Colligative property

Colligative property

1. Colligative Properties of Solution
B.Sc. SEM-V
Paper-II (Physical Chemistry)
Dr. N. G. Telkapalliwar
Associate Professor
Department of Chemistry
Dr. Ambedkar College, Nagpur

2. Solution:Asolutionisahomogeneousmixtureoftwoormore
substances.
Solute
Solvent
Concentrationofthesolution:Itisdefinedastheamountofsolute
presentinagivenamountofsolution.
Concentrationisgenerallyexpressedasthequantityofsoluteina
unitvolumeofsolution.
Dilutesolution
Concentratedsolution.
Introduction

3. WAYS OF EXPRESSING CONCENTRATION
Percentage by weight
Mole fraction
Strength
Molarity
Molality
Normality
Parts per million (ppm)

4. 1) Percentage by Weight
It is the weight of the solute as a per cent of the total weight of the solution.
Example:
If a solution of HClcontains 36 per cent HClby weight, it has 36 g of HClfor 100 g of
solution.

5. 2) Mole fraction
A simple solution is made of two substances : one is the soluteand the other solvent.
Mole fraction (X) of solute is defined as the ratio of the number of moles of solute
and the total number of moles of solute and solvent.
Thus,

7. 3) Strength of Solution
Strength of a solution: It is defined as the number of grams of solute present
per liter of the solution.
Molarity(M) is defined as the number of moles of solute per litreof
solution.
4) Molarity

9. 5) Molality
Molality(m) of a solution is defined as the number of moles of solute dissolved in one
kilogram of the solvent.

10. 6) Normality
Normality (N) is defined as the number of gram equivalent of a solute present per
litreof solution.
If 40 g of NaOH(eq. wt. = 40) be dissolved in one litreof solution, normality of
the solution is one and the solution is called 1N (one-normal).
A solution containing 4.0 g of NaOHis 1/10 N or 0.1 N or decinormal.

12. 7) Parts per million (ppm)
Parts per million (ppm) : It is defined as the number of parts of the solute present in
one million parts of solution.
In case of solution of solid in liquid,
“It is defined as the number of milligrams of a solid solute present in one liter of
solution”.
Ex:A350mLsampleofdrinkingwaterwasanalyzedandfoundtocontain0.0046gof
sulphatesalts.Calculatetheconcentrationofsulfatesaltsinthiswatersample?

13. Colligative Properties
The properties which depends on the number of particles in solution and
not on the nature of the particles are called as colligativeproperties.
Dilutesolutionscontainingnon-volatilesoluteexhibitthefollowing
colligativeproperties:
(1) Lowering of the Vapour Pressure
(2) Elevation of the Boiling Point
(3) Depression of the Freezing Point
(4) Osmotic Pressure

14. Compare the properties of 1.0 M Sucrose solution to a
0.5 M solution of NaCl.
Despitetheconc.ofNaCl=½theconc.ofsucrosebothsolutions
havepreciselythesamenumberofdissolvedparticles,why?
BecauseeachNaClunitcreatestwoparticlesupondissolutiona
Na
+
andaCl
-
.
Bothsolutionshavethesamefreezingpoint,boilingpoint,vapor
pressure,andosmoticpressurebecausethosecolligativeproperties
ofasolutiononlydependonthenumberofdissolvedparticles.
Othernon-colligativepropertiesincludinge.g.viscosity,surface
tension,andsolubilityaredifferent.

15. Lowering of the Vapour Pressure (Raoults’sLaw)
Raoult’slaw:Thepartialpressureofanyvolatilecomponentof
asolutionatanytemperatureisequaltotheproductofvapour
pressureofthepurecomponentandmolefractionofthat
componentinthesolution.
P
s= X
A. P
P
s= X
B. P
Where,
P
s=Vapourpressureofasolution
P=Vapourpressureofpuresolvent
X
A=Molefractionofsoluteonsolution
X
B=Molefractionofsoluteonsolution

16. Thevapourpressureofapuresolventisdecreasedwhenanon-
volatilesoluteisdissolvedinit.
Ifpisthevapourpressureofthesolventandp
sisthevapourpressure
ofthesolution,thenloweringofvapourpressureis(p–p
s).
Thisloweringofvapourpressurerelativetothevapourpressureof
thepuresolventistermedtheRelativeloweringofVapourpressure.
Thus,
Lowering of the Vapour Pressure

17. Raoults’slaw:Therelativeloweringofthevapourpressureofa
dilutesolutionisequaltothemolefractionofthesolutepresent
indilutesolution.
Raoult’sLawcanbeexpressedmathematicallyintheform:
Where,
n = number of moles or molecules of solute
N = number of moles or molecules of solvent.

18. Thevapourpressureofthepuresolventiscausedbythenumberofmolecules
evaporatingfromitssurface.
Whenanonvolatilesoluteisdissolvedinsolution,thepresenceofsolutemoleculesin
thesurfaceblocksafractionofthesurfacewherenoevaporationcantakeplace.
Derivation of relative lowering of the Vapour Pressure (Raoult’sLaw)
Thiscausestheloweringofthevapourpressure.Thevapourpressureofthesolution
is,therefore,determinedbythenumberofmoleculesofthesolventpresentatany
timeinthesurfacewhichisproportionaltothemolefraction.
Thatis,
Where,N=molesofsolventandn=molesofsolute.
Where, k being proportionality constant .
---------(1)

21. Determination of Molecular Mass

24. Experimental determination of lowering of vapour pressure
Ostwald and Walker’s Dynamic Method (Gas Saturation Method)
Procedure:
The apparatus used by Ostwald and Walker is shown in Fig.
It consists of two sets of bulbs :
(a) Set A containing the solution
(b) Set B containing the solvent
Each set is weighed separately.
A slow stream of dry air is then drawn by suction pump through the two sets of bulbs.
At the end of the operation, these sets are reweighed.
From the loss of weight in each of the two sets, the lowering of vapour pressure is
calculated.
The temperature of the air, the solution and the solvent must be kept constant
throughout.

25. Calculations:
AstheairbubblesthroughsetAitissaturateduptothevapourpressurep
sofsolution
andthenuptovapourpressurepofsolventinsetB.Thustheamountofsolventtaken
upinsetAisproportionaltop
sandtheamounttakenupinsetBisproportionalto
(p–p
s).
KnowingthelossofmassinsetB(w
2)andthetotallossofmassinthetwosets
(w
1+w
2),wecanfindtherelativeloweringofvapourpressurefromequation(4).

26. Osmosis and osmotic pressure

27. Letusconsiderapuresolventandsolutionseparatedbyamembrane
whichpermitsthepassagetosolventmoleculesbutnottosolutemolecules.
Onlythesolventwilldiffusethroughthemembraneintosolution.
Amembranewhichispermeabletosolventandnottosolute,iscalleda
semipermeablemembrane.
Thenetflowofthesolventmoleculesfromlowerconcentrationtohigher
concentrationregionthroughsemipermeablemembraneiscalledas
Osmosis.
WHAT IS OSMOSIS?

28. What is osmotic pressure?
Theminimumpressuretobeappliedonthesolutiontostoptheprocessof
osmosisiscalledasosmoticpressure.
Or
Osmoticpressuremaybedefinedastheexternalpressureappliedtothe
solutioninordertostoptheosmosisofsolventintosolutionseparatedby
asemipermeablemembrane.
External pressure

29. What is Reverse Osmosis?
ReverseOsmosisworksonthesameprincipleasosmosis,butinthereverse
direction.
InthisReverseosmosisprocessdirectionoftheflowofthesolvent
moleculesarereversedbyapplyinggreaterpressure.
Theflowofthesolventmoleculesfromhigherconcentrationtolower
concentrationregionthroughsemipermeablemembraneunderhighpressureis
calledreverseosmosis.
Osmosis Reverse Osmosis

30. Measurement of Osmotic pressure
Berkeley and Hartley’s Method
Berkeley and Hartley (1904-1909) employed the technique of applying external
pressure on the solution just enough to prevent osmosis.
Procedure:
Theinnerporcelaintubeisfilledwithpuresolventandthejacketwiththesolution
whoseosmoticpressureistobedetermined.
Thelevelofthesolventmeniscusinthecapillarytubewilltendtomovedownas
solventflowsintothesolutionacrossthemembrane.
Pressureisthenappliedthroughthepistonsothatthemeniscusbecomesstationary.
Itindicatesthatosmosishasbeenstoppedandnowthepressurerecordedbythe
pressuregaugegivestheosmoticpressureofthesolution.
Advantages
(a)Itisquickandaccurate.
(b)Itcanbeusedfordetermininghigh
osmoticpressures.
(c)Theosmoticpressurebeingbalanced
bytheexternalpressure,thereisnostrain
leftonthemembraneandthedangerofits
burstingiseliminated.

31. Determination of molecular mass from osmotic pressure
Themethodofdeterminationofmolarmassonthebasisofosmotic
pressureisbasedonvantHoff’stwolaws:-
1)Boyle-vantHoff’slaw:Atconstanttemperature,osmoticpressureofa
dilutesolutionisdirectlyproportionaltomolarconcentrationofthe
solution.
2)Charle-vantHoff’slaw:Foragivenconcentrationofasolution,the
osmoticpressureisdirectlyproportionaltotheabsolutetemperature.
That is, π ∝C at constant temperature, Where π is osmotic pressure
That is, π ∝T at constant concentration, Where π is osmotic pressure

32. By combining the two laws we get,
π ∝C T or π = R C T
But Concentration is expressed as number of moles of solute per unit
volume of the solution.
That is, C = n/V
Therefore, π = nRT/V
πV= nRT
πV= wRT/M
M = wRT/πV
Where,
M = molecular mass of the solute
w = amount of solute in grams
R = 0.0821 litre-atmosphere
T = (t°C+ 273) K
π = osmotic pressure in atmospheres
V = volume of solution in litres

35. Elevation of boiling point
Whenaliquidisheated,itsvapourpressurerisesandwhenitequalsthe
atmosphericpressure,theliquidboils.
Theadditionofanonvolatilesolutelowersthevapourpressureand
consequentlyelevatestheboilingpointasthesolutionhastobeheatedtoa
highertemperaturetomakeitsvapourpressurebecomeequalto
atmosphericpressure.
IfT
bistheboilingpointofthe
solventandTistheboilingpointof
thesolution,thedifferenceinthe
boilingpoints(ΔT)iscalledthe
elevationofboilingpoint.
T–T
b=ΔT
Thevapourpressurecurvesofthe
puresolvent,andsolutions-1and2
withdifferentconcentrationsof
soluteareshowninFigure.

36. Ostwald-Walker method of measuring the relative lowering of vapour pressure.
Fordilutesolutions,thecurvesBDandCE
areparallelandstraightlines
approximately.Thereforeforsimilar
trianglesACEandABD,wehave
Where,p–p
1andp–p
2areloweringof
vapourpressureforsolution1and
solution2respectively.
Hence,theelevationofboilingpointisdirectlyproportionaltothelowering
ofvapourpressure.
ΔT∝p–p
s ...(1)

37. Determination of Molecular Mass from Elevation of Boiling Point
Theelevationofboilingpointisdirectlyproportionaltotheloweringofvapourpressure.
ΔT∝p–p
s ...(1)
Since p is constant for the same solvent at a fixed temperature, from (1) we can write
But from Raoult’sLaw for dilute solutions,
Since M (mol mass of solvent) is constant, from (3)
From (2) and (4)

38. WhereK
bisaconstantcalledBoilingpointconstantorEbulioscopicconstantof
molalelevationconstant.Ifw/m=1,W=1,K
b=ΔT.
Thus,
Molalelevationconstantmaybedefinedastheboiling-pointelevation
producedwhen1moleofsoluteisdissolvedinonekg(1000g)ofthesolvent.
Ifthemassofthesolvent(W)isgiveningrams,ithastobeconvertedinto
kilograms.Thustheexpression(5)assumestheform
Where,ΔT=elevationofboilingpoint;K
b=molalelevationconstant;w=massof
soluteingrams;m=molmassofsolute;andW=massofsolventingrams.

39. TheconstantK
b,whichischaracteristicofaparticularsolventused,can
alsobecalculatedfromthermodynamicallyderivedrelationship.
Where,
R=gasconstant;
T
b=boilingpointofsolvent;
L
v=molarlatentheatofvaporization.

42. Depression of freezing point
Thevapourpressureofapureliquidchangeswithtemperatureasshownby
thecurveABC.ThereisasharpbreakatBwhere,infact,thefreezing-point
curvecommences.ThusthepointBcorrespondstothefreezingpointofpure
solvent,T
f.
Thevapourpressurecurveofa
solution(solution1)ofanonvolatile
soluteinthesamesolventisalso
showninFig.Itissimilartothevapour
pressurecurveofthepuresolventand
meetsthefreezingpointcurveatF,
indicatingthatT
1isthefreezingpoint
ofthesolution.Thedifferenceofthe
freezingpointofthepuresolventand
thesolutionisreferredtoasthe
Depressionoffreezingpoint.Itis
representedbythesymbolΔTorΔT
f.
T
f–T
1= ΔT

43. Relation between lowering of vapour-pressure and depression of freezing point
Whenmoreofthesoluteisaddedtothesolution1,wegetamoreconcentrated
solution(solution2.)Thevapourpressureofsolution2meetsthefreezing-pointatC,
indicatingafurtherloweringoffreezingpointtoT
2.
FordilutesolutionsFDandCEareapproximatelyparallelstraightlinesandBCisalsoa
straightline.
Since the triangles BDF and BEC are similar,
Where p
1and p
2are vapour pressure of
solution 1 and solution 2 respectively.
Hence depression of freezing point is directly
proportional to the lowering of vapour
pressure.
or ΔT ∝p –p
s ...(1)

44. Determination of Molecular Weight from Depression of Freezing point
Depression of freezing point is directly proportional to the lowering of vapour pressure.
or ΔT ∝p –p
s ...(1)
Since p is constant for the same solvent at a fixed temperature, from (1) we can write
But from Raoult’sLaw for dilute solutions,
Since M (mol wt) of solvent is constant, from (3)
from (2) and (4)

45. whereK
fisaconstantcalledFreezing-pointconstantorCryoscopicconstant
orMolaldepressionconstant.Ifw/m=1andW=1,K
f=ΔT.
Thus,Molaldepressionconstantmaybedefinedasthefreezing-point
depressionproducedwhen1moleofsoluteisdissolvedinonekg(1000g)of
thesolvent.
Ifthemassofsolvent(W)isgiveningrams,ithastobeconvertedinto
kilograms.Thustheexpression(5)assumestheform;
Where, m = molecular mass of solute ; K
f= molaldepression constant ; w = mass
of solute; ΔT = depression of freezing point ; W = mass of solvent.

46. The constant K
f, which is characteristic of a particular solvent, can
also be calculated from the relation,
Where,
T
f= freezing point of solvent in K;
L
f= molar latent heat of fusion;
R = gas constant.

49. Abnormal Molar Masses and Van’t Hoff Factor
Colligative properties depends upon the number of particles (molecules or
ions) of the solute present in the solution. If the solute undergoes dissociation
or association in the solution, the molar mass obtained is abnormal.
Van’t Hoff introduced a correction factor‘i’ which is defined as follows;
Since colligative properties are inversely proportional to the molecular mass
of solute,
Such abnormal molar masses are due to two reasons
1) Association
2) Dissociation

50. Degree of association:
Degreeofassociationmeansthefractionofthetotalnumberofmoleculeswhich
combinetoformbiggermolecules.
Consideronemoleofasolutedissolvedinagivenvolumeofasolvent.
Suppose,nsimplemoleculescombinetoformanassociatedmolecule,
i.e.,nA (A)
n
Let,bethedegreeofassociation
Thenumberofunassociatedmoles=1-
Thenumberofassociatedmoles=/n
Therefore,Thenumberofeffectivemoles=1-+/n
Since,Colligativeeffectisproportionaltothenumberofmoles,therefore,van’tHoff
factor‘i’isgivenby,
Thus,knowingthevaluesof‘i’and‘n’thedegreeofassociationcanbecalculated

51. Degree of Dissociation:
Degreeofdissociationmeansthefractionofthetotalnumberofmoleculeswhich
dissociates,thatis,breakintosimplemoleculesorions.
Consideronemoleofuni-univalentelectrolyte(KCl)dissolvedinagivenvolumeof
solvent(water).
Suppose,nsimplemoleculescombinetoformanassociatedmolecule,
i.e.,A nP
Let,bethedegreeofdissociation
Thenumberofundissociatedmoles=1-
Thenumberofdissociatedmoles=n
Therefore,Thenumberofeffectivemoles=1-+n
Since,Colligativeeffectisproportionaltothenumberofmoles,therefore,van’tHoff
factor‘i’isgivenby,
Thus,knowingthevaluesof‘i’and‘n’thedegreeofdissociationcanbecalculated

52. References
•Essentials of Physical Chemistry, S. ChandPublication by ArunBahl, B.S. Bahl, G. D. Tuli.
•University Chemistry, Volume I, Alliance & Co., by R. M. Jugade.
•A Text Book of Chemistry, B.sc. SEM-I, Himalaya Publishing House.
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