# Honors Chemistry Unit 6 � Chemical Equations

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```					 AP Chemistry
Unit 6 – Solutions
Lesson 7 – Colligative Properties
Book Section: 13.5
Colligative Properties
   Changes in colligative properties
depend only on the number of solute
particles present, not on the identity of
the solute particles.
   Among colligative properties are
   Vapor pressure lowering
   Boiling point elevation
   Melting point depression
   Osmotic pressure
Vapor Pressure

Because of solute-
solvent intermolecular
forces, higher
concentrations of
solutes make it harder
for solvent to escape
to the vapor phase.
Vapor Pressure

Therefore, the vapor
pressure of a solution
is lower than that of
the pure solvent.
Raoult’s Law
PA = XAPA
where
   XA is the mole fraction of compound A, and
   PA is the normal vapor pressure of A at
that temperature.

NOTE: This is one of those times when you
want to make sure you have the vapor
pressure of the solvent. (Since solvent is
evaporating)
Practice
   Glycerin (C3H8O3) is a nonvolatile
nonelectrolyte with a density of 1.26
g/mL at 25 ºC. Calculate the vapor
pressure at 25 ºC of a solution made by
adding 50.0 mL of glycerin to 500.0 mL
of water. The vapor pressure of pure
water at 25 ºC is 23.8 torr, and its
density is 1.00 g/mL.
Practice
   Glycerin (C3H8O3) is a nonvolatile
nonelectrolyte with a density of 1.26
g/mL at 25 ºC. Calculate the vapor
pressure at 25 ºC of a solution made by
adding 50.0 mL of glycerin to 500.0 mL
of water. The vapor pressure of pure
water at 25 ºC is 23.8 torr, and its
density is 1.00 g/mL.
   23.2 torr
Boiling Point Elevation and
Freezing Point Depression
Solute-solvent
interactions also
cause solutions to
have higher boiling
points and lower
freezing points than
the pure solvent.
Boiling Point Elevation
   The change in boiling point is proportional to the molality
of the solution:
Tb = Kb  m

where Kb is the molal boiling point elevation constant, a
property of the solvent.

   The change in freezing point can be found similarly:
Tf = Kf  m

   Here Kf is the molal freezing point depression constant of
the solvent.
Boiling Point Elevation and
Freezing Point Depression
   Note that in both
equations, T does
Tb = Kb  m
not depend on what
the solute is, but only
on how many
particles are             Tf = Kf  m
dissolved.
   Dissociation affects
the number of
particles.
Sample Problem
   Automotive antifreeze consists of ethylene
glycol, CH2(OH)CH2(OH), a nonvolatile
nonelectrolyte. Calculate the boiling point and
freezing point of a 25.0 mass % solution of
ethylene glycol in water.
   Kb for water = 0.51 ºC/m
   Kf for water = 1.86 ºC/m
Sample Problem
   Automotive antifreeze consists of ethylene
glycol, CH2(OH)CH2(OH), a nonvolatile
nonelectrolyte. Calculate the boiling point and
freezing point of a 25.0 mass % solution of
ethylene glycol in water.
   Kb for water = 0.51 ºC/m
   Kf for water = 1.86 ºC/m
   Boiling point = 102.7 ºC
   Freezing Point = -10.0 ºC
Sample Problem
   List the following aqueous solutions in order of
their expected freezing point: 0.050 m CaCl2,
0.15 m NaCl, 0.10 m HCl, 0.050 m CH3COOH,
0.10 m C12H22O11.
Sample Problem
   List the following aqueous solutions in order of
their expected freezing point: 0.050 m CaCl2,
0.15 m NaCl, 0.10 m HCl, 0.050 m CH3COOH,
0.10 m C12H22O11.

   Lowest: 0.15 m NaCl
   0.10 m HCl
   0.050 m CaCl2
   0.10 m C12H22O11
   0.050 m CH3COOH
Colligative Properties of
Electrolytes
   A 1 M solution of NaCl shows twice the effect on
boiling/freezing point as a 1 M solution of methanol,
because NaCl dissociates and doubles its particle count.
Osmotic Pressure

In osmosis, there is net movement of solvent
from the area of higher solvent
concentration (lower solute concentration)
to the are of lower solvent concentration
(higher solute concentration).
Osmotic Pressure
The pressure required to stop osmosis,
known as osmotic pressure, , is

n
=(       )RT = MRT
V
where M is the molarity of the solution.

When solutions have the same pressure
on both sides of the membrane, it is said
to be isotonic.
Sample Problem
   The average osmotic pressure of blood is 7.7
atm at 25 ºC. What molarity of glucose
(C6H12O6) will be isotonic with blood?
Sample Problem
   The average osmotic pressure of blood is 7.7
atm at 25 ºC. What molarity of glucose
(C6H12O6) will be isotonic with blood?
   0.31 M
HW: 13.58, 60, 66, 68, 70, 72, 74,
78
   Next week:
   Tuesday: Begin Unit 7 – The First Law of
Thermodynamics (5.1-5.2)
   Wednesday: Solutions Exam
   Thursday: Enthalpy Changes (5.3-5.4)

   Problem Set 5 Due Jan. 3

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