Section 13.5 Colligative Properties
COLLIGATIVE PROPERTY = a physical property of a solution that depends on the quantity (ie
concentration) but not on the kind (or identity) of the solution particles.
Examples of colligative properties include:
1. Vapour pressure
2. Boiling point
3. Freezing point
4. Osmotic pressure
1. Lowering the Vapour Pressure
The vapour pressure of a solution is always LOWER than that of the pure solvent (the
solute must be a non-volatile substance).
The extent to which the VP is lowered is proportional to its concentration
RAOULT’S LAW for non-volatile solutes expresses this relationship by
PA = XA ·P0A ,where PA = the solvent’s VP above the solution
XA = mole fraction of solvent
P0A = vapour pressure of pure solvent
Ideal solutions obey Raoult’s Law. Real solutions follow the law best when the solute
concentration is low and when the solute and solvent have similar molecular size and
types of intermolecular attractions
Example: Calculate the vapour pressure of water above a solution prepared by adding 22.5
grams of lactose (C12H22O11) to 200.0 grams of water at 338 K. (Vapour pressure data for
water are given in Appendix B.)
Sample Exercise 13.8 Calculation of Vapour Pressure Lowering p. 547
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/ml at 25 0C.
Calculate the vapour pressure at 250C of a solution made by adding 50.0 ml if glycerin to
500.0 ml of water. The vapour pressure of pure water at 250C is 23.8 torr and its density is
1.00g/ml. Answer: 23.2 torr
Raoult’s Law for Volatile Solutes
If you have a volatile solute, such as an alcohol, there are two vapour pressures
which are additive.
Psolvent = Xsolvent + P0 solvent
Psolute = Xsolute + P0 solute
Vapour pressures of
the pure liquids
In the case of a volatile solute, the vapour pressure could goo up or down.
Ideal Solutions:
Ptotal = Psolvent + P solute
Solutions act most ideally when the solute and solvent have similar structures.
(Intermolecular forces between the solute and solvent molecules are similar to those
between the solvent molecules.)
Non-ideal Solutions
Ptotal < Psolvent + P solute
Occurs when intermolecular forces between the solute and solvent are vwey strong.
(intermolecular forces between the solute and solvent are stronger than those between
sovent molecules)
Example: At 63.50C the vapour pressure of water is 175 torr, and that of ethanol is 400
torr. A solution made by mixing 50.0 grams of water and 50.0 grams of ethanol. What is
the vapour pressure of the solution at 63.50C? Answer: 238 torr
2. Boiling Point Elevation
The boiling point of a solution is always HIGHER than that of the pure solvent (when
using a non-volatile solute)
The difference in the boiling point (between the solution and the solvent) is directly
proportional to the concentration of the solution expressed by its molality.
∆ Tb = Kb ∙ m ∙ i
∆ Tb = change in boiling point temperature
m = concentration of solution in molality
i = van’t Hoff Factor (= the number of ions)
Kb = the MOLAL BOILING POINT ELEVATION CONSTANT
The magnitude of Kb, which is called the MOLAL BOILING POINT ELEVATION
CONSTANT, depends only on the solvent (it does not matter what solute gets
dissolved). The values for the Kb are given in tables (p. 549).
The boiling point elevation depends only on the concentration of the solute particles
(in molality). If NaCl dissolves in water it will break into 2 ions giving it twice the
concentration of solute particles thus doubling the effect of elevating the boiling
point. (ie 2 x Kb). NOTE: for ideal solutions the effect of the NaCl would be doubled
using the van’t Hoff Factor, but in real solutions the value is usually lower because of
ion attractions.
Sample Problems:
1.
Calculate the boiling point of a solution made by adding 0.240 moles of naphthalene
(C10H8) to 2.45 moles of chloroform (CHCl3).
2.
Calculate the boiling point of a solution made by adding 2.04 grams of KBr to188 grams
of water.
3.
List the following aqueous solutions in order of increasing boiling points: 0.120 m
glucose; 0.050 m LiBr; and 0.050 m Zn(NO3)2.
3. Freezing Point Depression
The freezing point of a solution is always LOWER than that of the pure solvent.
The change in freezing point temperature is directly proportional to the molality so,
∆ Tf = Kf • m• i
Kf = MOLAL FREEZING POINT DEPRESSION CONSTANT
Again, the more particles that are present, the greater the freezing point
depression. So, 1 molal of NaCl has twice the effect of 1 molal of sucrose.
Sample Problems:
1.
Calculate the freezing point of a solution made by adding 0.240 moles of naphthalene
(C10H8) to 2.45 moles of chloroform (CHCl3).
2.
Calculate the freezing point of a solution made by adding 2.04 grams of KBr to188 grams
of water.
3.
List the following aqueous solutions in order of decreasing freezing point: 0.120 m
glucose; 0.050 m LiBr; and 0.050 m Zn(NO3)2.
Challenge Question: Try at home
Sample Exercise 13.9
Calculation of Boiling Point Elevation and Freezing
Point Lowering p. 550
Automotive antifreeze consists of ethylene glycol (C 2H6O2) a nonvolatile nonelectrolyte.
Calculate the boiling point and freezing point of a 25.0 mass % solution of ethylene
glycol in water. (Find the answer solved on p.550 in your text)
Sample Exercise 13.12 Determination of Molar Mass from Boiling Point Elevation p.555
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 grams
of the substance in 40.0 grams of CCl4. The boiling point of the resultant solution was 0.3570C
higher than that of the pure solvent. Calculate the molar mass of the solute.
Practice Exercise Determination of Molar Mass from Freezing Point Depression p. 555
Camphor melts at 179.80C, it has a particularly large freezing-point depression constant, Kf =
40.00C/m. When 0.186 grams of an organic substance of unknown molar mass is dissolved in
22.01 grams of liquid camphor, the freezing point of the mixture is found to be 176.70C. What
is the molar mass of the solute. Answer: 110 g/mol
4. Osmotic Pressure
OSMOTIC PRESSURE – is the pressure required to prevent osmosis
OSMOSIS_ the movement of solvent molecules through a semi-permeable membrane toward
the solution with higher concentration. (many membranes are semi-permeable meaning that
some molecules can pass through tiny pores while others cannot).
** ASIDE: Use u-tubes on p 552 explain the pressure needed. Eventually the pressure
difference resulting from the unequal heights of the liquids becomes so large that the net flow
of solvent ceases. Pressure can be applied to the apparatus to halt the net flow of solvent. The
pressure required to prevent osmosis of the solvent is called the OSMOTIC PRESSURE.
Osmotic pressure obeys the law of:
∏V = nRT • i
or
∏= MRT • i
Where ∏ = osmotic pressure
V = volume (L) of solution
n= number of moles of solute
R = gas constant = 0.08206 L∙ atm/mol∙K
T = Kelvin temperature
i = van’t Hoff Factor
M= molarity
 If two solutions of identical osmotic pressure are separated by a semi-permeable
membrane, no osmosis will occur. The two solutions are said to be ISOTONIC.
 If one solution is of lower osmotic pressure, it is called HYPOTONIC (it has lower
concentration) with respect to the more concentrated solution.
 If one solution is of higher osmotic pressure , it is called HYPERTONIC (it has higher
concentration) with respect to the dilute solution.
 Read p 553-555
Sample Exercise 13.11 Calculations involving Osmotic Pressure p. 553
The average osmotic pressure of blood is 7.7 atm at 250C. What molarity of glucose will be
isotonic with blood?
Sample Exercise 13.13 Molar Mass from Osmotic Pressure p 556
The osmotic pressure of an aqueous solution of a certain protein was measured to determine
the protein’s molar mass, The solution contained 3.50 mg of protein dissolved in sufficient
water to form 5.00ml of solution. The osmotic pressure of the solution at 250C was found to be
0.002026 atm. Treating the protein as a nonelectrolyte, calculate the molar mass.