J . S o p o u š e k ú l o h a
/5/
1.a-c
1.a. Determination of molar mass and dimerization constant of
benzoic acid
Molar mass of some substances determined by cryoscopic method may deviate
from the tabulated molar mass, even though the solution behaves ideally.
Deviation can be caused by dissociation, association, and by others effects that change
the number of moles of constituent entities in the solution. For example, benzoic acid in
the benzene solution partially dimerizes:
22 AA dK
→← (1.1.)
In this case, the cryoscopic method can be used to determine both the molar mass M
and the thermodynamic dimerization constant 𝐾𝐾𝑑𝑑, which can be defined by
( ) ( )2
1
2
*
1
*
2
1 n
mmn
m
m
m
m
a
a
K Rd
d
d
d
⋅⋅
=






== * (1.2.)
where symbols 𝑎𝑎1 (𝑎𝑎𝑑𝑑) and 𝑚𝑚1 (𝑚𝑚𝑑𝑑) are activity and molarity of the solute in the form of
monomer (respectively dimer), 𝑚𝑚∗ is a standard molarity whose value is 1 mol kg-1
. 𝑛𝑛1
and 𝑛𝑛𝑑𝑑 are molar amounts of monomer and dimer in the solution, 𝑚𝑚𝑅𝑅 is the weight of the
solvent (i.e. benzene).
Thus, the difference between the freezing-point point of the benzoic acid solution and
benzene is:
∆ 𝑇𝑇 = 𝐾𝐾𝑓𝑓 ∙
� 𝑛𝑛1+𝑛𝑛𝑑𝑑�
𝑚𝑚𝑅𝑅
(1.3.)
where 𝐾𝐾𝑓𝑓 is the cryoscopic constant of benzene (xxxxxx Table I.).
TASK: Determine molar mass and dimerization constant of benzoic acid in
freezing benzene. Measure the freezing-point depressions for at least three
solutions at concentrations.
LABORATORY AIDS AND CHEMICALS: Apparatus for cryoscopy (see xxxxx Fig. 2),
digital thermometer, burette for measuring volumes of volatile liquids (25 cm3
),
analytical balance, 3 small weighting bottle (25 cm3), spoon, benzene, naphthalene, ice,
stopwatch.
INSTRUCTIONS: Determine the benzene cooling curve (XXXXFIG.3) as instructed in
the introductory chapter. Apply 20 cm3
of pure benzene. Weight benzene liquid in
small weighting bottle (need for 4 significant digits). Repeat this curve measurement
three times using new benzene liquid.
Weigh about (0,2±0,02) g (4 significant digits) of benzoic acid in small weighting bottle
and add it in the cryoscopic tube with benzene liquid. Dissolve the benzoic acid and
determine the freezing-point of the solution twice.
In the same way, measure the freezing-point temperature of the other solutions at
solute concentrations corresponding to the total additions of (0.2, 0.4, 0.6, ...) g of
benzoic acid in 20 cm3
benzene.

?


ú l o h a J . S o p o u š e k
/6/
1.a-c
DATA ANALYSIS: METHOD 1 (SOLVING TWO EQUATIONS OF TWO UNKNOWNS):
We compile the equation of the matter conservation law for the benzoic acid in the
solution:
(1.4.)
where 𝑚𝑚𝐵𝐵 is weight of benzoic acid and 𝑀𝑀𝐵𝐵 is tabular molar mass of benzoic acid in
monomer form.
Equations (1.3.) and (1.4.) we can write as follow:
∆T
m
K
n nR
K
d⋅ = +1 (1.5.)
(1.6.)
These two equations involve experimental data ∆T, 𝑚𝑚𝐵𝐵, 𝑚𝑚𝐵𝐵 , and two unknowns 𝑛𝑛1 and
𝑛𝑛𝑑𝑑. Unknowns 𝑛𝑛1 and 𝑛𝑛𝑑𝑑 can be obtained and the result values can be inserted into
expression (1.2.), which give the dimerization constant of the benzoic acid 𝐾𝐾𝑑𝑑 in
benzene solution.
Use the results obtained with various benzoic acid concentration to determine the
confidence interval of the dimerization constant 𝐾𝐾𝑑𝑑.
METHOD 2 (THE NON-LINEAR LEAST SQUARE METHOD):
The molar amounts of the dimer 𝑛𝑛𝑑𝑑 can be obtained from eqn (1.2.) and inserted in eqn
(1.3.) and eqn (1.4.). Thus, we obtain
( ) ( )
R
Rd
K
m
mmnKn
KT
⋅+
⋅=∆ *
2
11 /
(1.7.)
and
(1.8.)
The eqn (1.8.) is quadratic equation with respect to 𝑛𝑛1, which can be rewritten in
quadratic formula:
(1.9.)
the value of unknown 𝑛𝑛1 can be obtained as a positive root:
(1.10.)
This equation can be inserted in eqn (1.7.) and we obtain model function ∆𝑇𝑇 = 𝑓𝑓(𝑚𝑚𝐵𝐵)
(the complete function is not listed here for its large size). ∆𝑇𝑇 can be treated as
dependent value 𝑦𝑦, 𝑚𝑚𝐵𝐵 as independent value 𝑥𝑥. 𝑚𝑚∗ and 𝑚𝑚𝑅𝑅 are constants. The model
m
M
n nB
B
d= + ⋅1 2
m
M
n nB
B
d= + ⋅1 2
( ) ( )Rd
B
B
mmnKn
M
m
⋅⋅⋅+= *
2
11 2
( )
( ) 0
2
1
2
1
*
=−+⋅
⋅
⋅
B
B
R
d
M
m
nn
mm
K
( )
( )R
d
RB
Bd
mm
K
mmM
mK
n
⋅
⋅
⋅
++−
=
*
*
1
4
8
11

J . S o p o u š e k ú l o h a
/7/
1.a-c
function have two parameters and 𝑀𝑀𝐵𝐵 , which can be optimized as a best match of
model function on experimental data set [𝑥𝑥, 𝑦𝑦] (i.e. [𝑚𝑚𝐵𝐵, ∆𝑇𝑇] ) using least square method.
The data analysis can be done by MS-EXCEL using “Solver“ tool for example. The
parameters and 𝑀𝑀𝐵𝐵 can be obtained if the experimental error was low. The more
experimental cooling curves we obtain, the better. Otherwise, we use tabulated value of
molar mass 𝑀𝑀𝐵𝐵 for data analysis. Thus the dimerization constant should be obtained
with better accuracy.
REPORT: Systematic deviation of the digital thermometer, tabulated molar mass of
benzoic acid 𝑀𝑀𝐵𝐵. Table 1: for each experiment: weight of benzene 𝑚𝑚𝑅𝑅 as used,
weight of benzoic acid 𝑚𝑚𝐵𝐵, experimental benzene/solution freezing point temperature T,
mean benzene freezing-point temperature 𝑇𝑇0 or solution freezing-point temperature 𝑇𝑇𝑓𝑓.
Table 2: for each solution experiment: weight of benzene 𝑚𝑚𝑅𝑅 in solution, weight of
benzoic acid 𝑚𝑚𝐵𝐵, freezing-point depression ∆𝑇𝑇, apparent molar mass of benzoic acid in
benzene calculated by means eqn (xxxxxx1.3). Common graph 1: cooling curves of
pure benzene liquids and benzoic acid solutions.
UNDER THE METHOD I: TABLE 3: for each solution experiment: unknowns 𝑛𝑛1 and 𝑛𝑛𝑑𝑑,
dimerization constant , coincidence interval of . Graph 2: symbols of the
experimental dependence ∆𝑇𝑇 on 𝑚𝑚𝐵𝐵.
UNDER THE METHOD II: TABLE 3: for each solution experiment: weight of benzoic acid 𝑚𝑚𝐵𝐵,
experimental freezing-point depression ∆𝑇𝑇, freezing-point depression given by model
function after parameter optimization, least squares of the difference between
experimental and theoretical freezing-points, sum of least squares, parameters 𝑀𝑀𝐵𝐵 and
. Graph 2: symbols of experimental values [𝑥𝑥, 𝑦𝑦] and curve of model function
∆𝑇𝑇 = 𝑓𝑓(𝑚𝑚𝐵𝐵) after optimisation.
Graph 3: the same as Graph 2 but add two limiting lines. First line for (i.e no
dimerization). The second line for (i.e dimer only in solution). Use union of eqns
xxxx (1.1) and (1.2), suppose 𝑚𝑚 = 𝑚𝑚𝐵𝐵.
dK
dK
dK dK
dK
0=dK
∞=dK
