Honors Chemistry Unit 6 � Chemical Equations

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Honors Chemistry Unit 6 � Chemical Equations Powered By Docstoc
					 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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