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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.
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.
Vapor pressure of the solution
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 x 0.99707 g x 1 mol mL 18.015 g
mol sugar = 226 g x 1 mol 342.30 g
Xwater = 6.559 = 0.9085 7.2192
= 6.559 mol
= 0.660 mol
What is the vapor pressure at 25°C (room temperature) of a solution made by adding 226 g (1 cup) sugar (C12H22O11) to ½ cup of water?
Pwater over sugar solution = XwaterP°water
Pwater over sugar solution = (0.9085) (23.76 torr)
Adding sugar to the water lowered its vapor pressure. Adding more sugar would lower it still more.
= 21.59 torr
from Appendix B
Raoult’s Law – Example: Nonvolatile Solute
What is the vapor pressure at 25°C of 80-proof alcohol (40% alcohol by volume)?
Alcohol: C2H5OH, MM = 46.07 g, v.p.(25°C) = 54.68 torr, density (25°C) = 0.786 g/mL
Water: MM = 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 Xwater:
60 mL water x 0.997 g x 1 mol mL 18.015 g
40 mL alcohol x 0.786 g x 1 mol mL 46.07 g
Xwater = 3.321 3.321+0.682
Raoult’s Law – Example: Two Volatile Components
= 3.321 mol
= 0.682 mol
= 0.8296
Pwater over water/alcohol solution = XwaterP°water
Pwater over w/alc soln = 0.8296 (23.76 torr) = 19.71 torr
Palcohol over water/alc solution = XalcoholP°alcohol
Xalcohol = 0.682 4.003
Palcohol over w/alc soln = 0.170 (54.68 torr) = 9.32 torr
Raoult’s Law – Example: Two Volatile Components
= 0.170
Pwater over water/alcohol solution = 19.71 torr
Palcohol over water/alcohol 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
Raoult’s Law – Example: Two Volatile Components
Boiling Point of a Solution• For water in an open container, the boiling
point 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
ΔTb = (0.51°C/m) (5.586m)
ΔTb = Tbp(solution) - Tbp(pure solvent) = 2.8°C
Tbp(solution) = 100.00°C + 2.8°C = 102.8°C
= 5.586 m
= 2.8°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 salt = 1.32 mol ions 0.11815 kg water 0.11815 kg water
= 11.18 m
Boiling Point Elevation - Electrolytes
Electrolytes dissolve in water to form ions. Each ion is a solute particle.ΔTb = Kbm and m = 11.18
ΔTb = (0.51°C/m) (11.18 m)
ΔTb = Tbp(solution) - Tbp(pure solvent) = 5.7°C
Tbp(solution) = 100.00°C + 5.7°C = 105.7°C
= 5.7°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
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.
note the difference!!
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 kg water 0.11815 kg water
ΔTf = (1.86°C/m) (5.586m)
ΔTf = Tfp(pure solvent) - Tfp(solution) = 10.4°C
Tfp(solution) = 0.00°C – 10.4°C = -10.4°C
= 5.586 m
= 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
ΔTf = (1.86°C/m) (11.17m)
ΔTf = Tfp(pure solvent) - Tfp(solution) = 20.8°C
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.
= 11.17 m
= 20.8°C
Freezing Point Depression
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)
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/mΔTb = 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) m
Molar Mass from Boiling Point Elevation Data – Example
= 2.908°C
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) - 2.908 = -22.3 - 2.908 = -25.2°C
Molar Mass from Boiling Point Elevation Data – Example
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 SolutionsWhen 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 . 0.08206 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.
= 0.30 M