#### Transcript 2-3-2011

```2-3-2011
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Change in Freezing
Point
Common
Applications of
Freezing Point
Depression
Propylene glycol
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Ethylene
glycol –
small
animals
Freezing point depression
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Freezing-Point Depression
0
DTf = T f – Tf
T
0
is the freezing point of
the pure solvent
Tf is the freezing point of
the solution
f
0
T f > 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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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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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.
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dilute
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more
concentrated
Osmotic Pressure (p)
Higher P
Lower 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
A cell in an:
Dilute Soln
Conc. Soln
Conc
isotonic
solution
hypotonic
solution
hypertonic
solution
Dilute Soln
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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
Molecular mass from colligative
properties
From Molality
From Molarity
1. Get molality
2. Get no. of moles
Mol = m * kg solvent
3. FW = (g solute/no. mol)
1. Get molarity
2. Get no. of moles
Mol = M * VL
3. FW = (g solute/no. mol)
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Molecular Weight of Unknown
What is the molecular mass of a sample if
250 grams of the sample is placed into
1000 grams of water and the temperature
rose by 3.5°C?
DTb = Kb m
3.5 oC = (0.52 oC/m)* m
m = 6.73
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Find the number of moles in 1000 g
No. mol = 6.73 mol
No. mol = mass/FW
FW = 250/6.73 = 37 g/mol
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A 5.50 g of a newly synthesized compound
was dissolved in 250 g of benzene) kf =
5.12oC/m) and the freezing point
depression was found to be 1.2 oC .Find
the molecular mass of the compound.
DTf = kf m
1.2 oC = 5.12 oC/m * m
Molality = 0.199 m
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Mol = m * kg solvent
Mol solute = (0.199 mol/kg) * (0.250 kg)
mol solute = 0.0498 moles
mol solute = Wt solute/FW
FW = 5.50 g/0.0498 mol = 111 g/mol
It should be realized that a solvent with high
kf is an advantage for such experimental
calculations of molecular masses.
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Example
A solution of 0.85 g of a compound in 100 g
of benzene has a freezing point of 5.16 oC.
What are the molality and the molar mass
of the solute? The normal freezing point of
benzene is 5.5 oC, kf = 5.12 oC/m.
DTf = kf m
m = (5.5 – 5.16)/5.12 = 0.066 mol/kg
benzene
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Mol = m * kg solvent
Number of moles of compound = (0.066
mol/kg benzene)* 0.100 kg benzene =
6.6*10-3
FW = g compound/mol
FW = 0.85 g/(6.6*10-3 mol) = 128 g/mol
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Molecular mass from osmotic
pressure
A solution is prepared by dissolving 2.47 g of
an organic polymer in 202 mL of benzene.
The solution has an osmotic pressure of
8.63 mmHg at 21 0C. Find the molar mass
of the polymer.
p = MRT
Can calculate the molarity, where:
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Get the number of moles of polymer where 1 L
of polymer solution contains 4.53*10-4 mol
Mol polymer = (4.53*10-4 mol/1L soln) * 0.202 L
soln = 9.15*10-5 mol
Molar mass = g polymer/no. mol
Molar mass = (2.47 g / 9.15*10-5 mol) = 26,992
g/mol
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Remember that osmotic pressure can be
easily and accurately measured in mmHg.
However, it is not as easy to measure
freezing point depression when the
concentration of solute is very small.
Therefore, osmotic pressure is better
suited for calculation of molar mass of
solutes dissolved in solutions.
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Colligative Properties of Electrolyte Solutions
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
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nonelectrolytes
NaCl
CaCl2
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2
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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
Freezing Point Depression
At what temperature will a 5.4 molal
solution of NaCl freeze? Assume i
=2
Solution
∆Tf = Kf • m • i
∆Tf = (1.86 oC/m) • 5.4 m • 2
∆Tf = 20.1 oC
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FP = 0 – 20.1 = -20.1 oC
What is the freezing point depression of a 0.15 m
aqueous solution of Al2(SO4)3? (kf = 1.86 oC/m)?
Solution
Al2(SO4)3 = 2 Al3+ + 3 SO42-
Therefore, the overall effective molality of the
solution is:
m = 0.15 * 5 = 0.75, neglecting interionic
attractions
DTf = {1.86 oC/m}*0.75 m = 1.4 oC
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Vapor pressure lowering
A solution of CaCl2 (FW = 111 g/mol) was prepared
by dissolving 25.0 g of CaCl2 in exactly 500 g of
H2O. What is the expected vapor pressure at 80
oC (Po
oC = 355 torr)? What would the
at
80
water
vapor pressure of the solution be if CaCl2 were
not an electrolyte?
Psoln = xsolvent Posolvent
Therefore, calculate the number of moles of water
and CaCl2
nwater = 500g/{18.0 g/mol} = 27.8
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nCaCl2 = 25.0 g/{111 g/mol} = 0.225
nspecies = 3 * 0.225 = 0.675
xwater = 27.8/{27.8 + 0.675} = 0.975
Psoln = 0.975 * 355 torr = 346 torr
If CaCl2 were not an electrolyte, CaCl2 will not
dissociate and the number of moles of all
species of solute will be just 0.225
Xwater = 27.8/{27.8 + 0.225} = 0.993
Psoln = 0.993 * 355 torr = 352 torr
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Boiling point elevation
Find the boiling point elevation of a 0.100 m
MgSO4 aqueous solution (kb = 0.51 oC/m, i = 1.3).
DTb = ikb * m
DTb = 1.3 * 0.51 oC/m * 0.100m = 0.066 oC
If MgSO4 was not an electrolyte
DTb = 0.51 oC/m * 0.100 m = 0.051 oC.
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Selected Problems: 6, 7 – 13, 15 - 17, 20 24, 27, 29, 31, 35, 39 – 44, 49, 50 - 52, 54
– 56, 60 - 64, 70, 71, 73, 75 -77, 79, 80.
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