Lecture 7.4 - Colligative Properties
Transcript of Lecture 7.4 - Colligative Properties
States Of Matter III:
Colligative
Properties
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Solution Types • Gas in gas
• Gas in liquid
• Gas in solid
• Liquid in liquid Miscible - refers to 2 or more liquids that are
infinitely soluble in one another. Immiscible - refers to 2 liquids that are not
soluble in one another and if mixed separate into 2 layers.
• Liquid in solid
• Solid in liquid
• Solid in solid2
Key terms
• Solution - A general term for a solute dissolved in a solvent. A homogeneous mixture of 2 or more components in which particles intermingle at the molecular level.
• Solvent - The component of a solution that is the greater quantity.
• Solute - The component of a solution that is the lesser quantity. 3
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“like dissolves like”
Two substances with similar intermolecular forces are likely to be soluble in each other.
• non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
• polar molecules are soluble in polar solvents
C2H5OH in H2O
• ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
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Energetics of Dissolving Process
H can be either + or -, it depends on– the enthalpy to break the crystal apart– the enthalpy of disrupting solvent structure– the enthalpy change for hydrating solute.
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Three types of interactions in the solution process:• solvent-solvent interaction• solute-solute interaction• solvent-solute interaction (hydration)
Molecular view of the formation of solution
Hsoln = H1 + H2 + H3
Energetics of Dissolving Process
Costs energy to disrupt solvent and solute
structure.H = +
Formation of solvent-solute interactions releases energy.
H = − 9
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Concentration UnitsThe concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass
% by mass = x 100%mass of solutemass of solute + mass of solvent
= x 100%mass of solutemass of solution
Mole Fraction (X)
XA = moles of A
sum of moles of all components
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Concentration Units Continued
M =moles of solute
liters of solution
Molarity (M)
Molality (m)
m =moles of solute
mass of solvent (kg)
Example Problem
1.36 g of MgCl2 are dissolved in 47.46 g of water to give a solution with a final volume of 50.00 mL. Calculate the concentration of the solution in mass %, ppm, mole fraction, molarity, and molality.
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mass percent
%78.2%10046.4736.1
36.1100
solution mass
solute mass% mass
gg
g
p a r t p e r m i l l i o n
278001046.4736.1
36.110
solution mass
solute massppm 66
gg
g
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mole fraction
OH mol 63.2OH g 8.021
OH mol 1OH g 46.47
MgCl mol1043.1MgCl g 5.29
MgCl mol 1MgCl g 36.1
1039.5OH mol 63.2MgCl mol 0.0143
MgCl mol 0.0143
moles total
MgCl moleX
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22
2
22
3
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2
2MgCl2
and
where
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Molality
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MgCl m 301.0kg 04746.0
MgCl mol 0143.0
solvent of kg
MgCl molMgCl m
molarity
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MgCl M 286.0L 0500.0
MgCl mol 0143.0
solution of L
MgCl molMgCl M
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What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL?
m =moles of solute
mass of solvent (kg)M =
moles of solute
liters of solution
Assume 1 L of solution:5.86 moles ethanol = 270 g ethanol927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
m =moles of solute
mass of solvent (kg)=
5.86 moles C2H5OH
0.657 kg solvent= 8.92 m
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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 Tb = Kb m
Freezing-Point Depression Tf = Kf m
Osmotic Pressure () = MRT
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Raoult’s LawThe presence of a nonvolatile solute lowers the
vapor pressure of the solvent.
0solventsolventsolution PP
Psolution = Observed Vapor pressure of
the solution
P0solvent = Vapor pressure of the pure solvent
solvent = Mole fraction of the solvent
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Liquid-liquid solutions in which both components are volatile
Modified Raoult's Law:Modified Raoult's Law:
00BBAABATOTAL PPPPP
P0 is the vapor pressure of the pure solvent
PA and PB are the partial pressures
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Raoult’s Law – Ideal SolutionRaoult’s Law – Ideal SolutionA solution that obeys Raoult’s Law is called an
ideal solution
•When Hsoln = 0•Solvent-Solvent, Solute-Solvent, and Solute-Solute interactions are similar
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Negative Deviations from Raoult’s LawNegative Deviations from Raoult’s Law
Strong solute-solvent interaction results in a vapor pressure lower than predicted
Exothermic mixing = Negative deviation
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Positive Deviations from Raoult’s LawPositive Deviations from Raoult’s Law
Weak solute-solvent interaction results in a vapor pressure higher than predicted
Endothermic mixing = Positive deviation
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PA = XA P A0
PB = XB P B0
PT = PA + PB
PT = XA P A0 + XB P B
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Ideal Solution
Why does Raoult’s law work?
• Nature favors disorder
- A solution is more disordered than pure solvent
- Solvent molecules have less tendency to leave solution
• Solute particles interfere with solvent molecules
- solute particles occupying surface of solution lower the probability of high KE solvent molecules reaching and escaping from surface 27
• What is the vapor pressure of a solution made of 10.0 g of glucose and 100.0 g water at 37.0oC? (Vapor pressure of water at 37oC is 47.1 torr.)
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M o l e H 2 O = 1 0 0 g x 1 m o l / 1 8 g H 2 O = 5 . 5 6 m o l H 2 O
M o l e C 6 H 1 2 O 6 = 1 0 . 0 g C 6 H 1 2 O 6 x 1 m o l / 1 8 0 g C 6 H 1 2 O 6
= 0 . 0 5 5 6 m o l e C 6 H 1 2 O 6
torr46.6 torr)47.1)(990.0(
990.0mol 0.056)(5.56
OH mol 5.56
moles total
OH mol
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Pure
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OHOHOH
OH
PXP
X
o r t h e v a p o r p r e s s u r e l o w e r i n g i s4 7 . 1 t o r r – 4 6 . 6 t o r r = 0 . 5 t o r r
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Boiling point elevation
• A non-volatile solute raises the boiling point of a solvent.
Tb = Kb m where
Tb = boiling point elevation
– Kb = a constant
– m = molality
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Boiling-Point Elevation
Tb = Tb – T b0
Tb > T b0 Tb > 0
T b is the boiling point of the pure solvent
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T b is the boiling point of the solution
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point elevation constant (0C/m) for a given solvent
Freezing point depression
• A non-volatile solute depresses the freezing point of a solvent.
Tf = Kf m where
Tf = freezing point depression
– Kf = a constant
– m = molality
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Freezing-Point Depression
Tf = T f – Tf0
T f > Tf0 Tf > 0
T f is the freezing point of the pure solvent
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T f is the freezing point of the solution
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point depression constant (0C/m) for a given solvent
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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.
Tf = Kf m
m =moles of solute
mass of solvent (kg)= 2.41 m=
3.202 kg solvent
478 g x 1 mol62.01 g
Kf water = 1.86 oC/m
Tf = Kf m = 1.86 oC/m x 2.41 m = 4.48 oC
Tf = T f – Tf0
Tf = T f – Tf0 = 0.00 oC – 4.48 oC = -4.48 oC
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Osmotic Pressure ()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 () is the pressure required to stop osmosis.
dilutemore
concentrated
Semipermeable Membrane Up Close
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Osmotic PressureOsmotic Pressure
The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution
Initially At Equilibrium39
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A cell in an:
isotonicsolution
hypotonicsolution
hypertonicsolution
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HighP
LowP
Osmotic Pressure ()
= MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
solvent solution
time
R = 0.0821 Latm/molK
Example Problem
For a solution containing 3.00 g of pepsin in 10.0 mL of solution π = 0.213 atm at 25oC. What is the molecular mass of pepsin?
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Chemistry In Action: Reverse Osmosis
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Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution 0.2 m ions in solution
van’t Hoff factor (i) = actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
nonelectrolytesNaCl
CaCl2
i should be
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Dissociation Equations and Dissociation Equations and the Determination of the Determination of ii
NaCl(s)
AgNO3(s)
MgCl2(s)
Na2SO4(s)
AlCl3(s)
Na+(aq) + Cl-(aq)
Ag+(aq) + NO3-(aq)
Mg2+(aq) + 2 Cl-(aq)
2 Na+(aq) + SO42-(aq)
Al3+(aq) + 3 Cl-(aq)
i = 2
i = 2
i = 3
i = 3
i = 445
Ideal vs. Real van’t Hoff FactorIdeal vs. Real van’t Hoff Factor
The ideal van’t Hoff Factor is only achieved in The ideal van’t Hoff Factor is only achieved in VERY DILUTEVERY DILUTE solution. solution.
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Boiling-Point Elevation Tb = i Kb m
Freezing-Point Depression Tf = i Kf m
Osmotic Pressure () = iMRT
Colligative Properties of Electrolyte Solutions
Suspensions and Colloids
Suspensions and colloids are NOT solutions.
Suspensions: The particles are so large that they settle out of the solvent if not constantly stirred.
Particle size > 1000nmBlood, paint, aerosols, muddy water
Colloids: The particles intermediate in size between those of a suspension and those of a
solution.Particle size ~ 2-1000nm
Milk, fog, butter
Solution – smallest particles < 2nm 48
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A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.
Colloid versus solution
• collodial particles are much larger than solute molecules
• collodial suspension is not as homogeneous as a solution
• colloids exhibit the Tyndall effect
The Tyndall EffectColloids scatter light,
making a beam visible. Solutions do
not scatter light.
Which glass contains a colloid? solutioncolloid
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Types of colloids
Aerosol – liquid in gas
Sol -- solid in liquid like protein particles in milk
Solid Aerosol – solid in gas
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Gel – a solid emulsion which is soft but holds its shape like Jell-O
Emulsion – liquid in liquid like oil droplets in mayonnaise.
Foams – gases in liquids like whipped cream
Solid emulsion – liquid in a solid like milk in butter
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Homework
0solution solvent solventP P
Tb = iKb m
Tf = iKf m
• π = iMRT (R = 0.0821 L atm mol(R = 0.0821 L atm mol-1-1KK-1-1))
p. 548 # 25 - 32, 37- 40p. 549 #45 – 49, 53p. 550 # 58, 62, 64, 66, 69, 70p. 551 # 74, 75, 79AP Questions – IMF, Solids, Solutions (on Learning Point)
van’t Hoff factor (i) = actual number of particles in soln after dissociation
number of formula units initially dissolved in soln53
• Tartaric acid can be produced from crystalline residues found in wine vats. It is used in baking powders and as an additive in foods. Analysis show that it is 32.3%C, 3.97% H, and the remainder O. When 1.161 g tartaric acid is dissolved in 11.23 g water, the solution freezes at –1.26oC. Determine the empirical and molecular formula for tartaric acid.
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• Calculate the FP and BP of a solution containing 100 g of ethylene glycol (C2H6O2) in 900 g H2O.
• For water Kb = 0.52 oC/m
• Kf = 1.86 oC/m
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• The freezing point depression constants for the solvents cyclohexane and naphthalene are 20.1oC/m and 6.94oC/m respectively. Which would give a more accurate determination by freezing point depression of the molar mass of a substance that is soluble in either solvent? Why?
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