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Colligative Properties
On adding a solute to a solvent, the properties of
the solvent are modified.
• Vapor pressure
decreases
• Freezing point
decreases
• Boiling point
increases
• (osmotic pressure) increases
These changes are called COLLIGATIVE
PROPERTIES..
PROPERTIES
They depend only on the NUMBER of solute particles relative to
solvent particles, not on the KIND of solute particles.
Understanding
Colligative Properties
To understand colligative properties,
study the LIQUID
LIQUID--VAPOR
EQUILIBRIUM for a solution.
Understanding Colligative Properties
VP = Vapor Pressure
VP of H2O over a solution depends on the
number of H2O molecules per solute molecule.
Psolvent proportional to Xsolvent
Psolvent = Xsolvent • Posolvent
VP of solvent over solution
= (Mol fraction solvent)•(VP pure solvent)
RAOULT’S LAW
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Raoult’s Law
An ideal solution is one that obeys
Raoult’s law.
PA = XA • PoA
Because mole fraction of solvent, XA, is always
less than 1, then PA is always less than PoA.
The vapor pressure of solvent over a solution is
always LOWERED!
Raoult’s Law
Assume the solution containing 62.1 g of glycol
(C2H6O2) in 250. g of water is ideal. What is the
vapor pressure of water over the solution at 30 oC?
(The VP of pure H2O is 31.8 mm Hg data
must be given or use a chart)
Solution
Xwater = .933
Pwater = Xwater • Powater = (0.933)(31.8 mm Hg)
Pwater = 29.7 mm Hg
Vapor
Pressure
Lowering
Figure 14.12
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Changes in Freezing and
Boiling Points of Solvent
VP Pure solvent
1 atm
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
T
See Figure 14.12
Boiling Point Elevation and
Freezing Point Depression
∆t = K•m•i
A generally useful equation
i = van’t Hoff factor = number of particles produced
per formula unit.
Compound
i (expected) i (observed)
glycol
1.0
1
NaCl
2.0
1.9
MgCl2
3.0
2.7
See Table 11.6 on
p.542 for observed
The boiling point of a
solution is higher than
that of the pure solvent.
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Elevation of Boiling Point
Elevation in BP = ∆tBP = KBP•m · i
(where KBP is characteristic of solvent)
VP Pure solvent
1 atm
VP solvent
after adding
solute
P
BP solution
BP pure
solvent
T
Change in Boiling Point
Dissolve 62.1 g of glycol (C2H5O2H) in 250. g of
water. What is the BP of the solution?
kBP = 0.512 oC/molal for water
i = 1 since glycol is molecular
(see Table 11.5 p. 532)
Solution
1.Calculate solution molality = 4.00 m
2.∆tBP = kBP • m · i
∆tBP = (0.512 oC/m)(4.00 m)(1)
∆tBP = 2.05 oC
BP = 102.05 oC
Change in Freezing Point
Pure water
Ethylene glycol/water
solution
The freezing point of a solution is
LOWER than that of the pure solvent.
FP depression = ∆tFP = kFP•m · i
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Freezing Point Depression
Calculate the FP of a 4.00 molal glycol/water
solution.
kFP = 1.86 oC/molal (constants on table 11.5 p. 532)
Solution
∆tFP = kFP • m ·i
= (1.86 oC/molal)(4.00 m)(1)
∆tFP = 7.44 oC therefore new f.p. = -7.44 oC
Recall that ∆tBP = +2.05 ˚C for this solution.
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg
of water to lower FP to -10.00 oC?.
Solution
Calc. required molality
∆tFP = kFP • m · i
Recall that 1 mol NaCl(
NaCl(aq
aq))
-->
--> 1 mol Na+(
Na+(aq
aq)) + 1 mol Cl
Cl--(aq
aq))
Therefore i = 2 ions
10.00 oC = (1.86 oC/molal)
molal) • m · 2
Freezing Point Depression
How much NaCl must be dissolved in 4.00 kg of water to
lower FP to -10.00 oC?.
Solution
Concentration required = 2.69 molal
This means we need 2.69 mol of dissolved particles
per kg of solvent.
2.69 mol NaCl = x mol NaCl
1 kg water 4.00 kg H2O
X = 10.8 mol NaCl 58.5 g NaCl = 629 g NaCl
1 mol
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