AP Chemistry Notes
Solutions & Colligative Properties
Solutions…

A solution is a homogenous mixture of two or more
substances.
 The solute is(are) the substance(s) present in the
smaller amount(s).
 The solvent is the substance present in the larger
amount.
Solutions…


An electrolyte is a substance that, when dissolved in
water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved,
results in a solution that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
Solutions…

A saturated solution contains the maximum
amount of a solute that will dissolve in a given
solvent at a specific temperature.

An unsaturated solution contains less solute than
the solvent has the capacity to dissolve at a
specific temperature.

A supersaturated solution contains more solute
than is present in a saturated solution at a specific
temperature.
Solutions…

Three types of interactions in the solution process:

solvent-solvent; solute-solute; and solvent-solute
DHsoln = DH1 + DH2 + DH3
Solutions…

Two substances with similar intermolecular
forces are likely to be soluble in each other.

‘LIKE DISSOLVES LIKE’
 Non-polar

CCl4 in C6H6
 Polar

molecules are soluble in polar solvents
C2H5OH in H2O
 Ionic

molecules are soluble in non-polar solvents
compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
Concentration…

The concentration of a solution is the
amount of solute present in a given quantity
of solvent or solution.

Percent by Mass:
mass of solute x 100%
=
mass of solution
Concentration…

Mole Fraction (X)
moles of A
XA =
sum of moles of all components
Concentration…

Molarity:
moles of solute
M =

liters of solution
Molality:
m =
moles of solute
mass of solvent (kg)
Concentration…
What is the molality of a 5.86 M ethanol (C2H5OH) solution
whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution: 5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
=
5.86 moles C2H5OH
0.657 kg solvent
= 8.92 m
Solubility…
Solid solubility
and
temperature
Solubility…
Gas solubility
and
temperature
Solubility…

Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional
to the pressure of the gas over the solution
(Henry’s law).
 Mathematically expressed: c = kP

c
is the concentration (M) of the dissolved gas
 P is the pressure of the gas over the solution
 k is a constant (mol/L•atm) that depends only on
temperature
Henry’s Law:
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
low P
high P
low c
high c
Colligative Properties
NON-Electrolytic Solutions
Colligative Properties…

Colligative properties are properties that depend
only on the number of solute particles in solution
and not on the nature of the solute particles.




Vapor-Pressure Lowering
Boiling Point Elevation
Freezing Point Depression
Osmosis
Vapor Pressure Lowering…

Raoult’s law:
P1 = X1 P 10
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = DP = X2 P 10
X2 = mole fraction of the solute
Vapor Pressure Lowering…

At a given temperature water has a vapor
pressure of 22.80 mmHg. Calculate the
vapor pressure above a solution of 90.40 g
of sucrose (C12H22O11) in 350.0 mL of water,
assuming the water to have a density of
1.000 g/mL.
Vapor Pressure Lowering…
90.40 g C12H22O11  0.26 mol
 350.0 ml H2O  19.4 mol
 X = (19.4/(19.4+0.26)) = 0.9868
 VP = XP
 VP = (0.9868)(22.80)
 VP = 22.50 mmHg

Vapor Pressure Lowering…

23.00 g of an unknown substance was
added to 120.0 g of water. The vapor
pressure above the solution was found to be
21.34 mmHg. Given that the vapor pressure
of pure water at this temperature is 22.96
mmHg, calculate the Molar Mass of the
unknown.
Vapor Pressure Lowering…












23.0 g X
120.0 g H2O  6.65 mol H2O
VP = 21.34
P = 22.96
VP = XP
21.34 = X (22.96)
X = 0.929
X = mol H2O / total mol
0.929 = 6.65/(6.65 + x)
X mol = 0.506
M = g/mol = 23.0 / 0.506
M = 45.46 g/mol
Vapor Pressure Lowering…
PA = XA P A0
PB = XB P 0B
PT = PA + PB
PT = XA P A0 + XB P 0B
Ideal Solution
Vapor Pressure Lowering…

At 20.0 oC the vapor pressures of methanol
(CH3OH)and ethanol (C2H5OH) are 95.0 and
45.0 mmHg respectively. An ideal solution
contains 16.1 g of methanol and 92.1 g of
ethanol. Calculate the vapor pressure.
Vapor Pressure Lowering…
16.1 g CH3OH  0.5 mol
 92.1 g C2H5OH  2 mol
 Total mol = 2.5
 VP = XaPa + XbPb
 VP = (0.5/2.5)(95) + (2/2.5)(45)
 VP = 55 mmHg

Boiling Point Elevation…
DTb = Tb – T b0
T b is the boiling point of
the solution
T b0 is the boiling point of
the pure solvent
Tb > T b0
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
Freezing Point Depression…
DTf = T 0f – Tf
T f is the freezing point of
the solution
T 0f is the freezing point of
the pure solvent
T 0f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
Constants…
What is the freezing point of a solution containing 478g
of ethylene glycol (antifreeze) in 3202 g of water?
GIVEN: The molar mass of ethylene glycol is 62.01 g.
DTf = Kf m
Kf water = 1.86 0C/m
478 g x
moles of solute
1 mol
62.01 g
=
m =
mass of solvent (kg)
= 2.41 m
3.202 kg solvent
DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
DTf = T 0f – Tf
Tf = T 0f – DTf = 0.00 0C – 4.48 0C = -4.48 0C
Summary…

Colligative properties are properties that
depend only on the number of solute
particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 10
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
p = MRT
Colligative Properties
Electrolytic Solutions
Colligative Properties…

Colligative properties are properties that depend only on
the number of solute particles in solution and not on the
nature of the solute particles.

0.1 m NaCl solution
0.1 m NaCl solution

van’t Hoff factor (i) =
0.1 m Na+ ions & 0.1 m Cl- ions
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
1
Nonelectrolytes
2
NaCl
3
CaCl2
Colligative Properties…
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iMRT
Colligative Properties…
At what temperature will a 5.4 molal solution of NaCl
freeze?
Solution:
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
Colligative Properties…
Osmotic Pressure (p):
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
more
concentrated
Colligative Properties…
High
P
Low
P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
Colloids…
 A colloid is a dispersion of particles of one substance
throughout a dispersing medium of another substance.

Colloid versus solution:


Collodial particles are much larger than solute molecules
Collodial suspension is not as homogeneous as a solution