Chapter 11: Properties of Solutions
Solution Composition
Molarity, M = #moles of solute
volume of solution in dm3
Mass percent = mass of solute
mass of solution
x
Mole fraction = χA=
# moles A
of A
#moles A + #moles B
(no. units)
Molality
100 %
for a two component system A + B
= #moles of solute
kilogram of solvent
Example:- A sample of 0.892g KCl is dissolved in 54.6g of water. What is the percent by
mass of KCl in this solution?
Mass percent of KCl =
=
Example:- Calculate the molality of H2SO4 solution containing 24.4 g of sulfuric acid in
198g of water. The molar mass of sulfuric acid is 98.08 g/mol.
Molality = moles of solute =
mass of solvent (kg)
Factors Affecting Solubility
1. Structure effects – molecular structure determines a substance’s solubility e.g.
vitamins can be divided into two groups: fat soluble vitamins that are non-polar
e.g. vitamin A, and water soluble vitamins (vitamins B and C) polar molecules.
2. Effect of pressure. An increase in pressure increases the solubility of a gas in a
solvent.
Henry’s Law: p
= kC
p = partial pressure of gas
C = molar concentration (mol/L)
k = a constant that depends only on temperature.
See example exercise 11.4 page 524.
Raoult’s Law
eqn  Psoln
χsolvent Posolvent
Po = v. pressure of pure solvent
χ = mole fraction solvent
Psoln = observed vapor pressure of solution
Example: At 25oC, the vapor pressure of pure water is 23.76 mmHg and that of an
urea solution is 22.98 mmHg. Estimate the molality of the solution.
=
Raoult’s Law can be used in a second form:The relative lowering of vapor pressure (i.e. lowering divided by vapor pressure
of pure solvent is equal to the ratio of the number of moles of solute to the total
number of moles of solvent and solute.
Po – Psoln=
Po
n
N+n
this law holds strictly only in dilute solutions
n=moles solute, N=moles solvent
Since in dilute solutions n will be very small compared to N and the expression
can be written as:
ΔP
=
Po-Psoln =
n
o
P
Po
N
ΔP
P
=
# moles water in 1 kg of water =
# moles urea present in 1kg of water
concentration or molality of urea solution is
See sample exercise 11.5 p.529. Try Ex 11.47 end of chapter.
Note: If a solution obeys Raoult’s Law it is called an IDEAL SOLUTION when
solute-solute, solvent-solvent and solute-solvent interactions are very similar. If a
strong solute-solvent interaction is present this makes the vapor pressure lower
than that predicted by Raoult’s Law, i.e. it produces a negative deviation. A
positive deviation is observed when two liquids have extremely weak interactions
( e.g. ethanol and hexane).
Question of Lowering of Vapor Pressure
Psoln
=
Work out χsolvent =
χsolvent Posolvent
NH2O
NH2O + n urea
Qu. To do 47, 48*, 49, 51,53, 55. All lowering of vapor pressure questions.
Effect of Temperature on Solubility
See figure 11.6, page 525
The dependence of the solubility of a solid on temperature varies considerably. The
solubility of NaNO3, for example, increases sharply with temperature, while that of NaBr
changes very little. This wide variation provides a means of obtaining pure substances
from mixtures. Fractional crystallization is the separation of a mixture of substances into
pure components on the basis of their differing solubilities.
Suppose we have a sample of 90g of KNO3, that is contaminated with 10g of NaCl. To
purify the sample, we dissolve the sample in 100mL of water at 60oC and then gradually
cool the solution to 0oC. At this temperature the solubilities of KNO3 and NaCl are
12.1g/100g H2O and 34.2g/100g H2O, respectively. Therefore, (90-12)g or 78g of KNO3
will crystallize out of the solution but all of the NaCl will remain dissolved. By this
method we can obtain about 90% of the original amount of KNO3.
Colligative Properties of Non-Electrolyte Solutions
These are properties that depend only on the number of solute particles in solution,
regardless of whether they are atoms, ions or molecules. [ Note: we are concerned only
with relatively dilute solutions about 0.2M.]
1. Lowering of Vapor Pressure by a non-volatile solute.
Boiling Point Elevation and Freezing Point Depression
Boling Point Elevation
The boiling point of a liquid is the temperature at which its vapor pressure equals the
external atmospheric pressure. Since the presence of a non-volatile solute lowers the
vapor pressure of a solution, then such a solution must be heated to a higher temperature
that the normal boiling point of the pure solvent to reach a vapor pressure of 1
atmosphere.
The change in boiling point can be represented by:-
Look at sample ex. 11.8
Try Questions 60 and 59
Example
18.00g glucose was dissolved in 150.0g water. The resulting solution was found to have a
boiling point of 100.34oC. Calculate the molar mass of glucose.
Chapter 11 Questions
Lowering of Vapor Pressure
Question 47:
Question 60: Molality of biomolecule =
Question 59:
Freezing Point Depression
Adding NaCl or CaCl2 to ice on frozen roads melts the ice by lowering the
freezing point.
Sample Ex 11.9, 11.10
Worked Example:- Calculate the freezing point of a solution containing 651g of
ethylene glycol [(EG) CH2(OH)CH2(OH) – b.pt 197oC] in 2505g of water. The
molar mass of ethylene glycol is 61.02g.
# moles of EG =
Qu. 61, 63, 65, 71
Calculation of Molar mass from Freezing Point Depression
A sample of human hormone weighing 0.564g was dissolved in 15.0g benzene
and the freezing point depression was found to be 0.240oC. Calculate the molar
mass of the hormone.
Given:- Kf for benzene = 5.12oC.kg/mol
Since Tf = Kfm
Then
m
of hormone
But
molality of
hormone
=
ΔT
Kf
=
= # mol hormone
kg benzene
=
#mol hormone
0.0150kg benzene
Osmosis
Is the selective passage of solvent molecules through a semi-permeable membrane
from a dilute solution to a more concentrated one. Eventually the system reaches
an equilibrium position.
Osmotic Pressure is the pressure required to stop osmosis.
Note: only solvent molecules pass through the semi-permeable membrane.
Initially the rate of solvent transfer is greater in the direction of solvent
solution. However, eventually an equilibrium position is reached where the rate
of solvent transfer in both directions is equal. As the solution level rises, the
resultant pressure forces some solvent molecules back across the s.p.m.
Relationship between osmotic pressure and the solution concentration.
Osmotic
π
Pressure in
Atmospheres
= MRT
R = gas law constant
T = Kelvin temperature
M = Molarity
Look at sample ex. 11.11, page 541
A solution is prepared by dissolving 35.0g of hemoglobin (Hb) in enough water to
make up 1L in volume. If the osmotic pressure of the solution is found to be
10.00mmHg at 25oC, calculate the molar mass of hemoglobin.
First, calculate the molarity of the solution
The volume of the solution is 1L, so it must contain 5.38x10-4 mol of Hb. We can
use this quantity to calculate the molar mass
Practice Exercise:- A 202mL benzene solution containing 2.47g of an organic
polymer has an osmotic pressure of 8.63 mmHg at 21oC. Calculate the molar
mass of the polymer.
Osmotic Pressure Practice Problem.
Solutions that have identical osmotic pressures are said to be ISOTONIC.
Colligative Properties of Electrolyte Solutions
Colligative properties depend on the total concentration of solute particles. For
example, a 0.10m glucose solution lowers the freezing point of water by 0.186oC
where 0.1m NaCl lowers it by 0.37oC. Since electrolytes dissociate into ions in
solution they produce more than one particle. For example each unit of NaCl
dissociates into two ions – Na+ and Cl-. Thus, the colligative properties of NaCl
are twice that of a non-electrolyte, like glucose. Similarly, CaCl2, which produces
3 ions, would depress the freezing point three times as much as glucose. This
relationship can be expressed using the van’t Hoff factor i:
i=
moles of particles in solution
moles of solute dissolved
i is 1 for all non-electrolytes. Equations for colligative properties must be
modified to:ΔTb = i Kbm
ΔTf
= i Kfm
π
= i MRT
In reality, the colligative properties of electrolyte solutions are usually smaller
than anticipated, because at higher concentrations, electrostatic forces come into
play and bring about the formation of ion pairs. An ion pair is made up of one or
more cations and one or more anions held together by electrostatic forces. The
presence of an ion pair reduces the number of particles in solution, causing a
reduction in colligative properties.
Example:
The osmotic pressure of 0.010 M potassium iodide solution KI at 25oC is 0.465
atm. Calculate the van’t Hoff factor for KI at this temperature and concentration.
If KI dissociated completely to K+ and I- ions then O.P. would be:
π
= i MRT =
However, observed π = 0.465 atm. This means that there must be some ion pair
formation.
i=