New Basic Chemistry 2 Solution ( Ideal Solution and Colligative Properties

7/28/2019 New Basic Chemistry 2 Solution ( Ideal Solution and Colligative Properties

http://slidepdf.com/reader/full/new-basic-chemistry-2-solution-ideal-solution-and-colligative-properties 1/83

BASIC CHEMISTRY 2

Lecturer : Irma Ratna K., M.Sc.Tech.SOLUTION ( IDEAL SOLUTIONAND COLLIGATIVE PROPERTIES)

Group 2:

1. Nurul Hanifah Aldi (3315126598)

2. Retno Ayu Puspita (3315126600)

3. Sela Devi Anggraeni (3315126605)

Bilingual Education Of Chemistry

2012



WHAT WE WILL BE DISCUSSSED?

Solution

Ideal Solution

Colligative Properties



SOLUTION

Homogenous mixture fromtwo or more substances

SoluteSubstance

SolventSubstance

A smaller numberof substances

A bigger number ofsubstances



SOLUTION

Saturated Solution

Unsaturated Solution

Supersaturatedsolution





Two substances with the forces betweenmolecules of the same is likely to be mutually

dissolved:

•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 )



IDEAL SOLUTION

• Ideal Solution is solution where the

attractive forces between the solute

and solvent molecules is equal to theforce of attraction between

molecules of solute or solvent

molecules, respectively.



IDEAL SOLUTION

Another feature is that the

ideal solution would be

the sum appropriate

volume volume itscomponents. In the non-

ideal solution, the sum of

the volume of pure solute

and pure solvent is not thesame as the volume of the

solution.

When the interaction

between the molecular

components of the

solution is equal to theinteraction between the

molecular components

are in a pure state, then

formed an idealization of the so-called ideal

solution



• Actually there is no mixture that could be

considered ideal. But some solution mixed

condition actually approached the ideal

situation. Here is an example:

*hexane dan heptane

* benzene dan methylbenzene

IDEAL SOLUTION



• In a solution, several large-energy molecules

can use energy to beat the appeal of

intermolecular liquid surface and then breakaway to become vapor.

• The smaller the intermolecular energy, the

more molecules that can escape at a certaintemperature.



• If you have a second solution, the same thing

happened. At a given temperature, most of the existing molecules will have enough

energy to escape from the surface of the

solution.



• In an ideal mixture of the two, the tendency of

the two kinds of molecules in it to escape

unchanged.



Raoult’s LawDescribes vapor pressure lowering mathematically.

• The lowering of the vapour pressure when a non-volatile solute is dissolved in a volatile solvent (A)can be described by Raoult’s Law:

PA = cAP°A

PA = vapour pressure of solvent A above the solutioncA = mole fraction of the solvent A in the solution

P°A = vapour pressure of pure solvent A

only the solvent (A) contributes to

the vapour pressure of the solution



What is the vapor pressure of water above a sucrose (MW=342.3 g/mol)

solution prepared by dissolving 158.0 g of sucrose in 641.6 g of water at

25 ºC?

The vapor pressure of pure water at 25 ºC is 23.76 mmHg.

mol sucrose = (158.0 g)/(342.3 g/mol) = 0.462 mol

mol water = (641.6 g)/(18 g/mol) = 35.6 mol

X water =mol water

(mol water)+(mol sucrose)

=35.6

35.6+0.462

= 0.98

P sol’n

= X water P water = (0.987)(23.76 mm Hg)

= 23.5 mm Hg



Mixtures of Volatile LiquidsBoth liquids evaporate & contribute to the vapor pressure



Raoult’s Law: Mixing Two Volatile Liquids

• Since BOTH liquids are volatile and contribute to thevapour, the total vapor pressure can be representedusing Dalton’s Law:

PT = PA + PB

The vapor pressure from each component follows Raoult’s Law:

PT = cAP°A + cBP°B

Also, cA + cB = 1 (since there are 2 components)



Benzene and Toluene

Consider a two solvent (volatile) system

– The vapor pressure from each component follows

Raoult's Law.

– Benzene - Toluene mixture:

• Recall that with only two components, cBz + cTol = 1

• Benzene: when cBz = 1, PBz = P°Bz = 384 torr &

when cBz = 0 , PBz = 0

• Toluene: when cTol = 1, PTol = P°Tol = 133 torr &

when cTol = 0, PBz = 0



384 torr

133 torr

X Benzene

X Toluene

0 1

1 0

P (Total)

P (Benzene)

P (Toluene)

133 torr

384 torr



COLLIGATIVE PROPERTIES

• Solution properties who does not

depend on the type of dissolvedsubstances but depending on the

number of particles



COLLIGATIVE PROPERTIES

• There are four common types of

colligative properties:

1. Vapor pressure lowering

2. Freezing point depression

3. Boiling point elevation

4. Osmotic pressure

• Vapor pressure lowering is the key to all

four of the colligative properties.



Lowering of Vapor Pressure and

Raoult’s Law

• Addition of a nonvolatile solute to a solution

lowers the vapor pressure of the solution.

– The effect is simply due to fewer solvent

molecules at the solution’s surface.

– The solute molecules occupy some of the spaces

that would normally be occupied by solvent.

• Raoult’s Law models this effect in ideal

solutions.




Raoult’s Law

• Derivation of Raoult’s Law:

P P

where P vapor pressure of solvent

P vapor pressure of pure solvent

mole fraction of solvent

solvent solvent solvent0

solvent

solvent0

solvent

X

in solution

X in solution



23


Raoult’s Law

• Lowering of vapor pressure, Psolvent, is defined

as:

0solventsolvent

0

solventsolvent

0

solvent

solvent

0

solventsolvent

)P1(

)P)((-PPPP

X

X



24

• Lowering of vapor pressure, Psolvent, is defined

as:

0

solventsolvent

0

solventsolvent

0

solvent

solvent

0

solventsolvent

)P1()P)((-P

PPP

X

X

Lowering of Vapor Pressure andRaoult’s Law



25

• Remember that the sum of the mole fractions

must equal 1.

• Thus X solvent + X solute = 1, which we can substituteinto our expression.

LawsRaoult'iswhich

PP

-1

0

solventsolutesolvent

solventsolute

X

X X

Lowering of Vapor Pressure andRaoult’s Law



26

• This graph shows how the solution’s vapor

pressure is changed by the mole fraction of

the solute, which is Raoult’s law.


Raoult’s Law



27

The vapor pressure of water is 17.5 torr at 20°C.

Imagine holding the temperature constant while

adding glucose, C6H12O6, to the water so that the

resulting solution has XH2O

= 0.80 and XGlu

= 0.20.

What is , the vapor pressure of water over the

solution 0

A A A P X P

torr X P X P A A A 5.1780.00

= 14 torr

Examples



h f ° i



The vapor pressure of pure water at 110°C is 1070torr. A solution of ethylene glycol and water has avapor pressure of 1.00 atm at 110°C. Assuming that

Raoult's law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290

P°H2O =1070 torr

PH2O

= 1 Atm = 760 torr

XH2O = ---------PH2O

P°H2O = ---------

760 torr

1070 torr= 0.71028

XH2O + XEG = 1

0.7103 + XEG = 1

1- 0.7103 = XEG

XEG =0.28972 = 0.290

M E l



30

More Examples

Sucrose is a nonvolatile, nonionizing solute inwater. Determine the vapor pressure lowering,at 27°C, of a solution of 75.0 grams of sucrose,C

12H

22O

11, dissolved in 180. g of water. The

vapor pressure of pure water at 27°C is 26.7torr. Assume the solution is ideal.

mol Suc g

Sucmol gSucnSuc 219.0

3.342

10.75

mol Watyer g

Water mol gWater nWater 99.9

18

1180

978541.02191.0991.9

991.9

UcWater

water Water

nn

n X

13.2697854.07.260 X torr X P P Water Water Water

Vapor Pressure Lowered = 26.7-26.1= 0.6



31

solution is made by mixing 52.1 g of propyl chloride,C

3H8Cl, and 38.4 g of propyl bromide, C

3H8Br. What

is the vapor pressure of propyl chloride in the

solution at 25°C? The vapor pressure of pure propylchloride is 347 torr at 25°C and that of pure propyl

bromide is 133 torr at 25°C. Assume that thesolution is an ideal solution.

6633.054.78

11.52 CP g CP mol CP g nCP

312.099.122

14.38

CB g

CBmol CB g nCB

67996.03122.06633.0

6633.0

PB PC

PCr

PC nn

n X

Torr X X P P PC PC PC 23695.235679964.03470

At 25°C l ti i t f 0 450 l f



32

. At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane,C5H

10. What is the mole fraction of cyclopentane in

the vapor that is in equilibrium with this solution?The vapor pressure of the pure liquids at 25°C are451 torr for pentane and 321 torr for cyclopentane.Assume that the solution is an ideal solution.

95.202451450.0 X X P P Pen Pen Pen

25.80321250.00 X X P P CPenCPenCPen

RT

V P n

RT

V P n

RT

PV n CPen

CPen Pen

Pen ;;

PenCPen

CPen

PenCPen

CPen

PenCPen

CPen

CPen P P

P

RT

V P

RT

V P RT

V P

nnn X

283.095.20225.80

25.80

CPen P



33

Boiling Point Elevation

• Addition of a nonvolatile solute to a solutionraises the boiling point of the solution abovethat of the pure solvent.

– This effect is because the solution’s vaporpressure is lowered as described by Raoult’s law.

– The solution’s temperature must be raised tomake the solution’s vapor pressure equal to theatmospheric pressure.

• The amount that the temperature is elevatedis determined by the number of moles of solute dissolved in the solution.



34


• Boiling point elevation relationship is:

solventfor the constantelevation point boilingmolalK

solutionof ionconcentratmolal

elevation point boilingT:where

K T

b

b

b b

m

m



35


• Example 14-4: What is the normal boiling

point of a 2.50 m glucose, C6H12O6, solution?

C101.28=C28.1+C100.0=solutiontheof PointBoiling

C28.1T

)50.2)(C/512.0(T

K T

000

0

b

0

b

b b

mm

m



Th increase in boilin i t l ti t th t f th



37

The increase in boiling point relative to that of thepure solvent, Tb, is directly proportional to thenumber of solute particles per mole of solvent

molecules.Molality expresses the number of moles of soluteper 1000 g of solvent, which represents a fixednumber of moles of solvent m K T bb

Solvent B.Point (°C) Kb (°C/m)

Freezing P.(°C)

Kf (°C/m)

Water, H2O 100.0 0.52 0.00 1.86

Benzen, C6H6 80.1 2.53 5.5 5.12

Ethanol, C2H6O 78.4 1.22 -114.0 1.99

Carbon tetrachloride, CCl4 76.8 5.02 -22 29.8

Chloroform,

CHCl3 61.2 3.63 -63.5 4.68

Automotive antifreeze consists of ethylene glycol



38

Automotive antifreeze consists of ethylene glycol,C2H6O2, a nonvolatile nonelectrolyte. Calculate theboiling point of a 25.0 mass percent solution of ethylene glycol in water.

Calculate the boiling point of a solution of 2 0 molal



39

Calculate the boiling point of a solution of 2.0 molal

of NaCl. Kb, water= 0.52 °C /mola.

t = Kbm

t = (0.52 °C/molal)(4.0 molal) =2.08 °C

NaCl(aq) Na+ + Cl-

2.0 m 2.0 m 2.0 m2.0 m + 2.0 m = 4.0m

BP = NBP +t = 100.00°C +2.08 °C = 102.08° C



40

Freezing Point Depression

• Addition of a nonvolatile solute to a solution

lowers the freezing point of the solution

relative to the pure solvent.

• See table 14-2 for a compilation of boiling

point and freezing point elevation constants.

http://d/Media/PowerPoint_Lectures/PowerPoint%20Media/Movies/14M08AN3.MOV






41


• Relationship for freezing point depression is:

T K

where: T freezing point depression of solvent

molal concentration of soltuion

K freezing point depression constant for solvent

f f

f

f

m

m



42


• Notice the similarity of the two relationships

for freezing point depression and boiling pointelevation.

• Fundamentally, freezing point depression and boiling point

elevation are the same phenomenon.

– The only differences are the size of the effect which is reflected in

the sizes of the constants, Kf & Kb.

• This is easily seen on a phase diagram for a solution.

mm b bf f K Tvs.K T



43




44


• Example 14-5: Calculate the freezing point of a

2.50 m aqueous glucose solution.

C4.65-=C4.65-C0.00=solutionof PointFreezing

C65.4T

)50.2)(C/(1.86T

K T

000

0

f

0

f

f f

mm

m



45


• Example 14-6: Calculate the freezing point of a

solution that contains 8.50 g of benzoic acid

(C6H5COOH, MW = 122) in 75.0 g of benzene,

C6H6.

You do it!



46


C0.72=C4.76-C5.48=F.P.

C76.4)929.0)(C/12.5(T

K T

solution.for thisdepressiontheCalculate.2

929.0COOHHCg122

COOHHCmol1

HCkg0.0750

COOHHCg50.8

HCkg

COOHHCmol?

molality!Calculate.1

000

00

f

f f

56

56

66

56

66

56

mm

m

m



47


• The size of the freezing point depressiondepends on two things:

1. The size of the Kf for a given solvent, which are wellknown.

2. And the molal concentration of the solution whichdepends on the number of moles of solute and themass of solvent.

• If Kf and mass of solvent are known, as is often

the case in an experiment, then we candetermine # of moles of solute and use it todetermine the molecular weight..

Example Problem 4



What is the freezing point of a solution containing 1499

g of ethylene glycol (antifreeze) in 3255 g of water? The

molar mass of ethylene glycol is 62.0 g\mol.

T f = K f m i

m =

moles of solute

mass of solvent (kg)= 7.43 m =

3.255 kg solvent

1499 g x1 mol

62.0 g

K f water = 1.86 0C/m

T f = K f m = 1.86 0C/m x 7.43 m = 13.8 0C

Solution FP = Normal FP - Tf

= 0.00 0C – 13.8 0C = -13.8 0C

i = 1.0 for all non-electrolytes (molecules that don’t ionize)



49

Osmotic Pressure• Osmosis is the net flow of a solvent between

two solutions separated by a semipermeablemembrane.

– The solvent passes from the lower concentrationsolution into the higher concentration solution.

• Examples of semipermeable membranesinclude:

1. cellophane and saran wrap

2. skin

3. cell membranes



50

Osmotic Pressure

H2O 2O

semipermeable membrane

H2O H2O

sugar dissolved

in water H2O

H2O

H2O

H2O

net solvent flow



51

Osmotic Pressure





http://d/Media/PowerPoint_Lectures/PowerPoint%20Media/Movies/14Z01AN1.MOV

http://d/Media/PowerPoint_Lectures/PowerPoint%20Media/Movies/14Z01AN1.MOV



52

Osmotic Pressure• Osmosis is a rate controlled phenomenon.

–The solvent is passing from the dilute solution intothe concentrated solution at a faster rate than inopposite direction, i.e. establishing an equilibrium.

• The osmotic pressure is the pressure exerted

by a column of the solvent in an osmosisexperiment.

M

M

RT

where: = osmotic pressure in atm

= molar concentration of solution

R = 0.0821L atm

mol K

T = absolute temperature



53

Osmotic Pressure

For very dilute aqueous solutions, molarity

and molality are nearly equal. M m

m

for dilute aqueous solutions only

RT



54

Osmotic Pressure• Osmotic pressures can be very large.

–

For example, a 1 M sugar solution has an osmoticpressure of 22.4 atm or 330 p.s.i.

• Since this is a large effect, the osmotic pressuremeasurements can be used to determine the

molar masses of very large molecules such as:1. Polymers

2. Biomolecules like

• proteins

• ribonucleotides



55

Osmotic Pressure

• Example 14-18: A 1.00 g sample of a biologicalmaterial was dissolved in enough water to give1.00 x 102 mL of solution. The osmotic pressure

of the solution was 2.80 torr at 25o

C. Calculatethe molarity and approximate molecular weightof the material.

You do it!



56

Osmotic Pressure

M M

M M

RTRT

atm = 2.80 torr 1 atm

760 torr atm =

=atm

0.0821 K L atmmol K

? .

..

0 00368

0 00368

298150 10 4



57

Osmotic Pressure

M M

M M

M

RT RT

atm = 2.80 torr 1 atm

760 torr atm =

=atm

0.0821 K

g

mol

1.00 g

0.100 L

L

typical of small proteins

L atmmol K

g mol

? .

..

?

. .

0 00368

0 00368

298150 10

1

150 10 6 67 10

4

4 4



Colligative properties of

electrolyte solution



before we enter

the material

van't Hoff factor

first we will first

identify the

characteristics of non-electrolyte and

electrolyte

compounds

l l d



non-electrolyte compounds

solution that can not

conduct electricity

example:

carbohydrate family

alcohol family

Ether family

urea (CO(NH2)2)ethanol (C2H5OH)



electrolyte compounds

solution that can conduct electric current

Example:

Acid solution

Base solution

Saline solution



electrolyte compounds



.1. The solution will be more difficult to evaporate than pure

solvent because the solvent vapor pressure decreased as a result

of the dissolved particles.

2. If the solution is connected with the pure solvent through the

membrane semipermeable, then the solution will experience

volume due to osmotic pressure.

If we dissolve a solute in a pure solvent, then it

most likely will be the following



3. If boiled, the solution will boil at a higher temperature if

compared with pure solvent. As a result of the dissolved

particles will increase the boiling point.

4. If frozen, the solution would freeze at a temperature less

than or below the freezing temperature of the pure solvent.

As a result of the dissolved particles will decrease the

freezing point.

C12H22O11(s) → C12H22O11(aq)

NaCl(s) → Na+(aq) + Cl –(aq)



Colligative properties of an electrolyte solution is generally

greater than the colligative properties of non-electrolyte

solution when the two solutions that have the same

concentration. to explain this difference using the Van't Hoff

factor i is known as the Van't Hoff factor.

This factor is the ratio of colligative properties of electrolyte

solutions with a particular concentration divided by thecolligative properties of a non-electrolyte solution with the

same concentration.



Electrolyte VS Non-Electrolyte

• In the same concentration, colligative

properties of electrolyte solutions is greater

than the colligative properties of solutions of

non-electrolytes.

• WHY?

Electrolyte Solution



Electrolyte Solution

•

The decomposition of the electrolyte solutioninto ions is due to the equilibrium reaction

force of attraction ion-ion opposite.



The degree of Dissociation ()

• To express a lot or at least the ionized substances used term

electrolyte ionization degree or degrees of dissociation.

• Strong electrolytes as easily ionized then the degree of

ionization rates approaching one.

• Electrolyte solution of weak prices ionization degree is very

small due to difficult ionized.



The number of solute particles ionize the

reaction of electrolyte solution formulated in the

Van't Hoff factor. Calculation of colligativeproperties of electrolyte solutions are always

multiplied by a factor of Van't Hoff

Colligative Properties and Dissociation



70


of Electrolytes

• Electrolytes have larger effects on boiling pointelevation and freezing point depression thannonelectrolytes.

– This is because the number of particles released insolution is greater for electrolytes

• One mole of sugar dissolves in water to produceone mole of aqueous sugar molecules.

• One mole of NaCl dissolves in water to producetwo moles of aqueous ions: – 1 mole of Na+ and 1 mole of Cl- ions




71


of Electrolytes

•

Remember colligative properties depend on thenumber of dissolved particles.

– Since NaCl has twice the number of particles we can

expect twice the effect for NaCl than for sugar.



• Calculate the vapor pressure of a solution of

NaOH 0.2 mol in 90 grams of water if the

vapor pressure at a given temperature is 100mmHg!

A



• Answer :

X NaOH :

: 0,2 mol =0,038

0,2 mol + 90g/18g/mol

Because NaOH is strong electrolyte (α=1) andn=2, so:

∆P : Po . Xt . i {1+(n-1)α}

: 100 mmHg . 0,038 . {1+ (2-1) 1}:7,6 mmHg

P : 100 mmHg – 7,6 mmHg

: 92,4 mmHg

Example Problem 2



What is the freezing point of anaqueous solution containing 100. g of

NaCl dissolved in 600. g of water?

T f = K f m i

Kf = 1.86 C°/m and i = 2.0 (two ions)

Molal concentration must be calculated.

mol solutem =

Kg solvent

1 mol NaCl100. g NaCl

58.44 g=

0.600 Kg

= 2.85 m

f f ΔT = K m i oCm= 1.86 2.85 m 2.0

0= 10.6 C

Solution FP = Normal FP – Tf

= 0.0 – 10.6 = - 10.6 °C

Na+/Cl-

A 0 1 molal H2SO4 solution the substance is a strong



A 0.1 molal H2SO4 solution, the substance is a strong

acid with a degree of ionization of α = 1. if the

solvent water, and the value of kb water = 0.52 ° C /

molal.

Determine the boiling point of the solution



Example of calculating matter colligative properties

electrolyte solution

What is the freezing point depression of the solution

in the solvent water?

Kf water = 1,86°C m –1.a. 0,1 m Glucose solution

b. 0,1 m NaCl solution

Answer:l l l b h b f



a. Because glucose is a non-electrolyte substances, the number of

molecules of glucose in a glucose solution into 0.1 m 0.1 m.

C6H12O6(s) → C6H12O6(aq)

0,1 molal 0,1 molalLowering freezing point:

ΔTf =m × Kf

ΔTf =1 × 0,1 m × 1,86 °C m –1.

ΔTf =0,186°C(b) Because NaCl is the electrolyte substance in 0.1 molal NaCl

solution there will be 0.2 molal ions:

NaCl(s) → Na+(aq) + Cl –(aq)

0,1 molal 0,1 molal 0,1 molalThe lowering freezing point of the solution is :

ΔTf=2 × 0,1 m × 1,86°C m –1

=0,372°C



Example Determining Electrolyte Solution Osmotic Pressure

Sea water containing 0,5 M NaCl. Calculate the osmotic

pressure at 25°C and the percent deviation of the phase of the

van’t hoff factor. Known Van’t Hoff factor for NaCl = 1,9.



Answer:

Because NaCl is an electrolyte solution, there would

theoretically ions 2 times the salt concentration. Thecalculation of the theoretical osmotic pressure:

π = (2) M RT =(2) (0,5 mol L –1) (0,082 L atm mol –1 K –

1) (298 K)

π = 24,436 atmCalculation of the osmotic pressure of the

experiment:

π = (i) M R T = (1,9)(0,5 mol L –1)(0,082 L atm mol –1K –

1)(298 K)

π = 23,214 atm




80

g p

of Electrolytes

• Ion pairing or association of ions prevents the

effect from being exactly equal to the number

of dissociated ions

Example Problem 2

Wh t i th f i i t f



What is the freezing point of anaqueous solution containing 100. g of

NaCl dissolved in 600. g of water?

T f = K f m i

Kf = 1.86 C°/m and i = 2.0 (two ions)

Molal concentration must be calculated.

mol solutem =

Kg solvent

1 mol NaCl100. g NaCl

58.44 g=

0.600 Kg

= 2.85 m

f f ΔT = K m i oCm= 1.86 2.85 m 2.0

0= 10.6 C

Solution FP = Normal FP – Tf

= 0.0 – 10.6 = - 10.6 °C

Na+/Cl-

Osmotic Pressure Example Problem 6



The Jubail desalination plant in Saudi

Arabia is the largest in the world. The

plant, located on the Persian Gulf,produces 800 million gallons per day (50%

of Saudi Arabia’s fresh water supply).

If the concentration of salt in the seawater

is 0.60 M, what pressure in psi is requiredto start reverse osmosis (used in

desalination)?Assume 25.0°C, i = 2.0, and 1atm=14.7psi

MRTi mol L atmL mol K 0.60 0.0821 298 K (2.0) = 29.36 atm

14.7 psi29.36 atm

1.00 atm

= 432 psi



Documents

New Basic Chemistry 2 Solution ( Ideal Solution and Colligative Properties