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Colligative Properties. Test Chapter 15 Tuesday May 7, 2002. Colligative Properties. Colligative Properties are those properties of a liquid that may be altered by the presence of a solute. Examples of these properties are: the vapor pressure the freezing and boiling points and - PowerPoint PPT Presentation
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Colligative Properties
Test Chapter 15
Tuesday May 7, 2002
Colligative Properties
Colligative Properties are those properties of a liquid that may be altered by the presence of a solute.
Examples of these properties are: the vapor pressure the freezing and boiling points and the osmotic pressure.
All of these properties ultimately relate to the vapor pressure.
The vapor pressure of a solvent depends on how pure it is.
Raoult’s Law
Psolvent = Xsolvent Posolvent
Pure solvent
Mole fraction
Solution
Boiling Point Elevation
Pure water boils at 100ºC
A solution of water and sucrose boils at MORE than 100 ºC
The more sucrose we add the HIGHER the boiling point.
Freezing Point Depression
Pure water freezes at 0ºC
A solution of water and sucrose freezes at LESS than 0ºC
The more sucrose we add the LOWER the freezing point.
Impurities in a substance cause a change in its phase diagram by making the liquid region bigger
The addition of solute RAISES the boiling point and LOWERS the freezing point of a solvent.
MOLALITY
Molarity (M) number of moles per liter of solution
MOLALITY (m) number of moles per kilogram of solvent.
number of moles m = ---------------------------------- mass of solvent (kg)
Molality: Problem 1
You pour 12 g of KBr into a beaker that contains 600 mL of water. What will be the molality of the resulting solution?
M KBr = 39.098 + 79.904 = 119.002 g
n = 12/119.002 = 0.10084 mol KBr
mass (H2O) = 600 mL = 600 g = 0.60 kg
m = 0.10084 mol / 0.60 kg = 0.168 m
Molality: Problem 2
How many grams of MgF2 would you need to prepare a 0.75m solution of MgF2 using 220g of water?
M MgF2 = 24.305 + 2 x 18.996 = 62.297 g mass of Mg = 0.75m x 0.220kg x 62.297 g = 10.2790 g
Colligative Properties: Computations
Boiling Point elevation:
ΔTb = ٭ kb m Molality Coefficient of BP elevation
Number of particles
Change in BP (ºC)
etiin
Colligative Properties: Computations
Freezing Point depression:
ΔTf = ٭ kf m Molality Coefficient of FP depression
Number of particles
Change in FP (ºC)
etiin
Number of particles in solution When sucrose is dissolved in water, the molecules
remain as one particle.
C12H24O12(s) C12H24O12(aq) When sodium chloride is dissolved in water, it
dissociates into 2 particles: NaCl(s) Na+(aq) + Cl– (aq) When calcium chloride is dissolved in water, it
dissociates into 3 particles
CaCl2(s) Ca+ + 2 Cl–(aq) NaCl and CaCl2 are called electrolytes because
their solution conduct electricity.
The larger the number of particles released by a solute the greater its effect on the BP and FP.
So we would expect: NaCl to be twice as effective as sucrose CaCl2 to be thrice as effective as sucrose
The size of the particles is unimportant In real life ionic compounds do not
dissociate completely
Boltzmann curve for pure solvent
OSMOSIS is the movement of solvent through a membrane to equalize the concentration on both sides.