28-2-2011
1
Colligative Properties of solutions
A colligative property is that which depends
on the relative number of solute and
solvent molecules. The number of solute
molecules relative to solvent, would reflect
a change in some physical properties.
Physical properties which will be studied
include vapor pressure, boiling point,
melting point, as well as osmotic pressure.
2
Vapor pressure lowering
The presence of solutes in liquid solutions
has great effects on the physical
properties of the resulting solution. The
vapor pressure of the solution will vary
depending on the amount and nature of
dissolved solute. Raoult's law discusses
the effects of solutes on the vapor
pressure of solutions provided that these
solutes are non dissociable (non ionics)
and nonvolatile.
3
Raoult’s law states that the vapor pressure of
solution (Psoln) is directly proportional to the mole
fraction of the solvent. This can be represented
by the relation:
Psoln = xsolvent * Posolvent
Where xsolvent is the mole fraction of the solvent
and Posolvent is its vapor pressure. Therefore, it is
obvious that:
4
1.
Dissolving a nonvolatile solute will result in a
decrease in the vapor pressure of the solution.
2.
When the vapor pressure of the solute
equals the vapor pressure of the solvent, no
change in the vapor pressure of the solution
would be observed.
3.
When the vapor pressure of the solute is
greater than the vapor pressure of the solvent,
the vapor pressure of the solution would be
increased.
5
Example
10.0 g of a paraffin (FW = 282 g/mol) oil, a
nonvolatile solute, was dissolved in 50.0 g
of benzene (C6H6, FW = 78.1 g/mol). At 53
oC, the vapor pressure of pure benzene is
300 torr. What is the vapor pressure of the
solution at that temperature?
Psoln = xbenzene * Pobenzene
6
Solution
nsolvent = 50.0 g/(78.1 g/mol) = 0.640
nsolute = 10.0 g/(282 g/mol) = 0.035
xsolvent = 0.640/(0.640 + 0.035) = 0.948
Psoln = xsolvent * Posolvent
Psoln = 0.948 * 300 torr = 284 torr
7
Example
At 25 oC, the vapor pressure of pure water is
23.76 mmHg and that of a urea (FW = 60
g/mol) solution is 22.98 mmHg. Find the
molality of solution.
We only have nonvolatile urea and water.
Therefore:
Psoln = xwater * Powater
8
Xurea = 1 – xwater
Psoln = (1 – xurea) Powater
Powater - Psoln = xurea Powater
DP = xurea Powater
Xurea = (23.76 – 22.98)/23.76 = 0.033
Xurea = nurea/(nurea + nwater)
9
Number of moles of water in 1 kg = 1000/18
= 55.6, therefore:
0.033 = nurea /(55.6 + nurea)
nurea = 1.9
Therefore, the molality of the urea soln is
about 1.9 m
10
Solutions with more than One
Volatile Component
When solutions contain more than one
volatile component, the vapor pressure of
the solution is the sum of the partial vapor
pressures of all components. Assume a
situation where a solution is composed
from three volatile components A, B, and
C, the vapor pressure of the solution is:
Psoln = xA * PoA + xB * PoB + xC * PoC
11
Example
A solution containing 50.0 g of CCl4 (FW = 153.8
g/mol) and 50.0 g of CHCl3 (FW = 119.4 g/mol).
At 50 oC, if the vapor pressure of pure CCl4 and
CHCl3 is 317 and 526 torr, respectively, find the
partial pressure of each substance and the
vapor pressure of the solution.
Solution
nCCl4 = 50.0 g /(153.8 g/mol) = 0.325 mol
nCHCl3 = 50.0 g /(119.4 g/mol) = 0.419 mol
12
xCCl4 = 0.325/(0.325 + 0.419) = 0.437
xCHCl3 = 1 – 0.437 = 0.563
PCCl4 = xCCl4* PoCCl4
PCCl4 = 0.437 * 317 torr = 139 torr
PCHCl3 = xCHCl3* PoCHCl3
PCHCl3 = 0.563 * 526 torr = 296 torr
Psoln = PCCl4 + PCHCl3
Psoln = 139 torr + 296 torr = 435 torr
13
The Psoln can also be calculated directly form
the relation below without calculation of
partial pressures in separate steps.
Psoln = xA * PoA + xB * PoB
Psoln = 0.437 * 317 torr + 0.563 * 526 torr =
435 torr
14
It should be emphasized here that Raoult's
law is valid for dilute solutions only while
concentrated solutions usually show
deviations from that law. Solutions obeying
Raoult's law are referred to as ideal
solutions while non ideal solutions show
either negative or positive deviations from
Raoult's law. Three situations can be
stated:
15
1.
16
The situation where the intermolecular
forces between solute-solute and solvent
solvent are of the same strength (DHsoln =
0). No deviations from Raoult's law as this
is an ideal solution (like benzene-carbon
tetrachloride solution).
17
2. The situation where the intermolecular
forces between solute- solvent molecules
are larger than solute-solute and solventsolvent intermolecular forces (DHsoln = -ve,
exothermic process). Negative deviations
from Raoult's law is observed (like
acetone-water solution).
18
19
3. The situation where the intermolecular
forces between solute-solvent molecules
are less than solute-solute and solventsolvent intermolecular forces (DHsoln = +ve,
endothermic process). Positive deviations
from Raoult's law is observed (like
ethanol-hexane solution).
20
21
Boiling point elevation
22
Boiling-Point Elevation
DTb = Tb – T b
0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
0
Tb > T b
DTb > 0
DTb = Kb m
m is the molality of the solution
23
Kb is the molal boiling-point
elevation constant (0C/m)
24