Colligative Properties of Solutions
are properties of solutions that depend solely
on the number of particles of solute and NOT
on their chemical identity.
• vapor pressure
• boiling point
• freezing point
• osmotic pressure
Vapor Pressure of a Solution
A solute that is nonvolatile is
one that has no measurable
vapor pressure.
We will study the effects of
nonvolatile solutes on the
properties of solutions.
The presence of a nonvolatile
solute causes the vapor
pressure of the solution to be
lower than the vapor pressure
of the pure solvent.
Vapor Pressure of a Solution –
Raoult’s Law
The extent to which the vapor pressure of a
solvent is lowered by a nonvolatile solute is
given, for an ideal solution, by Raoult’s Law.
Vapor pressure of
the solution
Raoult’s Law
Psolvent over solution = XsolventP°solvent
Xsolvent is the mole fraction of the solvent.
P°solvent is the vapor pressure of the pure solvent at
the solution temperature.
Raoult’s Law – Example:
Nonvolatile Solute
What is the vapor pressure at 25°C (room
temperature) of a solution made by adding 226
g (1 cup) sugar (C12H22O11) to 118.5 mL (½ cup)
of water?
Pwater over sugar soln = XwaterP°water
𝒎𝒐𝒍 𝒘𝒂𝒕𝒆𝒓 = 𝟏𝟏𝟖. 𝟓 𝒎𝑳
𝐦𝒐𝒍 𝒔𝒖𝒈𝒂𝒓 = 𝟐𝟐𝟔 𝒈
𝑷𝒘𝒂𝒕𝒆𝒓 𝒐𝒗𝒆𝒓 𝒔𝒖𝒈𝒂𝒓 𝒔𝒐𝒍𝒏
𝟎. 𝟗𝟗𝟕𝟎𝟕 𝒈 𝟏 𝒎𝒐𝒍
= 𝟔. 𝟓𝟓𝟗 𝒎𝒐𝒍
𝒎𝑳
𝟏𝟖. 𝟎𝟏𝟓𝒈
𝟏 𝒎𝒐𝒍
= 𝟎. 𝟔𝟔𝟎 𝒎𝒐𝒍
𝟑𝟒𝟐. 𝟑𝟎𝒈
from Appendix B
𝟔. 𝟓𝟓𝟗 𝒎𝒐𝒍
=
𝟐𝟑. 𝟕𝟔 𝒕𝒐𝒓𝒓 = 𝟐𝟏. 𝟓𝟗 𝒕𝒐𝒓𝒓
𝟔. 𝟓𝟓𝟗 + 𝟎. 𝟔𝟔𝟎 𝒎𝒐𝒍
Raoult’s Law – Example: Two
Volatile Components
What is the vapor pressure at 25°C of 80-proof
alcohol (40% alcohol by volume)?
Alcohol: C2H5OH, molar mass = 46.07 g,
v.p.(25°C) = 54.68 torr, density (25°C) = 0.786 g/mL
Water: molar mass = 18.015 g, v.p.(25°C) = 23.76 torr,
density (25°C) = 0.997 g/mL
Apply Raoult’s Law to each volatile component.
By convention, the liquid component present in
larger volume is the “solvent.”
Raoult’s Law – Example: Two
Volatile Components
What is the vapor pressure at 25°C of 80-proof
alcohol (40% alcohol by volume)?
Raoult’s Law for the water:
Pwater over water/alcohol solution = XwaterP°water
Find Xwater: We need a volume for the solution, don’t we?
Any volume will do! 100 mL is convenient, though.
100 mL – 40 mL alcohol = 60 mL water
Raoult’s Law – Example: Two
Volatile Components
Pwater over water/alcohol solution = XwaterP°water
Find both mol fractions:
𝟎. 𝟗𝟗𝟕 𝒈 𝟏 𝒎𝒐𝒍
𝟔𝟎 𝒎𝑳 𝑯𝟐 𝑶
= 𝟑. 𝟑𝟐𝟏 𝒎𝒐𝒍 𝑯𝟐 𝑶
𝒎𝑳 𝟏𝟖. 𝟎𝟏𝟓 𝒈
𝟎. 𝟕𝟖𝟔 𝒈 𝟏 𝒎𝒐𝒍
𝟒𝟎 𝒎𝑳 𝑬𝒕𝑶𝑯
= 𝟎. 𝟔𝟖𝟐𝟒 𝒎𝒐𝒍 𝑬𝒕𝑶𝑯
𝒎𝑳 𝟒𝟔. 𝟎𝟕𝒈
𝑿𝒘𝒂𝒕𝒆𝒓
𝟑. 𝟑𝟐𝟏 𝒎𝒐𝒍
=
= 𝟎. 𝟖𝟐𝟗𝟓
𝟑. 𝟑𝟐𝟏 + 𝟎. 𝟔𝟖𝟐𝟒 𝒎𝒐𝒍
𝑿𝑬𝒕𝑶𝑯
𝟎. 𝟔𝟖𝟐𝟒 𝒎𝒐𝒍
=
= 𝟎. 𝟏𝟕𝟎𝟓
(𝟑. 𝟑𝟐𝟏 + 𝟎. 𝟔𝟖𝟐𝟒) 𝒎𝒐𝒍
Raoult’s Law – Example: Two
Volatile Components
Now find the vapor pressures of the two components:
Pwater over soln = XwaterP°water
Pwater over soln = 0.8295 (23.76 torr) = 19.71 torr
Palcohol over soln = XalcoholP°alcohol
Palcohol over soln = 0.1705 (54.68 torr) = 9.32 torr
Raoult’s Law – Example: Two
Volatile Components
Pwater over solution = 19.71 torr
Palcohol over solution = 9.32 torr
Now use Dalton’s Law of Partial Pressures to find
the total vapor pressure of the solution:
Ptot = 19.71 + 9.32 = 29.0 torr
Note that only the final answer is rounded.
Boiling Point of a Solution
The boiling point of water in an open container
is the temperature at which the vapor pressure
of water equals the prevailing atmospheric
pressure.
• Our sugar solution at 25°C has a lower vapor
pressure than water at 25°C.
• This means the temperature at which the sugar
solution boils will be higher (102.8°C) than the
temperature at which water boils (100.0°C).
This is called boiling point elevation.
Boiling Point Elevation
The relationship between boiling point
elevation and the number of particles of
solute in the solution is given by
ΔTb = Kbm
where ΔTb = Tbp(solution) - Tbp(pure solvent)
Kb is the molal boiling-point-elevation constant
and is for the solvent.
m is the molality of particles from the solute.
Boiling Point Elevation
Now you can calculate the boiling point of our
sugar solution yourself (Kb of water is
0.51°C/m):
ΔTb = Kbm
The molality of solute particles in our sugar solution is
the same as the molality of the sugar itself.
m = mol sugar = 0.660 mol sugar
kg water
0.11815 kg water
= 5.586 m
ΔTb = (0.51°C/m) (5.586m)
= 2.8°C
ΔTb = Tbp(solution) - Tbp(pure solvent) = 2.8°C
Tbp(solution) = 100.00°C + 2.8°C = 102.8°C
Boiling Point Elevation - Electrolytes
Electrolytes dissolve in water to form
ions. Each ion is a solute particle.
ΔTb = Kbm
If we made our solution up with 0.660 mol of NaCl
instead of sugar, the boiling point elevation would
be different from that of sugar.
m=
0.660 mol NaCl =
0.11815 kg water
1.32 mol ions*
= 11.18 m
0.11815 kg water
*NaCl(aq)  Na+(aq) + Cl-(aq)
1 mol NaCl in solution gives rise to 2 mol ions.
Boiling Point Elevation - Electrolytes
Electrolytes dissolve in water to form
ions. Each ion is a solute particle.
ΔTb = Kbm
ΔTb = (0.51°C/m) (11.18 m)
= 5.7°C
ΔTb = 5.7°C = Tbp(solution) - Tbp(pure solvent)
5.7°C + 100.00°C = Tbp(solution) = 105.7°C
Boiling Point Elevation - Electrolytes
ΔTb = Kbm
If we made our solution up with 0.660
mol CaCl2 instead of sugar or salt, the
boiling point elevation would be even
more, because CaCl2 dissolves in water
to release 3 ions per mol.
Freezing Point Depression
The addition of a nonvolatile solute
to a solution lowers the freezing
point of the solution relative to that
of the pure solvent.
Freezing Point Depression
The relationship between freezing point
depression and the number of particles of
solute in the solution is given by
ΔTf = Kfm
note the
difference!!
where ΔTf = Tfp(pure solvent) - Tfp(solution)
Kf is the molal freezing-point-depression
constant and is for the solvent.
m is the molality of particles from the solute.
Freezing Point Depression
We will now calculate the freezing point of our
sugar solution (Kf of water is 1.86°C/m):
ΔTf = Kfm
m = mol sugar = 0.660 mol sugar = 5.586 m
kg water
0.11815 kg water
ΔTf = (1.86°C/m) (5.586m) = 10.4°C
ΔTf = Tfp(pure solvent) - Tfp(solution) = 10.4°C
Tfp(solution) = Tfp(pure solvent) - ΔTf
Tfp(solution) = 0.00°C – 10.4°C = -10.4°C
Freezing Point Depression
We will now calculate the freezing point of our
salt solution:
ΔTf = Kfm
m = mol ions = 1.32 mol ions
kg water 0.11815 kg water = 11.17 m
ΔTf = (1.86°C/m) (11.17m) = 20.8°C
ΔTf = Tfp(pure solvent) - Tfp(solution) = 20.8°C
Tfp(solution) = Tfp(pure solvent) - ΔTf
Tfp(solution) = 0.00°C – 20.8°C = - 20.8°C
If we had used CaCl2, Tfp(solution) would be even
lower. That’s why CaCl2 is sometimes used to salt
icy sidewalks.
Boiling Point Elevation and Freezing
Point Depression
Adding a
nonvolatile
solute to a
solvent
raises its
boiling point
and lowers
its freezing
point.
Boiling Point Elevation and Freezing
Point Depression
Another
way to look
at things:
Adding a
nonvolatile
solute to a
solvent
expands its
liquid
range.
Finding the Molar Mass of a Solute from
Boiling Point Elevation or Freezing Point
Depression Measurement
Using either ΔTb
= Kbm or ΔTf = Kfm
• If you know the mass of solute that is not an
electrolyte and the mass of solvent used to
make a solution, and
• you can measure the freezing point depression
or boiling point elevation of the solution,
• you can calculate the molar mass of the solute.
Finding the Molar Mass of a Solute from
Boiling Point Elevation or Freezing Point
Depression Measurement
ΔTf = Kfm = Kf mol solute
kg solvent
= Kf (mass solute)
(molar mass of solute)(kg solvent)
Rearranging the equation gives:
molar mass of solute = Kf (mass solute)
ΔTf (kg solvent)
Molar Mass from Boiling Point
Elevation Data – Example
A solution of 0.64 g of adrenaline in 36.0 g of CCl4
has a b.p. of 77.03°C. a) What is the molar mass
of adrenaline? b) What is the f.p. of the solution?
CCl4: b.p. (760 torr) = 76.54°C Kb = 5.02°C/m
m.p. (760 torr) = -22.3°C Kf = 29.8°C/m
ΔTb = Kbm
ΔTb = 77.03 – 76.54 = 0.49°C
m = ΔTb / Kb = 0.49°C
= 0.0976 m
5.02°C/m
Molar Mass from Boiling Point
Elevation Data – Example
A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a
b.p. of 77.03°C. a) What is the molar mass of adrenaline?
b) What is the f.p. of the solution?
CCl4: b.p. (760 torr) = 76.54°C Kb = 5.02°C/m
m.p. (760 torr) = -22.3°C Kf = 29.8°C/m
Molar mass of adrenaline = (5.02°C/m) (0.64 g)
0.49°C (0.0360 kg)
= 182 g/mol
(really 180)
Molar Mass from Boiling Point
Elevation Data – Example
A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a
b.p. of 77.03°C. a) What is the molar mass of adrenaline?
b) What is the f.p. of the solution?
CCl4: b.p. (760 torr) = 76.54°C Kb = 5.02°C/m
m.p. (760 torr) = -22.3°C Kf = 29.8°C/m
ΔTf = Kfm
ΔTf = 29.8°C (0.0976 m) = 2.908°C
m
Molar Mass from Boiling Point
Elevation Data – Example
A solution of 0.64 g of adrenaline in 36.0 g of CCl4 has a
b.p. of 77.03°C. a) What is the molar mass of adrenaline?
b) What is the f.p. of the solution?
ΔTf = 29.8°C (0.0976 m) = 2.908°C
m
ΔTf =Tf (CCl4) - Tf(soln) = 2.908°C
Tf(soln) = Tf(CCl4) - ΔTf = -22.3 - 2.908 = -25.2°C
Osmotic Pressure
• The last colligative property we
will study is osmotic pressure.
• It is based on the tendency of
solvent molecules to move
toward an area of lesser
concentration.
• This movement causes osmotic
pressure when the areas of
differing solvent concentration
are separated by a
semipermeable membrane.
Osmotic Pressure
Osmotic pressure is
the pressure that
must be applied to
the solution in order
to just stop the
movement of solvent
molecules into the
solution.
Osmotic Pressure
The equation relating
osmotic pressure (π) to
concentration is very
similar to the ideal gas
law
π = MRT
M = molarity particles in the
solution
R = gas constant
T = temperature in K
Hypertonic Solutions
Osmotic pressure plays an important role in living
systems. For example, the membranes of red blood
cells are semipermeable.
When we eat too
much salt, the high
concentration of
salt in our plasma
makes it
hypertonic relative
to the inside of the
red blood cell and
causes water to
diffuse out of the
red blood cells.
A red blood cell in a
hypertonic solution shrinks.
Hypotonic Solutions
When we perspire
heavily and then
drink a lot of water
(not gatorade), the
low concentration
of salt in our
plasma makes it
hypotonic relative
to the inside of the
red blood cell and
causes water to
diffuse into the red
blood cells.
A red blood cell in a
hypotonic solution expands
and may burst.
Isotonic Solutions
When we lose a lot of fluids and have to replace them,
the ideal situation is to receive fluids that are neither
hypertonic nor hypotonic. Fluids that have the same
osmotic pressure are said to be isotonic.
The osmotic pressure of blood is 7.7 atm at 37°C. What
concentration of saline solution (NaCl in sterile water) is
isotonic with blood at human body temperature?
Using M = π /RT,
M = molarity of solute particles =
7.7 atm
. = 0.30 M
0.08206 L-atm (310. K)
mol-K
Molarity of NaCl for isotonic saline = 0.15 M
In mass percent, an isotonic saline solution is 0.9% NaCl.