Colligative Properties (Continued):
Properties of a solution that differ from the pure solvent due to the number of particles, not type, that are dissolved.
11.4 Boiling Point Elevation / Freezing Point Depression
Normal boiling point occurs when the vapor pressure equals normal air pressure, 760 torr. A non-volatile
solute will lower the freezing point and raise the boiling point of a solution.
Boling Point Elevation:
∆T = iKbm
Kb= molal boiling point elevation constant, Ex. H2O = 0.51ºC/m (See table 11.5)
m= solution concentration expressed in molality, moles solute/kg solvent
i= number of ions produced from dissociation
Freezing Point Depression:
∆T = iKfm
Kf= molal freezing point depression constant, Ex. H2O = 1.86ºC/m (See table 11.5)
m= solution concentration expressed in molality, moles solute/kg solvent
i= number of ions produced from dissociation
In Class Ex. 1:
Determine the boiling point and freezing point of a solution that contains 25.0 g of CaCl2 (111 g/mol) in
200.g of H2O? (CaCl2 
→ Ca2+ + 2Cl-)
 25.0 gCaCl 2 1molCaCl 2 
 x3ions = 3.38m
m= 
111g 
 0.200kgH 2O
Boling point of solution:
∆T = 0.51ºC/m . 3.38m = 1.73ºC, 100ºC = 1.73ºC = 101.73ºC
Freezing point of solution:
∆T = 1.86ºC/m . 3.38 = 6.29ºC, 0-6.29ºC = -6.29ºC
In Class Ex. 2:
What mass of ethylene glycol (non-volatile, non-ionizing, 62.1g/mol) is needed to prepare 10.0L water
(D=1 g/ml) with a freezing point of –10.0ºF (-23.3ºC)?
m=
∆T
23.3C
=
= 12.5mol / kg , _ 12.5mol / kg = 10.0kg = 125mol , _ 125molx62.1g / mol = 7760 g = 7.76kg
Kf 1.86C / m
In Class Ex. 3:
Determine the molecular mass of a non-ionizing, non-volatile solute if 0.546g of it are dissolved in
15.0g of benzene ( Kf=5.12 ºC/, F.P. = 5.5ºC) and the freezing point depression was 0.240ºC.
m=
∆T
0.240C
mol
=
= 0.0469
Kf 5.12C / m
kg
0.0469mol
x0.0150kg = 7.03x10 − 4mol
kg
0.546 g
= 776 g / mol
0.000703mol
11.5Osmotic Pressure
Can you recognize the direction of movement of water across a semi-permeable membrane in the below
situations? A semi-permeable membrane only allows water to pass through.
Osmosis: the flow of solvent into a solution across a semi-permeable membrane.
In a U-tube with a semi-permeable membrane at the bottom of the “U”: A solution on the left and pure
water on the right. The water will pass across the membrane into the solution raising the height and
creating hydrostatic pressure on the solution that is greater than the pure solvent. The minimum pressure
that stops osmotic pressure, π, is called osmotic pressure.
Very useful because a small number of molecules can create a relatively large osmotic pressure.
Osmotic pressure can be calculated with:
π = iMRT,
π=osmotic pressure in atm,
M is the molarity of solution,
R=0.08206 L.atm/mol .K
T= Kelvin temperature
i= number of ions produced from dissociation
In Class Ex. 1:
Find the molar mass of a protein if 1.00x10-3g of it was dissolved into enough solution to make 1.00
ml. The osmotic pressure was 1.12 torr at 25ºC.
M =
π
RT
=
0.00100 g
1.12torr / 760
mol
= 0.0000603
x0.00100 L = 0.0000000603mol , __ MM =
= 16600 g / mol
0.08206 x 298
L
0.0000000603mol
Sorry, Equation editor doesn’t do superscripts!
Types of Solutions:
Isotonic: identical osmotic pressures. Ex. isotonic with cell fluid.
Hypertonic: higher osmotic pressure than cell fluids.
Hypotonic: lower osmotic pressure than cell fluids.
Conditions affecting cells:
Crenation; cells are bathed in a medium that is hypertonic: more solute outside of cell than inside, water
exits cell, cell shrivels.
Hemolysis (lysis): cells are bathed in hypotonic solution: more solute particles in the cell, water enters the
cell, cell ruptures.
Application: Why do we salt meat for storage? Answer: Bacteria on meat shrivels and dies.
Reverse Osmosis: An external pressure applied is larger than the osmotic pressure. Good method for
desalination, which allows for solvent to flow from the solution across a semi-permeable membrane.
11.6Colligative Properties of Electrolyte Solutions
Relationship between the moles of solute and moles of particles
moles _ particles _ in _ so ln
*van’t Hoff factor, i: i= m
moles _ of _ solute _ dissolved
Dutch chemist J.H. van’t Hoff received first Nobel prize in chemistry in 1901.
Ion pairing, comparing the expected dissociation value, example, MgCl2 i= 3, with the actual. The actual
observed is 2.7. Often this is less than expected due to the ion pairing which count as a single particle.
More important in concentrated solutions. Less ion pairing occurs in dilute solutions.
Ex.
The observed osmotic pressure for a 0.10M solution of Fe(NH4)2(SO4)2 at 25ºC is 10.8 atm. Compare
the expected and experimental values for i.
Expected = Fe(NH4)2(SO4)2 
→ Fe2+ + 2NH4+ + 2SO42- =5
π
10.8atm
Observed =
=
= 4 .4
MRT 0.1M (0.08206)298
11.7Colloids
Larger particles that remain suspended in solution, colloidal suspension. These particles can scatter a beam
of light, Tyndal effect, unlike a true solution. Particles are large molecules or aggregates of molecules or
ions that range in size from 1 – 1000nm.
Ex.’s
Gelatin
Smoke
Foam
Clay
Milk
Fog
Cheese
Practice Problems: (Set 3): P. 520 #21, 23, 24, 58, 60, 61, 62, 64, 65, 68, 69, 70, 73, 76,
79, 83, 84, (92)
Terms
Solute: substance being dissolved
Solvent: substance in larger amount that surrounds the solute. Water is the universal solvent
Homogeneous: single, uniform phase such as Windex, vinegar, or saline.