Transcript Slide 1

ว30231 ปริมาณสัมพันธ์ สถานะของสาร และเคมีไฟฟ้ า
สมบัติของสารละลาย
(Colligative properties)
นายศราวุทธ แสงอุไร
ครูวิชาการสาขาเคมี โรงเรียนมหิดลวิทยานุสรณ์
วันที่ 13 พฤศจิกายน 2552
Colligative properties ผูส้ อน: อ.ศราวุทธ แสงอุไร
สมบัติของสารละลาย (Colligative properties)
• Vapor and osmotic pressures, bp, and mp are
colligative properties
– Depend on relative of solute and solvent particles
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For Examples:
• Vapor Pressure Reduction
– Related to boiling point
• Freezing Point Depression
– Salt on the road
– Anti-freeze in your radiator
• Boiling Point Elevation
– Anti-freeze in your radiator
• Osmotic Pressure
– Membrane diffusion
– The Great Sugar Fountain!
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Vapor Pressure
• Remember:
– Equilibrium vapor pressure
Pressure of vapor when liq and vapor in equilibrium
at specific temp
• Vapor pressure of soln lower than pure solvent
vapor pressure
• Vapor pressure of solvent  relative # of
solvent molecules in soln
– i.e., solvent vapor pressure  solvent mole fraction
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Vapor Pressure
At equilibrium, the rate of evaporation (liquid to gas)
equals the rate of condensation (gas to liquid). The
amount of gas is the “vapor pressure”
T=K
P
P = atm
Surface of liquid
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Vapor Pressure
At equilibrium, the rate of evaporation (liquid to gas)
equals the rate of condensation (gas to liquid). The
amount of gas is the “vapor pressure”
T=K
P
P = atm
Surface of liquid
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Vapor Pressure
At equilibrium, the rate of evaporation (liquid to gas)
equals the rate of condensation (gas to liquid). The
amount of gas is the “vapor pressure”
T=K
P
P = atm
Surface of liquid
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Raoult’s Law
Psolution = Xsolvent  P°solvent
So if 75% of molecules in soln are solvent molecules
(0.75 = Xsolvent)
– Vapor pressure of solvent (Psolvent) = 75% of
P°solvent
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Colligative Properties of Nonelectrolyte Solutions
P1 = X1 P 10
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10
- P1 = DP = X2 P 10
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X2 = mole fraction of the solute
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Ideal Solution
PA = XA P A0
PB = XB P 0B
PT = PA + PB
PT = XA P A0 + XB P 0B
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PT is greater than
predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
Force
Force
< A-A & B-B
A-B
Force
Force
Force
> A-A & B-B
A-B
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Problem
The vapor pressure of pure acetone (CH3COCH3) at
30°C is 0.3270 atm. Suppose 15.0 g of benzophenone,
C13H10O (MW = 182.217 g/mol), is dissolved in 50.0 g
of acetone (MW = 58.09 g/mol).
– Calculate the vapor pressure of acetone
above the resulting solution.
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Solution
13
mol
solute:15.0g
 0.0823mol
182.217g
mol
solvent: 50.0g
 0.861mol
58.09g
0.861mol
X solvent 
 0.913
0.861mol 0.0823mol
Psolution  X solvent  P  solvent  0.913 0.3270atm  0.2986atm
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Problem
The vapor pressure of pure liquid CS2 is 0.3914 atm at
20°C. When 40.0 g of rhombic sulfur (a naturally
occurring form of sulfur) is dissolved in 1.00 kg of CS2,
the vapor pressure falls to 0.3868 atm.
– Determine the molecular formula of rhombic
sulfur.
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Psolution  X solvent  P  solvent
0.3868atm  X solvent  0.3914atm
X solvent  0.9882
mol
solvent:1.00kg 1.00 10 g 
 13.1mol
76.143g
13.1mol
0.9882
13.1mol molrhombic sulfur
3
molrhombic sulfur  0.156
40.0g
256g

0.156mol mol
256g
mol

 7.98  8
mol 32.066gsulfur
S8
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Boiling Pt. Elevation
Freezing Pt. Depression
ΔTb = T
boiling, solution
–T
boiling, pure solvent
= Kb m
ΔTf = T
freezing, solution
–T
freezing, pure solvent
= - Kf m
m = molality of the solution
Kb = boiling constant
Kf = cryoscopic constant
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Freezing is a dynamic equilibrium between
melting and freezing.
T=K
P
P = atm
Surface of liquid
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Freezing is a dynamic equilibrium between
melting and freezing.
T=0
oC
P
P = 1atm
Surface of liquid
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Boiling Point Elevation
At equilibrium, the rate of evaporation (liquid to gas)
equals the rate of condensation (gas to liquid)
P
T = 100 oC
P = 1 atm
Surface of liquid
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Boiling Point Elevation
P
T = 100 oC
P = 1 atm
Surface of liquid
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Boiling Point Elevation
T=K
P
P = atm
Surface of liquid
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Boiling-Point Elevation
DTb = Tb – T b
0
T b0is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b 0
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
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Freezing-Point Depression
DTf = T f 0– Tf
T
0
f
is the freezing point of
the pure solvent
T f is the freezing point of
the solution
T f0> Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
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Colligative properties ผูส้ อน: อ.ศราวุทธ แสงอุไร
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Boiling Pt. Elevation
Freezing Pt. Depression
ΔTb = T
boiling, solution
–T
boiling, pure solvent
= Kb m
ΔTf = T
freezing, solution
–T
freezing, pure solvent
= - Kf m
m = molality of the solution
Kb = boiling constant
Kf = cryoscopic constant
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What is the freezing point of a solution containing 478 g
of ethylene glycol (antifreeze) in 3202 g of water? The
molar mass of ethylene glycol is 62.01 g.
DTf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg)
478 g x
1 mol
62.01 g
=
= 2.41 m
3.202 kg solvent
DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
DTf = T 0f – Tf
Tf = T 0f – DTf = 0.00 0C – 4.48 0C = -4.48 0C
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Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 10
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
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p = MRT
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• How many grams of sucrose (C12H22O11) are
needed to lower the freezing point of 100 g of
water by 3° C?
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ΔTf = - Kf m
We want to decrease the freezing point by 3°C
-3° C = -(1.86 °C/molal) m
m=1.61 molal = 1.61 moles solute/kg solvent
NOTE: Kf is the WATER cryoscopic constant
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1.61 moles solute = x moles solute
1 kg solvent
0.100 kg water
0.161 moles sucrose x 342 g sucrose = 55.1 g sucros
1 mole sucrose
342 g/mol = 11*12.01 g/mol + 22x1.008 g/mol + 11x16 g/mol
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How many grams of NaCl are needed to lower the
freezing point of 100 g of water by 3 °C?
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The Answer
ΔTf = = - Kf m
We want to decrease the freezing point by 3°C
-3° C = -(1.86 °C/molal) m
m=1.61 molal = 1.61 moles solute/kg solvent
NOTE: Kf is the WATER cryoscopic constant
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1.61 moles solute * 0.100 kg water = 0.161 moles solute
1 kg solvent
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It’s all about the of particles
1.61 moles solute * 0.100 kg water = 0.161 moles solute
1 kg solvent
BUT NaCl is an electrolyte:
NaCl
Na+ + Cl-
You get 2 moles of solute per mole NaCl
0.161 moles solute * 1 mol NaCl * 58.45 g NaCl = 4.7 g NaCl
2 mol solute
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1 mole NaCl
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Colligative Properties of Electrolyte Solutions
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0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
Colligative properties are properties that depend only on the number of
solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution
van’t Hoff factor (i) =
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes
1
NaCl
2
3
CaCl2
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Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
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p = iMRT
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Freezing Point Depression
At what temperature will a 5.4 molal solution of
NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
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Osmotic Pressure
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Osmotic pressure is the “funky” colligative property, but it
is very important biologically
Osmotic pressure is the pressure required to overcome
the natural pressure exerted by a solution by virtue of
having a concentration.
Osmotic pressure looks just like the ideal gas law:
ΠV = nRT
where Π is osmotic pressure
Π= (n/V) RT = M RT
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What Happens?
1 M NaCl
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4 M NaCl
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The water moves from the 1 M side
to the 4 M side. Why?
1 M NaCl

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4 M NaCl
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OSMOTIC PRESSURE
Π= M RT
On the 1 M side, the pressure is:
Π= M RT = 1 M (0.0821 Latm/mol K)(298 K)
Π=24.4 atm
On the 4 M side, the pressure is:
Π= M RT = 4 M (0.0821 Latm/mol K) (298
K)
Π= 97.9 atm
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Both solutions “push” on the membrane
The bigger push wins!
24.4 atm
1 M NaCl

4 M NaCl
97.9 m
73.5 atm
Note the direction of the arrows. Osmotic pressure
is pushing AGAINST the solution.
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Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but blocks
the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
more
concentrated
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Osmotic Pressure (p) in air
High
P
Low
P
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
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p = MRT
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A cell in an:
isotonic
solution
hypotonic
solution
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hypertonic
solution
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Chemistry In Action:
Desalination
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แหล่งอ้างอิง
• Martin S. Silberberg, Chemistry: The Molecular Nature
of Matter and Change, McGraw-Hill Higher Education,
2004
• Raymond Chang, Chemistry, Williams College,
McGraw-Hill Higher Education, 2002
Colligative properties ผูส้ อน: อ.ศราวุทธ แสงอุไร