Chapter 1 Properties of Solution for Lecturers (2)

7/28/2019 Chapter 1 Properties of Solution for Lecturers (2)

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CHAPTER 1

Properties of Solutions

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Introduction to Solutions and Solubility

Factors Affecting Solubility:

Solute-solvent interactions Temperature Pressure Solutions

Quantitative Ways of Expressing Concentration Molarity, Molality Parts by mass and by volume, Mole Fraction

2

SCOPE


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3

Colligative Properties of Solutions Vapor Pressure Lowering

Boiling Point Elevation

Freezing Point Depression

Osmotic Pressure

SCOPE


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Able to distinguish the factors affectingsolubility.

Able to express different types ofsolution concentrations and conductcalculations involved.

Able to understand colligativeproperties effect and conductcalculations involved.

4

Learning Outcomes


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1.1 Introduction to solutions

and solubility Solution

Homogeneous mixture of two or more pure

substance

Solvent May be gaseous, liquid or solid

Liquid of a liquid solution

Solute Dissolved substance in liquid solution

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Solute

A substance thatdissolves in the

dissolving medium

(solvent)

Solvent

The dissolving

medium - most

abundant component

of a given solution Solution

A homogeneous

mixture of two or

more substance

Miscible: Substances (liquid) that are soluble in each other in anyproportion


and solubility


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1.1 Introduction to solutions and

solubility

Solubility Maximum amount of solute that dissolves

completely in a given amount of solvent at a

particular temperature, TFor NaCl, S = 39.12 g/100 mL H2O @ 100

oC

For AgCl, S = 0.0021 g/100 mL H2O @ 100oC

Dissolution Process of dissolving a solute in a solvent to give

a homogeneous solution7


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Water molecules

Undissolved NaCl

Dissolution process8


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Types of solution

Saturated solution: contains the maximum amount of dissolved

solute at a given temperature in the presence ofundissolved solute.

Solute (undissolved) solute (dissolved)

Unsaturated solution: contains less than the maximum amount of

dissolved solute, it has the capacity to dissolvemore solute


and solubility


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Supersaturated solution:

contains more than the equilibrium amount ofdissolved solute.

unstable relative to the saturated solution. If a seed crystal is added, the excess solute will

crystallize immediately, leaving a saturated solution.

1.1 Introduction to solutionsand solubility


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Solute - solvent interaction

Temperature

Pressure - for gases

1.2 Factors Affecting Solubility


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Solute-Solvent Interactions

Solute - solvent interaction

The stronger the attractions between solute and solventmolecules, the greater the solubility

Solute will dissolve in the solvent if the intermolecular forces ofthe molecules in the solutes is comparable compared to theintermolecular forces of the molecules of the solvent

Ionic bondHydrogen bond

Dipole-dipole

London dispersion force

Strength increases



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Example

Example:

Dissolving NaCl in H2O compared to dissolving

NaCl in benzene

NaCl experience ionic bond, H2O hydrogen

bond these bond is comparable in the

strength order can dissolve

Benzene experience LDF too much weak

compared to ionic bond cannot compensate

to the forces of solute NaCl cannot dissolve



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The number of carbon atoms effects its solubility in water.

As the length of carbon chain increases the polar OH group

become a smaller part of the molecule, the moleculebecome more likely hydrocarbon, thus the solubility

decreases

If the number of OH group along the carbon chain,increases more solute-water hydrogen bonding, thereforeincrease solubility

e.g. Glucose C6H12O6



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London Dispersion forces increases with increasing

molecular mass, leads to higher solubilities

N2

O2Ar

0.69 X 10-328

1.38 X 10-3

Gas Solubility in H2O

at 20oC (M)

MW

(g/mol)

321.50 X 10-3

Kr 2.79 X 10-3

40

83.8



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Effect of Temperature

Temperature

In general, the solubilityof solid solutes in waterincreaseswith increasing temperature.

(exceptions in the curve for Ce2(SO4)3)

In contrast to solid solutes, the solubility of gases in waterdecreases with increasing temperature.

The decrease in gas solubility as temperature increases isprimarily a function of kinetic energy

As temperature increases, the kinetic energy of dissolved

gas will increase, making it easier for the gas molecules to

escape the solution



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Effectof Pressure

Pressure

If the pressure of a gas increase (at constant T) more gasmolecules are striking the surface of the container in a given

amount of time (Kinetic Molecular Theory)

A gas in contact with a solution is "dissolved" when gasmolecules strike the surface of the solution (and are

surrounded and dispersed by the solvent).

Thus, increasing the pressure (at constant T) results in morecollisions of the gas molecules, per unit time, with the surface

of the solvent This results in greater solubility.



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The solubility ofgas in any solvent increased asthe pressure of the gas over the solventincreases(at constant T).

The solubilities of solids and liquids are notappreciably (clearly noticed) affected by pressure.



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Effect of Pressure

Dynamic equilibrium:

Rate gas molecules enter the solution(liq.phase) = the ratesolute molecules enter the gas phase.

If piston is pushed down, gas volume decreases and pressureincreases. More gas dissolves until equilibrium is established



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1.3 Liquid solutions

For a gas to dissolve in a liquid, the gas molecules

must be able to disperse themselves evenly

throughout the solvent 21

1.3 Gas-Liquid Solutions


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Solubility of gases vary significantly with pressure For gases that do not react with the solvent, Henrys

law gives the relationship between gas pressure

and gas solubility

Sgas is the concentration/solubility of the gas

(molarity) Pgas is the partial pressure of the gas above the

solution

kHis called the Henrys law constant and is unique

to each gas

TPkS constantgasHgas

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Henrys law

Example: Workbook

Calculate the concentration of CO2 in a soft drink:

bottle with pressure of CO2 of 4.0 atm over the liquid at25oC. The Henrys law constant : 3.1 X 10-2 mol/L.atm.


Sgas = kPgas

= (3.1 X 10-2 mol/L.atm)(4.0atm)

= 0.12 mol/L

= 0.12 M


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1.3 Liquid solutions Equation is only true at low concentrations

and pressures

An alternative expression of Henrys law is:

S1 and P1 refer to initial conditions

S2 and P2 refer to final conditions

2

2

1

1

P

S

P

S

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1.3 Liquid solutions Formation of a liquidliquid solution requires that

the attractive forces present between the

molecules of the two pure liquids is overcome

Two substances are MISCIBLE when they mix

completely in all proportions

Two substances are IMMISCIBLE when theyform two layers upon the addition of one to the

other

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1.4 Liquid-liquid Solutions


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C C O

H

HH

H

H H

Likedissolveslike (ethanol is dissolved in H2O)

EthanolBenzene

hydrogen bond

C

CC

C

CC

H

H

H

HH

H

H O

H

H O

C2H5

H

OC2H5

H O

H

O

H

H

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1.4 Liquid-liquid Solutions


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1.3 Liquid solutions Liquidsolid solutions

Basic principles remain

the same

Solvation is when a solutemolecule is surrounded by

solvent molecules

Hydration occurs whensolutes become

surrounded by water

molecules

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1.5 Liquid-solid Solutions


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Liquidsolid solutions

Like-dissolves-like

When intermolecular attractive forces within

solute and solvent are sufficiently different,

the two do not

form a solution

Temperature can have asignificant effect on the

solubility of a solid solute

in a liquid

1.3 Liquid solutions

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1.4 Quantification of solubility:

the solubility product Ionic salts are generally classified as

being either soluble or insoluble in water

AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

AgCl(s) Ag+(aq) + Cl(aq)

Ksp = [Ag+][Cl]

Ksp is called the solubility productMaXb(s) aM

c+(aq) + bXd(aq)

Ksp = [Mc+]a[Xd]b

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The relationship between Ksp and solubility

Molar solubility (s)

molar concentration of a salt in its saturatedsolution

Molar solubility can be used to calculate Ksp,

assuming that all of the salt that dissolves is100% dissociated into its ions

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Example:

The solubility of AgBr in water is 1.3 104 g L1 at

25 C. Calculate Ksp for AgBr at this temperature.

Solution:

AgBr(s) Ag+(aq) + Br(aq)

Ksp = [Ag+][Br]

mol109.6gmol77.187

g103.1 71

4

M

mn

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Solution (cont):

[Ag+] = [Br] = 6.9 107 mol L1

Ksp = [Ag+][Br] = (6.9 107 mol L1)(6.9 107 mol L1)

Ksp = 4.8 1013 at 25 C

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the solubility product

Example: Workbook

Calculate the molar solubility of lead iodide,

PbI2, given that Ksp(PbI2) = 7.9 109

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The common ion effect Any ionic salt is less soluble in the presence of

a common ion, an ion that is in the salt

PbCl2

(s) Pb2+(aq) + 2Cl(aq)

Ksp = [Pb2+][Cl]2

Add Pb(NO3)2(aq) to saturated solution of PbCl2

instantaneously increases [Pb2+

] and thereforeQsp (ionic product).

Qsp > Ksp

PbCl2 is precipitated

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the solubility productExample:What is the molar solubility of PbI2 in a 0.10 M NaI solution?

Pbl2(s) Pb2+(aq) + 2l(aq)

Ksp = [Pb2+][I]2 = 7.9 109

Ksp = s(0.10 + 2s)2 = 7.9 109

Ksp = s(0.10)2 = 7.9 109

Molar solubility of PbI2 in 0.10 M NaI solution is 7.9107

M

PbI2(s) Pb2+(aq) + 2I(aq)

Initial concentration (M) 0 0.10

Change in concentration (M) +s +2s

Equilibrium concentration (M) s 0.10 + 2s

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the solubility product Prediction of precipitation

Qsp > Ksp precipitate will form

Qsp < Ksp no precipitate will form

AgCl(s) Ag+(aq) + Cl(aq)

Ksp = [Ag+][Cl] = 1.8 1010

[Ag+

] = 5.0 107

mol L1

[Cl] = 5.0 105 mol L1

Qsp = 2.5 1011

Qsp < Ksp no precipitate will form 37



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Mass / mole fraction

Mass / mole percent

Part per million (ppm) Part per billion (ppb)

Molarity , M

Molality , m

1.6 Expressing Concentration


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Quantitative Ways of ExpressingConcentration

Mole fraction, x

The number of moles of a particular component dividedby the total number of moles of material in the solution

The mole fraction ofA,xA, in a solution containingsubstancesA, B and C

The sum of the mole fractions must equal 1.

Temperature independent

xA = nA

nA + n B +nC



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Mole percentage of a component

e.g. A solution of hydrochloric acid that is 36% HCl by mass

contains 36 g HCl for each 100 g of solution.

36% Vol or 36% wt

Mole % of Component

= Mole of component in solution X 100

total mole of solution



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Parts per million (ppm)

mg/Kg = ppm wtl/L = ppm vol

1 mg of solute per kilogram of solution = 1 ppm

if with respect to concentration of solute in water

1 ppm = 1 mg/L

ppm of component

= mass of component in solution X 106

total mass of solution



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Parts per billion (ppb)

1 ppb = 1g of solute per billion (109) grams of solution,

or 1 microgram (g) of solute per Liter of solution

if with respect to concentration of solute in water

1 ppb = 1 g/L

ppb of component

= mass of component in solution X 109

total mass of solution



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Example

Example:

What is the mass percentage of iodine (I2) in asolution containing 0.045 mol I2 in 115 g ofCCl4 ?



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Example

Example:

Seawater contains 0.0079 g Sr2+ per

kilograms of water. What is theconcentration of Sr2+ measured in ppm?



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Quantitative Ways of Expressing Concentration1.6 Expressing Concentration

Molarity, M

Amount of substance in a particular volume of solution

Solutions (usually) increase in volume with increasingtemperature

The molarity of a solution changes as the temperaturechanges

Molarity (M) = moles of solute

liters solution


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Molality, m

Preferred method of expressing solution composition

when colligative properties involved

Defined as the number of moles of solute per kilogram

of solvent:

Temperature independent

Note:

i. Molarity is defined in terms of the volume of solution

ii. Molality is in terms of the mass of solvent

Molality (m) = moles of solute

kilograms of solvent

Quantitative Ways of Expressing Concentration1.6 Expressing Concentration


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Example : (Workbook)

Ascorbic acid (Vitamin C, C6H8O6) is a water soluble vitamin. A

solution containing 80.5 g of ascorbic acid dissolved in 210 g of

water has a density of 1.22 g/ml at 55oC. Calculate:

a. mass percentage,

b. mole fraction,

c. molality,

d. molarity of ascorbic acid in this solution.


1 7 C lli ti P ti f


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Colligative Properties

Colligative properties : Depend only on the number of dissolved particles in solution

and not on their identity.

Example: Ethylene glycol added to water in car radiator Lowers the freezing point of solution Raises the boiling point of solution, so that car can operate at

high temperature

Colligative properties affect: vapor pressure, boiling point,

freezing point, and

osmotic pressure of a solution

1.7 Colligative Properties of

Solutions

1 6 Colligati e properties of

1 7 C lli ti P ti f


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1.6 Colligative properties of

solutionsVapor pressure lowering

Boiling point of a solution containing a nonvolatilesolute is higher than that of the pure solvent

Boiling point of a solvent is the temperature at which

the vapor pressure of the solvent is equal to the

atmospheric pressure

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Solutions

1 6 Colligative properties of

1 7 C lli ti P ti f


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solutions The vapor pressure of solvent above the solution is

expressed by Raoults Law:

Psolution =xsolvent P*solvent

Psolution vapor pressure of the solution

xsolvent mole fraction or solvent in the solution

P*solvent vapor pressure of pure solvent

For a simple two component system

Provided the solution is sufficiently dilute

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Solutions


1 7 C lli ti P ti f


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1.6 Colligative properties ofsolutions

How does amount of of solute affectthe magnitude of the vapor pressurelowering?

Psolution =xsolvent P*solventxsolvent = 1xsolute

Psolution = (1xsolute)P*solvent

Psolution = P*solventxsoluteP

*solvent

P=xsoluteP*solvent

P = P*solventPsolution

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Solutions


1 7 C lli ti P ti f


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Raoults law

A solution that obeys Raoults law is called an ideal

solution


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Solutions

These solutions are

generally dilute and have

only small interactionsbetween their constituent

molecules


1 7 C lli ti P ti f


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Solutions containing more than one volatilecomponent

For componentA

pA = XAp*A

For component B

pB = XBp*B

Total pressure

ptotal= XAp*A + XBp

*B


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Solutions

1 6

Colligative properties of

1 7 C lli ti P ti f


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Boiling point elevation and freezing point depression

boiling point elevation

Tb = Kbb

freezing point depression

Tf= Kfb

Kb , Kf - molal boiling point elevation and freezing point depressionconstant, respectively (K mol1 kg)

Kb , Kf are properties of the solvent only and independent of the identity

of the solute

b - molality of the solution (mol kg1)


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Solutions


1 7 C lli ti P ti f


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solutions

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Solutions

Blue line- pure H2O

Red- solution

1 7 C lli ti P ti f


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Solutions

1 7 Colligative Properties of


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Example: (Workbook)

Given that Kb = 2.53oC/m for benzene, what

mass of acetone (CH3COCH3) must bedissolved in 200 g of benzene to raise theboiling point of benzene by 3.00oC.


Solutions



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Osmotic Pressure

Osmosis: the movement of a solvent from low soluteconcentration to high solute concentration throughsemipermeable membrane

(or from high solvent concn to low solvent concn)

As solvent moves across the membrane the fluid levelsbecomes uneven.

The pressure difference between the arms stops osmosis.Osmotic pressure () is the pressure required to stoposmosis.


Solutions




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Osmosis and osmotic pressure

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Solutions




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solutions Osmosis and osmotic pressure

Osmotic pressure,

In dilute aqueous solution

= cRT,

V= nRT

This is the vant Hoff equation for osmotic

pressure

Vnc

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Solutions




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Osmosis and osmotic pressure

Osmometer

Isotonic

sameosmotic

pressure

Hypotonic

lowerosmotic

pressure

Hypertonic

higherosmotic

pressure


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Solutions



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Example: (Workbook)

When 0.200 g of a high molecular weight

compound is dissolved in water to form 12.5 mLof solution at 250C, the osmotic pressure of thesolution is found to be 1.10 X 10-3 atm. What isthe molar mass of the compound?


Solutions




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solutions Measurement of solute dissociation

Molal freezing point depression constant for

water is 1.86 K mol1

1.00 mol kg1 NaCl freezes at about 3.37 C

NaCl(s) Na+(aq) + Cl(aq)

Solution has a a total molality of dissolved

solute particles of 2 mol kg1 Theoretically, a 1.00 mol kg1 NaCl solution

should freeze at3.72 C

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Solutions

1 7

Colligative Properties of


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Boiling-Point Elevation,Tb = iKbm

Freezing-Point Depression,Tf= iKfm

Osmotic Pressure, = iMRTi

nonelectrolytes 1

NaCl 2

CaCl2 3

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Solutions




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CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO

(aq)

1.00 mol kg1

aqueous acetic acid solutionfreezes at1.90 C

Only a little lower than expected if no

ionisation occurred

1

1

kgmol02.1

kgmolK86.1

K90.1

m

K

Tm

f

f

%2ionisation%

%10000.1

02.0ionisation%

100%availableacidofmol

ionisedofmolionisation%

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Solutions




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Some molecular solutes produce smaller

colligative effects than their molal

concentrations would suggest These weak colligative properties are

evidence to solute molecule clustering or

associating

C6H5 C O H

O

C6H5 C

O H

O

C6H5C

OH

O

2

benzoic acid benzoic acid dimer66


Solutions


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Chapter 1 Properties of Solution for Lecturers (2)

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