropes
1.3.2 Cooling curve
“To be in a compony of a wise ma is like going into a perfumery
You may not buy a thing, but the sweet scent will cling to you for a day” –Abot de Nathan-
MEMORUNDUM
Question 1
Question 2
For weak electrolytes  

according to Eq. 20.
o
C 2
1 
In paragraph 3 we found Ka =
(Eq. 8)
Substitution of Eq. 20 into Eq. 8 gives us
2
Ka

  
C o 
 
 

1 o 
 


2 C
o o  



2C
 
o 2
(22)
 o 
By rearranging Eq. 22 to make C the subject of the formula:


Ka[(o)2 - o]
=
2C
2C = Ka (o)2 - Ka o 
1
(C) = [Ka(o)2]
- Kao


y =
(m)
x +
c
Eq. 23 is the equation of a straight line with y = C; m = Ka(o)2 and c = -Kao.
From the plot of C against
1

, we can determine m and c.
(23)
 K o 2 
m
slope
= + o

  a
o 
c
y  int ercept

K


a

With o, slope and intercept known, we can solve for Ka by substituting the answer for o into
preferably m = slope = Ka (o)2.
2.2 Initially we will have only HCl which, being a strong acid, will dissociate completely in water to
give a large value of L . Reason: There are many H+ ions from the dissociation of HCl according to
the reaction
HCl  H   Cl 
Although Cl  ions also carry charge and contribute to L, H+ ions contribute much more because it
moves much faster in solution than Cl  ions. As we add NaOH, which also dissociates completely
according to the reaction
NaOH  Na  OH
we find that we are replacing the fast moving, good conducting H  cations with slow moving and less
good conducting Na  cations, because the H+ reacts with OH- to give neutral non-conducting H2O.
As a result the conductance, L, will decrease.
A graph of L (Y-axis) against volume NaOH added for this process is shown below, left hand side of
the graph.
Fig. 5 Conductometric titration of HCl with NaOH
2.3 o1 / 2 Ba( NO3 )2  oHNO3l  o1 / 2 BaSO4 = 135.04 x10-4 S m2 eq-1 – 421.21 x10-4 S m2 eq-1 + 429.60 x10-4 S m2 eq-1

 

 o1 / 2 Ba  oNO   oH   oNO   oH   o1 / 2 SO4  = 143.43 S cm2 eq-1


3
 o1 / 2 Ba o1 / 2 SO4

3
 o1 / 2 BaSO4
Question 3
Phase Diagram of Sulphur (One component system)
= 143.43 S cm2 eq-1
= 143.43 S cm2 eq-1
1.2: Equilibrium lines: OP; PK Sublimation, KU Evaporation, KR; RW Melting, PR polymorphism
Triple points: P, K, R
Critical point: U
Phases: Rhombic , Monoclinic sulphur (s),
1.3.1: Allotrope: crystalline forms of an element (e.g., sulphur)
A cooling curve is a line graph that represents the change of phase of matter, typically from a gas to a
solid or a liquid to a solid. The independent variable (X-axis) is time and the dependent variable (Yaxis) is temperature.