CHAPTER NAME :SOLUTIONSMODULE NO :07/07(ABNORMAL

MOLECULAR MASSES)SUBJECT : CHEMISTRYCLASS : XIISCHOOL : AECS,MYSORE

PREPARED BY :G BALA KRISHNAIAHPGT(CHEMISTRY)

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LEARNING OBJECTIVES1.Abnormal molecular mass

2.Abnormal molecular masses-reasons

and explanation

3.van’t Hoff factor(i)

4.Modified Colligative properties

5.van’t Hoff factor(i) to calculate degree of

(a) Dissociation (b) Association

6.Numericlas/Homework

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When the molecular mass of a substance determined by

studying any of the colligative properties comes out to be

different than the theoretically expected value, the

substance is said to show abnormal molecular mass.

1.Abnormal molecular mass

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2. Abnormal molecular masses-reasons and explanation

Abnormal molecular masses are observed in any one of the following cases(reasons):

(a). When the solution is non-ideal, i.e, the solution is not dilute.(b). When the solute undergoes association in the solution(c). When the solute undergoes dissociation in the solution

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ABNORMAL MOLECULAR MASSES : EXPLANATION

Dissociation:

The various relations as derived above for the colligative properties are applicable only to the solutions of non-electrolytes which do not undergo any dissociation or association in the solution , in case of the aqueous solution of electrolytes , i.e., acids (HCl, H2SO4 ,CH3COOH, etc) , inorganic bases (NaOH, KOH, Ca(OH)2, etc.) and salts (NaCl , KCl, KNO3, BaCl2, etc .) which dissociate completely or to small extent in the solution, the number of particles of the solute , therefore the experimentally observed value of colligative property comes out to be higher than the theoretically value .Further, as molecular mass is inversely proportional to the colligative property (refer to the formulae derived for molecular masses) therefore the observed molecular mass comes out to be less than the theoretically value .

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for example , in case of KCl , in the aqueous solution , each molecule dissociates to give two ions (KCl K++ Cl-) thus as the number of particles becomes double then observed value of colligative property is double than expected value and molecular mass is half of the expected value.

Association:In case of certain substances, association takes place in the solution. For example, acetic acid, benzoic acid, etc. when dissolved in benzene exist as doubly associated molecules in benzene due to hydrogen bonding which may be represented as

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Thus the number of particles in the solution decreases. As a result, the experimentally observed value of the colligative property is lower and the molecular mass is higher than the expected value. If the association were complete, the observed value of colligative property will be half and the molecular mass will be double than the expected value.( Association generally takes place in the non-aqueous solvents(like benzene) because in the aqueous solution, the high dielectric constant of water helps in the dissociation of the associated molecules.)

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To calculate the extent of association or dissociation, van’t Hoff in 1880 introduced a factor ‘i’ called van’t Hoff factor.

Van’t Hoff factor is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property, i.e.,

Van’t Hoff factor (i)=Abnormal/Experimental/Observed value of the colligative property(C.PAbn/Exp/Obs)Normal/Calculated/Theoretical value of the colligative property(C.PNor/Cal/The)

(OR)

= Normal/Calculated/Theoretical value of the molecular mass(MNor/Cal/The)Abnormal/Experimental/Observed value of the molecular mass(MAbn/Exp/Obs)

(Since Colligative property is inversely proportional molecular mass)(OR)

= Total number of moles of particles after association/dissociation

Number of moles of particles before association/dissociation

3.van’t Hoff factor(i)

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4.Modified Colligative properties

Relative lowering of vapour pressure of solvent (p1

o-p1 ) =i x2

p1o

Elevation of Boiling point, Tb = i Kb m

Depression of Freezing point, Tf = i Kf m

Osmotic pressure of solution, = i n2 R T V

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5.van’t Hoff factor(i) to calculate degree of

(a) Dissociation (b) Association

(a) van’t Hoff factor to calculate (apparent)degree of dissociation

during dissociation

Degree of dissociation is defined as the fraction of the total

substance that undergoes dissociation, i.e.,

Degree of dissociation()= Number of moles dissociated

Total number of moles taken

Suppose a molecule of an electrolyte when dissolved in a

solvent dissociates to give n ions and is the degree of

dissociation, i.e.,

In general A--- n1B + n2C+…….

Initial moles 1 0 + 0 +……..

Moles after dissociation 1- n1 +n2+……

n1+n2+….=n(number of moles of ions/species formed) 11

If we start with 1 mole of the solute, at equilibrium, we have

(1-) moles of undissociated molecules and n moles of the

ions, so that the total number of moles of ions formed

=1-+n=1+(n-1)

Thus, observed colligative property = 1+(n-1) and

Theoretical value = 1(as 1 mole of A was taken).

Van’t Hoff factor,i=1+(n-1) or =i-1

1 n-1

=MThe/MObs-1 [i= MThe/MObs]

n-1

= MThe - MObs

MObs (n-1)

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(b) van’t Hoff factor to calculate (apparent)degree of association

during dissociation

Degree of association is defined as the fraction of the total substance which exists in the form of associated molecules,i.e.,Degree of association()= Number of moles associated

Total number of moles takenSuppose n(number of moles of molecules/species involved in

association) simple molecules of the solute A associate to

form the associated molecule An so that we have the equilibrium: nA ↔ An

Initial 1 0After asso 1-n./n=1- /nIf is the degree of association and we start with one mole of A, then at equilibriumNumber of moles of A=1- Number of moles of An = /n Total number of moles=1-+/n

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Since the colligative property is proportional to the

number of moles of the solute present in solution.

Therefore, van’t Hoff factor i=1-+/n

1

=1-i

1-1/n

=1- MThe/MObs [i= MThe/MObs]

1-1/n

=Mobs-MThe

Mobs(1-1/n)

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At very dilute solution dissociation is complete Values of van’t Hoff factor,i, at Various concentrations for NaCl,KCl,MgSO4 and K2SO4

Salt Values of i van’t Hoff factor i for complete dissociation of solute

0.1m 0.01m 0.001m

NaCl 1.87 1.94 1.97 2.00KCl 1.85 1.94 1.98 2.00MgSO4 1.21 1.53 1.82 2.00

K2SO4 2.32 2.70 2.84 3.00Hence at very dilute solution dissociation is complete

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Q1. Now you can guess 0.01m aqueous solution of which one will exhibit largest freezing point depression.(a) KCl, (b) C6H12O6 (c) K2SO4 (d) Al2(SO4)3

Q2. Why do doctors advise gargles by saline water in case of sore throat ? Saline water is hypertonic solution , therefore , fluids causing irritation in throat will come out .

Q3.What is the van’t Hoff factor for a compound which undergoes tetramerization in an organic solvent?In an organic solvent 1/4.

Q4.What would be van’t Hoff factor for a dilute solution of H2SO4

& Na2SO4.10H2O in water? 3 for each

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Q5.Which has the highest freezing point?1M Glucose, 1M NaCl,1MCaCl2,1M AlF3

(as number of ions increases boiling point will increase whereas freezing point decreases)

Q6.The molecular masses of polymers(macromolecules) are determined by osmotic pressure and not by other colligative properties” ExplainIt is because except for osmotic pressure, other colligative properties do not show sufficiently large measurable values.

Q7.Out of 1M solution of glucose and 1M solution of urea, which will have greater boiling point?Both will have equal boiling point

Q8.How is it that the boiling points of the following solutions in water are different?0.1M NaCl solution (b) 0.1M sugar solution0.1M NaCl solution due to its dissociation

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Q9. The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.256K. Kf for benzene is 5.12Kkgmol-1. What conclusion can you draw about the molecular state of acetic acid in benzene. To decide molecular state we have to calculate i value

Tf=iKfm Here Tf =0.256K, Kf=5.12 m=0.1molal i=?

0.256=iX5.12X0.1i=0.5(<1), hence molecules are in associated state

Q10.Equimolal solutions of NaCl and BaCl2 are prepared in water. Freezing point of NaCl is found to be -2oC. What freezing point do you expect for BaCl2 solution? i for NaCl=2, i for BaCl2=3

(Tf NaCl/Tf BaCl2)=3/2Tf BaCl2=3/2XTf NaCl

Tf BaCl2=3/2X2=3

Hence TfBaCl2=-3oC

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Q11. Calculate the amount of NaCl which must be added to 100g of water so that the freezing point is depressed by 2K. for water Kf=1.86K/m.Here i=2 From Tf=iKfm m= Tf/iKf =2/2X1.86= 0.538i.e 1000g of solvent contains 0.538mols of NaCl

100g of solvent contains=100X0.538 =0.0538mols 1000

Mass of NaCl=0.0538X58.5=3.1473g

Q12.Phenol associates in benzene to a certain extend to form dimer. A solution containing 2.0X10-2Kg of phenol in 1.0Kg of benzene has its freezing point decreased by 0.69K. Calculate the degree of association of phenol (Kf for benzene=5.12K kg mol-1)

M2(observed)=1000Kfw2 =1000X5.12X2X10-2 =148.4g/molW1Tf 1.0X0.69

M2(Calculated) for C6H5OH=94g/moli= M2(Calculated) = 94/148.4=0.633

M2(observed)2 C6H5OH ↔ (C6H5OH)n

Initial 1mol 0

After association 1-2X/2=1- /2 Total=1-+/2=1-/2

i=1-/2 or =2(1-i)=2(1-0.633)=0.734

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Q13.Calculate the boiling point of a solution containing 0.61g of benzoic acid in 50g of carbon disulphide assuming 84% dimerisation of the acid. The boiling point and Kb of CS2 are 46.2oC and 2.3K Kg mol-1 respectively.

2 C6H5COOH ↔ (C6H5COOH)n

Initial 1mol 0

After association 1-2X/2=1- /2 Total=1-+/2=1-/2

i=1-/2= 1-0.84 =0.581 1

From Tb=iKbm =0.58X2.3X0.61X1000

122X50=0.1334

Tb-Tbo =0.1334

Tb-(46.2+273)=0.1334Tb=0.1334+(46.2+273)=319.3334KHOMEWORK:

2.33.

19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing

point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation

constant of fluoroacetic acid.

2.40.

Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its

osmotic pressure is 0.75 atm at 27° C20

March 2020(AISSCE-2020)• THESE ARE THE QUESTION ASKED FROM THIS CHAPTER IN RECENT BOARD EXAM

1.What happens when :(i) A pressure greater than osmotic pressure is applied on the

solution side separated from solvent by a semipermeable membrane?

(ii) Acetone is added to pure ethanol? (2M)

• 2. State Henry’s law. Calculate the solubility of CO2 in water at298K under 760 mm Hg .(KH for co2 in water at 298 K is 1.25 x 106 mmHg). (2M)

3. The freezing point of a solution containing 5g of benzoic acid (M=122g mol—1) in 35g of benzene is depressed by 2.94K . What is the percentage association of benzoic acid if it forms adimer in solution ?(Kf for benzene = 4.9 K kg mol—1). (3M)

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REFERENCES1.CHEMISTRY NCERT TEXT BOOK PART I2. NEW COURSE CHEMISTRY PART I

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