Colligative Properties
(solutions)
• A. Definition
• Colligative Property
– property that depends on the concentration of
solute particles, not their identity
4 Colligative Properties
•
•
•
•
Vapor-pressure
Freezing point
Boiling point
Osmotic pressure
Vapor pressure
• vapor pressure is the measure of the tendency
of molecules to escape from a liquid
• volatile- high vapor pressure
• non-volatile- low vapor pressure
• vapor pressure lowering depends on the
concentration of a nonelectrolyte solute and is
independent of solute identity, it is a colligative
property
• because vapor pressure is lowered, this lowers
the freezing point and raises the boiling point.
Freezing Point Depression
•
(tf) - is the difference between the
freezing points of the pure solvent, and it
is directly proportional to the molal
concentration of the solution.
–
f.p. of a solution is lower than f.p. of the
pure solvent
• the freezing point of a 1-molal solution of any
nonelectrolyte solute in water is found by
experiment to be 1.86°C lower than the freezing
point of water.
• Molal freezing-point constant, K, is the freezing
point depression of the solvent in a 1- molal
solution of a nonvolatile, nonelectroyte solute
• Each substance has a different molal freezing
point constant (p. 438)
• Applications
– salting icy roads
– making ice cream
– antifreeze
• cars (-64°C to 136°C)
• ∆tf = Kf mi
∆tf = freezing point depression (°C)
Kf = °C/ m m = mol solute/kg of solvent
i = # of particles
Boiling Point Elevation
• (tb)
– b.p. of a solution is higher than b.p. of the pure
solvent
– When the vapor pressure is equal to the atmospheric
pressure boiling will occur.
– molal boiling point constant (kb ) is the boiling-point
elevation of the solvent in a 1-molal solution of a
nonvolitle, nonelectrolyte solute. (0.51 °C/m)
– Boiling-point elevation,∆tb , is the difference
between the boiling points of the pure solvent
and a nonelectrolyte solution of that solvent,
and is directly proportional to the molal
connection of the solution.
• ∆tb = Kb m i
∆tb = boiling-point elevation (°C)
Kb = °C/ m
m = mol solute/kg of solvent
i = # of particles
Osmotic pressure:
•
•
•
Semipermeable membranes allow the
movement of some particles while blocking the
movement of others.
Osmosis: the movement of solvent through a
semipermeable membrane from the side of
lower solute concentration to the side of higher
solute concentration.
Osmotic pressure is the external pressure that
must be applied to stop osmosis.
• Electrolytes: remember dissociation of ionic
compounds
• NaCl lowers the freezing point twice as much as
sucrose C12H22O11
– NaCl Na+ + Cl-
• CaCl2 lowers the freezing point three times as
much as C12H22O11 due to the dissociation of the
ionic compounds
– CaCl2 → Ca2 + + 2Cl-
C. Calculations
t = k · m · i
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
(different depending on freezing or boiling)
m:molality (m)
i: # of particles
• # of Particles
– Nonelectrolytes (covalent)
• remain intact when dissolved
• 1 particle
– Electrolytes (ionic)
• dissociate into ions when dissolved
• 2 or more particles
When we dissolve most ionic substances in water
they break up into their individual ions.
NaCl(s)

Cl- (aq) + Na+ (aq)
Most molecules don’t break up into ions. For
example sugar.
C12 H24O12 (s) + H2O (aq)  C12 H24O12 (aq)
But some molecules do break up into ions.
Acids and bases are examples.
HCl(l) + H2O (aq)
 Cl- (aq) + H3O + (aq)
Dissociation Equations ( ionic )
i = 2
NaCl(s)  Na+(aq) + Cl-(aq)
i = 2
AgNO3(s)  Ag+(aq) + NO3-(aq)
MgCl2(s)  Mg2+(aq) + 2 Cl-(aq)
i = 3
i = 3
Na2SO4(s)  2 Na+(aq) + SO42-(aq)
AlCl3(s)  Al3+(aq) + 3 Cl-(aq)
i = 4
C. Calculations
• At what temperature will a solution that is
composed of 0.73 moles of glucose in 225 g of
phenol boil?
GIVEN:
WORK:
b.p. = ?
m = 0.73mol ÷ 0.225kg
tb = ?
tb = (3.60°C·kg/mol)(3.2m)(1)
kb = 3.60°C·kg/mol
tb = 12°C
m = 3.2m
b.p. = 181.8°C + 12°C
i=1
b.p. = 194°C
tb = kb · m · i
C. Calculations
• Find the freezing point of a saturated solution of
NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = -1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (-1.86°C·kg/mol)(4.8m)(2)
m = 4.8m
i=2
tf = kf · m · i
f.p. = 0.00°C - 18°C
tf = -18°C
f.p. = -18°C
What is the boiling point of a solution made
by dissolving 1.20 moles of NaCl in 750 g of
water?
What is the boiling point?
Δtb = kb .m.i
Find molality! Molality=1.20moles/.750 kg
molality= 1.6 m
Δtb = (0.51°C/m)(1.6m)(2)
Δtb = 1.63°C
Boiling pt: 100 + 1.63= 102 °C
Ex: What is the freezing point depression of water in a
solution of 17.1 g of sucrose, C12 H22O11 , and 200. g
of water? What is the actual freezing point of the
solution?
• Δtf = kf .m.i
• Molar mass of sucrose is 342.34 g/mol
• Find moles:
 1m ol 
17.1g 

 342.34g 
• = 0.04995 mol
• Molality= 0.04995 moles/.200 kg
• molality= 0.2498 m
• Δtb = (-1.86°C/m)(0.2498m)(1)
• Δtb = - 0.465°C
• Ex: A water solution containing an
unknown quantity of a nonelectrolyte
solute is found to have a freezing point of
-0.23°C. What is the molal concentration
of the solution?
Δtf = kf .m.i
-0.23 °C = (-1.86 °C/m)(x)(1)
-0.23 °C = x
-1.86 °C
0.12 m = x
• Ex: What is the boiling point elevation of a
solution made from 20.0 g of a
nonelectrolyte solute and 400.0 g of
water? The molar mass of the solute is
62.0g/mol.
• Find molality first: 20.0 g / 62.0 g/mol
• Molality = 0.3226 mol/0.400 kg=0.806 m
• Δtb = kb .m.i
= (0.51 °C/m)(0.806 m)(1)
= 0.411 °C
• Ex: What is the expected change in the
freezing point of water in a solution of 62.5
g of barium nitrate, Ba(NO3 )2 , in 1.00 kg
of water?
• Molality= 62.5 g/261.36 g/mol =0.2487 m
• Δtf = kf .m.i
= (-1.86 °C/m)(0.2487 m)(3)
= 0.746 °C
Assignment:
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•
•
•
Do practice problems 1-4 on page 440
Do practice problems 1-4 on page 441.
Do practice problems 1-3 on page 445
Practice problems have answers off to the
right.