Outline
1/12/07
First Chem seminar today…
 Pick up CAPA #2 - in front
 ChemBoard Assignment = Monday
 Fill in keypad response unit #’s…

Today:
Surfactants
Colligative Properties
Calculating Colligative Effects
Surfactants:
 Anything
which alters behavior of
solute-solvent mixture….(e.g. soap)
Applications?
NaCl in cooking
Road salt for ice
Antifreeze
Saline drips
Gatorade
e.g. Demo of S on water
Effect of solutes on
solvent properties...
Vapor pressures
Boiling points
Freezing points
Osmotic pressure
“Colligative properties”
Calculations of
Colligative Property effects
 Freezing
Point Depression
Kf = -1.858°C/m
DT = Kf m
 Boiling Point Elevation
Kb = 0.512°C/m
DT = Kb m
where Kf and Kb are constants
and m is concentration of solute
in terms of molality (mols/kg solvent)
liquid
solid
Calculations of
Colligative Property effects
 Freezing
Point Depression
Kf = -1.858°C/m
DT = Kf m
 Boiling Point Elevation
Kb = 0.512°C/m
DT = Kb m
where Kf and Kb are constants
and m is concentration of solute
in terms of molality (mols/kg solvent)
Example #1:
 Estimate
the freezing point of wine
that is 12% by mass ethanol...
Freezing Point Depression:
DT = Kf m
Kf = -1.858 C kg/mol (for water)
Assume 100g of wine, then you
have 12g ethanol and 88g H2O...
12g C2H5OH  46 g/mol = 0.261 mol
0.261 mol EtOH/88 g H2O = x mol/1000g
x = 2.97 m
DT = -5.5°C
Example #2:
 Compute
the molar mass of A, if a
solution containing 35.0 g of A
dissolved in 135 mL of H2O freezes at
-1.75C.
Assume A does not ionize in water.
Freezing Point Depression:
DT = Kf m
Kf = -1.858 C kg/mol (for water)
-1.75 = -1.858 m
or
m = 0.942 mol/kg
Example #2 (cont’d):
m
= 0.942 mols solute/kg solvent
 How many kg of solvent?
135 mL  1 g/mL = 135 g = 0.135 kg
0.942 mol/kg  0.135 kg = 0.127 mol
Molar mass:
35.0 g / 0.127 mol = 275 g/mol
Another useful application:
Osmotic Pressure (p)
Passage of solute molecules through
a semi-permeable membrane
Start with something we know:
PV = nRT (ideal gases)
or P = (n/V)RT
Adapt to solutions:
(n/V)  (mol/L) = Molarity!
Osmotic Pressure (p)
p = M RT
p.107
p.107
Example:
Compute the molar mass of hemoglobin,
if a 25 mL solution containing 0.420 g of
hemoglobin has an osmotic pressure of
4.6 torr at 27C.
Osmotic pressure:
p = M RT
M = [m (g) / MM (g/mol)]/V
p = m (RT)/ (MM) V
= 68300 g/mol
Practice…Worksheet #1
 Remember
to keep solute & solvent
separate when answering questions!
Henry’s Law
 Gases
dissolved in a liquid:
Concentration = KH p
Examples: soda
CO2, NOx, SOx and acid rain
For Monday…
 Do
CAPA set #1
 Post a trial message on Chemboard
 Read through Chapter 12
…get ahead!
Have a great week-end!