Lecture 50 Solutions III

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Lecture 50 - Solutions III
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Today...

• Colligative properties
• vapor pressure of a solution
• freezing point depression
• boiling point elevation
• osmotic pressure

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Colligative Properties

• In a solution
• pvap is lower
• Tb is higher
• Tf is lower
• than in the pure solvent
• Depends only on the amount of solute, not on its
  identity

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Vapor Pressure Lowering

• I. Solute is non-volatile

surface of pure solvent
surface of solution - less evaporation
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Vapor Pressure Lowering

• DGvap DHvap - TDSvap

same in pure solvent and solution
smaller in solution (more disorder in a solution)
Thus, DGvap is more positive for a solution
(less favorable)
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Vapor Pressure Lowering

• I. Solute is involatile
• psoln posolv ½ Xsolv

Vapor pressure of Pure solvent
(1 - Xsolute)
(Raoults Law)
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Vapor Pressure Lowering

• e.g. find psoln for 5 m glucose at 25oC
• 1. posolv pvap,H2O (25oC) 23.76 Torr

3. nglucose 5 mol nH2O 1000g/18g mol-1
55.5 mol
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Vapor Pressure Lowering
nH2O
55.5
4. Xsolv

0.917
nH2O nglucose
55.5 5
5. psoln posolv ½ Xsolv
23.76 Torr x 0.917 21.79 Torr
(would be the same for any 5 m solute)
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Salt Solutions

• N.B. salts dissociate!
• Na2SO4 ? 2 Na SO4-2

1 m
? 3 m
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Salt Solutions
- strong intermolecular forces - lower pvap
than Raoults Law predicts - non-ideal solution
ions
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Vapor Pressure Lowering

• II. Solute is Volatile
• Solvent A
• Solute B
• pA pAo XA

vapor pressure of A above solution
vapor pressure of pure A
similarly, pB pBo XB
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Vapor Pressure Lowering

• ptotal pA pB

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Vapor Pressure Lowering
DHsolngt0
pBo
ideal
pAo
ptot
DHsolnlt0
XA
0
1
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Boiling Point Elevation
S
L
P
1
G
DTb
Tb
T
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Boiling Point Elevation

• DTb µ Dpvap
• Dpvap µ msolute
• thus, DTb µ msolute
• in fact, DTb Kb msolute

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Boiling Point Elevation

• DTb Kb msolute

ebullioscopic constant
Kb is a function of the SOLVENT e.g. Kb (H2O)
0.51 oC kg mol-1
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Boiling Point Elevation

• e.g. find Tb of 0.5 m NaCl(aq)
• DTb Kb msolute
• 0.51 oC kg mol-1 x 1.0 mol kg-1
• 0.51oC
• i.e. Tb 100.51oC

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Boiling Point Elevation

• e.g. 18.0 g glucose in 150 g H2O
• boils at 100.34oC.
• Find the molar mass of glucose
• 1. DTb Kb msolute or
• msolute DTb / Kb
• 0.34oC / 0.51 oC kg mol-1
• 0.67 mol kg-1

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Boiling Point Elevation
nglucose
2. 0.67 mol kg-1
0.15 kg solvent
solving, nglucose 0.10 mol
18.0 g
3. molar mass of glucose
0.10 mol
180 g mol-1
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Boiling Point Elevation
S
L
P
1
G
DTb
DTf
Tb
T
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Freezing Point Depression

• DTf Kf msolute

cryoscopic constant
Kf is a function of the SOLVENT e.g. Kf (H2O)
1.86 oC kg mol-1
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Freezing Point Depression

• e.g. antifreeze, ethylene glycol(aq)

H-O
H-O
HCCH
how much must be added to 10.0 L H2O if Tf lt
-40oC?
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Freezing Point Depression

• 1. DTf Kf msolute or
• msolute DTf / Kf
• 40oC / 1.86 oC kg mol-1
• 21.5 mol (kg solvent)-1
• 2. 10.0 L H2O 10.0 kg H2O

nglycol
3. 21.5 mol kg-1
10.0 kg H2O
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Freezing Point Depression

• solving, nglycol 215 mol
• 4. 215 mol 215 mol x 62.1 g mol-1
• 13,350 g
• 13.35 kg

13.35
wt glycol
x 100 57
13.35 10.0
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Freezing Point Depression

• e.g. melting ice
• DTf Kf mNaCl
• assume mNaCl 3.25 m
• we predict DTf 1.86 (6.5)
• 12.1oC
• actual 11.9oC

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Freezing Point Depression
Cl-
Na
Cl-
Cl-
Cl-
Actual (effectively fewer particles)
Expected
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Road salt is ineffective below -20.7oC...

• ...saturated NaCl(aq) freezes at this temperature

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Osmosis
Osmotic Pressure (P)
solvent flow
solvent flow
pure solvent
solution
semi-permeable membrane
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Osmotic Pressure

• P MRT

P osmotic pressure M molarity of solute R
gas constant T absolute temperature
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Osmotic Pressure

• e.g. 1 mg of a protein is
• dissolved in 1 mL H2O.
• P 1.12 Torr at 25oC
• Find the MW of the protein.
• 1. P 1.12 Torr 1.12/760 0.00147 atm
• 2. R 0.082 L atm K-1 mol-1
• 3. T 25oC 298 K

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Osmotic Pressure
P
4. M
RT
0.00147 atm

0.082 L atm K-1 mol-1 (298 K)
6.01 x 10-5 mol L-1
5. 1 mg/mL 1 g/L
thus, 1 g 6.01 x 10-5 mol
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Osmotic Pressure
1.0 g
6. MW
6.01 x 10-5 mol
16,600 g mol-1
(a SMALL protein)
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Reverse Osmosis
pgtP
solvent flow!
pure solvent
solution