Colligative Properties of Solutions
17.4) Vapor Pressure Lowering - Raoult’s Law
17.5) Boiling Point Elevation
and Freezing Point Depression
17.6) Osmotic Pressure
We’ll start by focusing on nonvolatile nonelectrolitic solutes
(e.g. sucrose in H2O)
Later: volatile solutes, electrolytic solutes
Figure 17.8, Zumdahl
Raoult’s Law:
Psolvent = Xsolvent . P°solvent
X is the mole fraction
See Figure 17.9
Example: Vapor Pressure Lowering
Problem: Calculate the vapor pressure lowering (∆P) when 175g of
sucrose is dissolved into 350.00 ml of water at 750C. The vapor
pressure of pure water at 75 0C is 289.1 mm Hg, and it’s
density is 0.97489 g/ml.
Plan: Calculate the change in pressure from Raoult’s law using the
vapor pressure of pure water at 75 0C. We calculate the mole
fraction of sugar in solution using the molecular formula of
sucrose and density of water at 75 0C.
Solution: molar mass of sucrose ( C H O ) = 342.30 g/mol
12 22 11
x 0.97489g H2O = 341.21g H2O
ml H2O
175g sucrose
= 0.51125 mol sucrose
342.30g sucrose/mol
350.00 ml
H2O
341.21 g H2O
= 18.935 mol H2O
18.02g H2O/mol
Vapor Pressure Lowering (cont)
Vapor Pressure lowering:
∆P = Xsolute . P°solvent
mole sucrose
moles of water + moles of sucrose
0.51125 mole sucrose
= 0.2629
18.935 mol H2O + 0.51125 mol sucrose
= Molal boiling point constant for given liquid
See Table 17.5
∆P = Xsucrose * PoH2O = 0.2629 x 289.1 mm Hg = 7.600 mm Hg
=
Xsucrose =
Kb
Boiling Point Elevation
Recall: The normal boiling point (Tb) of a liquid is the
temperature at which its vapor pressure equals atmospheric
pressure.
Nonvolatile solutes lower the vapor pressure of a liquid
Æ greater temperature required to reach boiling point
Boiling Point Elevation:
∆Tb = Kbmsolute
Kb = Molal boiling point constant for given liquid
msolute = molal solute concentration
Freezing Point
Depression:
∆Tf = Kfmsolute
See Figure 17.3
Kf = Molal freezing point constant for given liquid
See Table 17.5
Example: Boiling Point Elevation and Freezing Point
Depression in an aqueous solution
0.512oC
(2.313m)= 1.18oC
m
Problem: We add 475g of sucrose (sugar) to 600g of water. What will
be the Freezing and Boiling points of the solution?
Plan: Find the molality of the sucrose solution, and apply the equations
for FP depression and BP elevation using the constants from table 17.5.
Solution: Sucrose (C12H22O11) has molar mass = 342.30 g/mol
475g sucrose
= 1.388 mole sucrose
342.30gsucrose/mol
molality = 1.388 mole sucrose = 2.313 m
0.600 kg H2O
∆Tb = Kb . m =
BP = 100.00oC + 1.18oC
= 101.18oC
1.86oC
oC
∆Tf = Kf . m =
(2.313
m)
=
4.30
m
FP = 0.00oC -4.30oC= -4.30oC
See Figure 17.12
Example: Boiling Point Elevation and Freezing Point
Depression in a Non-Aqueous Solution
Kb
=
Kf
o
. m =4.70 C (4.01m) =18.85oC
m
normal FP = - 63.5oC
new FP = - 82.4oC
Problem: Calculate the Boiling Point and Freezing Point of a
solution having 257g of napthalene (C10H8) dissolved into 500.00g of
chloroform (CHCl3).
Plan: Just like the first example.
Solution: napthalene = 128.16g/mol
chloroform = 119.37g/mol
257g nap =2.0053 mol nap
molesnap =
128.16g/mol
2.0053 mol
molality = moles nap =
= 4.01 m
0.500 kg
kg(CHCl3)
o
= . m = 3.63 C(4.01m) = 14.56oC
normal BP = 61.7oC
m
new BP = 76.3oC
∆Tb
∆Tf
Osmosis: The flow of solvent through a
semipermeable membrane into a solution
The semipermeable membrane allows solvent
molecules to pass, but not solute molecules
See Figures
17.15, 17.16
Determining Molar Mass from
Osmotic Pressure
.
1 atm = 0.00475 atm
760 torr
Problem: A physician studying hemoglobin dissolves 21.5mg of the
protein in water at 5.0oC to make 1.5 ml of solution in order to
measure its osmotic pressure. At equilibrium, the solution has an
osmotic pressure of 3.61 torr. What is the molar mass (M) of the
hemoglobin?
Plan: We know Π, R, and T. We convert Π from torr to atm, and T
from oC to K, and then use the osmotic pressure equation to solve for
molarity (M). Then we calculate the number of moles of hemoglobin
from the known volume and use the known mass to find M.
Solution:
P = 3.61 torr
Temp = 5.00C + 273.15 = 278.15 K
Osmotic Pressure, Π = MRT
(similar to ideal gas law!)
R = gas constant
M = molar concentration of solute
L soln = 3.12 x10 - 7 mol
= 2.08 x10 - 4 M
Molar Mass from Osmotic Pressure
. 0.00150
Concentration from osmotic pressure:
Π
M=
= 0.00475 atm
RT
0.082 L atm (278.2 K)
mol K
2.08 x10 - 4 mol
L soln
Finding # moles of solute:
n = M.V =
0.0215 g
= 6.89 x104 g/mol
3.12 x10-7 mol
Calculating molar mass of Hemoglobin (after changing mg to g):
M =
Colligative Properties of Solutions
17.4) Vapor Pressure Lowering - Raoult’s Law
17.5) Boiling Point Elevation
and Freezing Point Depression
17.6) Osmotic Pressure
We started with nonvolatile nonelectrolytes (e.g. sucrose/H2O)
Now let’s look at volatile solutes and electrolytes
Colligative Properties of Volatile Nonelectrolyte Solutions
From Raoult’s law: Psolvent = Xsolvent . P0solvent and Psolute = Xsolute . P0solute
Consider a solution having equal molar quantities of acetone and
chloroform, Xacetone = XCHCl3 = 0.500. At 350C, the vapor pressure of
pure acetone = 345 torr and pure chloroform = 293 torr.
¾ Determine the vapor pressure of the solution and the partial
pressure of each component. What are the mole fractions, X, of each
component in the vapor phase?
Pacetone = Xacetone . P0acetone = 0.500 . 345 torr = 172.5 torr
PCHCl3 = XCHCl3 . P0CHCl3 = 0.500 . 293 torr = 146.5 torr
PA
From Dalton’s law of partial pressures we know that XA = P
Total
Pacetone
172.5 torr
Xacetone = P
=
= 0.541
172.5 + 146.5 torr
Total
Total Pressure = 319.0 torr
P
146.5 torr
= 0.459
XCHCl3 = CHCl3 =
PTotal 172.5 + 146.5 torr
Vapor is enriched in acetone!
Non-ideal solutions
ƒ Ideal behavior is approached when solute and solvent are
involved in similar intermolecular interactions (∆Hsoln = 0)
ƒ When weaker solvent-solute interactions occur, heat is removed
upon dissolving (∆Hsoln > 0)
Æ the observed vapor pressure is higher than ideal
ƒ When stronger solute-solvent interactions occur, heat is released
upon dissolving (∆Hsoln < 0)
Æ the observed vapor pressure is lower than ideal
Figure 17.11
Colligative Properties of Ionic Solutions
For ionic solutions we must take into account the number of ions present!
i = van’t Hoff factor = “ionic strength”, or the number of ions present
vapor pressure lowering:
Tb = i Kb m
P = i XsoluteP 0solvent
For Electrolyte Solutions:
boiling point elevation:
Tf = i Kf m
π = i MRT
freezing point depression:
osmotic pressure:
Non-ideal
Behavior of
Electrolyte
Solutions
Non-ideal Behavior
of Electrolyte
Solutions
(0.05m aqueous)
See Table 17.6