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Co

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Colligative Properties

• The properties of solvent that depend ONLY on the amount of solute added to it

•NOT depend on the identity of solute.

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Colligative Properties

•VAPOR PRESSURE

•BOILING POINT

•FREEZING POINT

•OSMOTIC PRESSURE

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Understanding Understanding Colligative PropertiesColligative Properties

To understand colligative properties, study To understand colligative properties, study the the LIQUID-VAPOR EQUILIBRIUMLIQUID-VAPOR EQUILIBRIUM for a for a solution.solution.

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1. Vapor 1. Vapor PressurePressure

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Understanding Understanding Colligative PropertiesColligative Properties

To understand To understand colligative colligative properties, properties, study the study the LIQUID-VAPOR LIQUID-VAPOR EQUILIBRIUMEQUILIBRIUM for a solution.for a solution.

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Vapor Vapor PressurPressur

e e LowerinLowerin

gg

Figure 14.12

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PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

PPsolventsolvent = X = Xsolventsolvent • P • Poosolventsolvent

Understanding Understanding Colligative PropertiesColligative Properties

VP of H2O over a solution depends on the

number of H2O molecules per solute

molecule.

Psolvent proportional to Xsolvent

VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)

RAOULT’S LAW

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PPAA = X = XAA • P • PooAA

PPAA = X = XAA • P • PooAA

Raoult’s LawRaoult’s LawAn ideal solution obeys Raoult’s law.

mole fraction A (XA): always less than 1, PA

is always less than PoA.

The vapor pressure of solvent over a

solution is always LOWERED!

11Raoult’s LawRaoult’s Law A solution contains 62.1 g of glycol (1 mol) in

250. g of water. What is the vapor pressure of water over the solution at 30 oC?

(The VP of pure H2O is 31.8 torr)Solution

Mole fraction Xg = 0.0672

Know Xg + XH2O = 1.000

XH2O = 1.000 - 0.0672 = 0.933

PH2O = XH2O • PoH2O = (0.933)(31.8 torr)

Pwater = 29.7 mm Hg

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Changes in Freezing and Changes in Freezing and Boiling Points of SolventBoiling Points of Solvent

See Figure 14.12See Figure 14.12

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

13Elevation of Boiling Point

∆TBP = KBP•m(where K(where KBPBP is characteristic of solvent) is characteristic of solvent)

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T

VP solventafter addingsolute

VP Pure solvent

BP puresolvent

BP solution

1 atm

P

T∆TBP

14

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Change in Boiling Point Change in Boiling Point What is the BP of the solution with 62.1 g of

glycol (1.00 mol) dissolved in 250. g of water?

KBP = +0.512 oC/molal for water Solution

1. Calculate solution molality = 4.00 m

2. ∆TBP = KBP • m

∆TBP = +0.512 oC/molal (4.00 molal)

∆TBP = +2.05 oC new BP = 100 + 2.05 oC = 102.05 oC

16Boiling Point Elevation Boiling Point Elevation and Freezing Point and Freezing Point

DepressionDepression ∆T = K•m•iii = van’t Hoff factor = # of particles produced / formula

unit. Compound Value of i glycol 1 NaCl 2

CaCl2 3

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Boiling Point D6C32

• The addition of solute raises the boiling point.

• The larger the “m” the greater the increase in temperature: NOT matter what you add

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WHY?

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3. Freezing Point:

•The addition of

solute lowers the

freezing point. •Salt is added to streets •ice cream freezes at different temp (depends

on flavor, more you add the lower the temp)

20Change in Freezing Change in Freezing Point Point

The freezing point of a solution is The freezing point of a solution is LOWERLOWER than that of the pure solvent.than that of the pure solvent.

FP depression = ∆TFP depression = ∆TFPFP = K = KFPFP•m•m

Pure waterPure water Ethylene glycol/water Ethylene glycol/water solutionsolution

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22Boiling Point Elevation Boiling Point Elevation and Freezing Point and Freezing Point

DepressionDepression ∆T = K•m•iii = van’t Hoff factor = # of particles produced / formula

unit. Compound Value of i glycol 1 NaCl 2

CaCl2 3

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Calculate the FP of a 4.00 molal glycol/water solution.

KFP = -1.86 oC/molal (Table 14.4)

Solution

∆TFP = KFP • m

∆TFP = (-1.86 oC/molal) (4.00 m)

∆TFP = -7.44 oC

FP = 0 - 7.44 oC = -7.44 oC∆TBP = +2.05 ˚C for this solution.

Freezing Point Depression

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How much NaCl (g) must be dissolved in 1.00 kg of water to lower FP to -10.00 oC?.

Calc. required molality

∆TFP = KFP • m

-10.00 oC = (-1.86 oC/molal) • m

Conc = 5.38 molal ions

Freezing Point Freezing Point DepressionDepression

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Conc req’d = 5.38 molal ionsWe need 5.38 mol dissolved particles / kg of

solvent.

NaCl(aq) --> Na+(aq) + Cl-(aq)need

5.38 mol / 2 = 2.69 mol NaCl / kg

2.69 mol NaCl / kg ---> 157 g NaCl / kg

Freezing Point Depression

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Vapor PressureVapor Pressure

Boiling PointBoiling Point

Freezing PointFreezing Point

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4. Osmosis• The movement of solvent

through a membrane from an area of higher to lower solvent concentration.

• Semipermeable Membrane- barrier allowing solvent through, but block larger solute particles.

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Osmosis Osmosis at the Particulate Levelat the Particulate Level

Figure 14.15

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4. Osmosis

• Osmotic Pressure- Pressure developed across a membrane that allows the diffusion of solvent particles.

• Chemistry behind your kidneys and making pickles

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OsmosisOsmosis

The semi-permeable membrane allows only the movement of solvent molecules.

Solvent Solution

Semipermeable membrane

Solvent Solution

Semipermeable membrane

Solvent molecules move from pure solvent to solution in an attempt to

make both have the same concentration of solute.

Driving force is entropy

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Process of OsmosisProcess of Osmosis

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Osmosic Pressure, ∏Osmosic Pressure, ∏Equilibrium is reached when Equilibrium is reached when

pressure — the pressure — the OSMOTIC OSMOTIC PRESSUREPRESSURE, , ∏∏ — — produced by extra solution produced by extra solution counterbalances pressure counterbalances pressure of solvent molecules of solvent molecules moving thru the moving thru the membrane.membrane.

∏ ∏ = cRT= cRT

((cc is conc. in mol/L) is conc. in mol/L)

Osmotic pressure

33Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 torr at 25 ˚C. Calc. molar mass of hemoglobin.

SolutionSolution

(a)(a) Calc. Calc. ∏∏ in atmospheres in atmospheres

∏∏ = 10.0 torr • (1 atm / 760 mmHg)= 10.0 torr • (1 atm / 760 mmHg)

= 0.0132 atm= 0.0132 atm

(b)(b) Calc. concentrationCalc. concentration

34Osmosis Osmosis Calculating a Molar MassCalculating a Molar Mass

c = 5.39 x 10-4 mol/L(c) Calc. molar mass

Molar mass = 35.0 g / 5.39 x 10-4 mol/L

Molar mass = 65,100 g/mol

SolutionSolution

(b)(b) Use ∏ = cRT

Conc = 0.0132 atm

(0.0821 L • atm/K • mol)(298K)Conc =

0.0132 atm(0.0821 L • atm/K • mol)(298K)

35OsmosiOsmosiss

• Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.

• Osmotic pressure in living systems: FIGURE 14.16

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Osmosis and Living Osmosis and Living CellsCells

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Reverse OsmosisWater

Desalination

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Colligative PropertiesColligative PropertiesOn adding a solute to a solvent, the

props. of the solvent are modified.

• Vapor pressure decreases• Melting point decreases• Boiling point increases

• Osmosis is possible (osmotic pressure)

They depend only on the NUMBER of solute particles relative to solvent

particles, not on the KIND of solute particles.