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

22

2

22

3

22

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

0

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

0

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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