Solutions & Colligative Properties Ipe

Solutions & Colligative properties

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SOLUTIONS Solution : A homogeneous mixture of two or more substances. E.g. Aqueous sugar solution formed by dissolving sugar in water. Solvent : The components which is present in large quantity in a solution. Solute : The component which is present in a lesser quantity in a solution. Solution = solvent + solute Binary solution : The solution which contains only two components.

Gas Gas Gas

Solid Liquid Gas

Sugar in water Alcohol in water Soda water (CO2 in water)

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Solid Liquid Gas

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Liquid Liquid Liquid

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* Based on the physical state of solvent and solute, the solutions are classified as Solvent Solute Example Solid Solid Copper in Gold Solid Liquid Mercury in Gold Solid Gas Occlusion of H2 in Pd

Camphor in air Water vapour in air Air (O2 in N2 )

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* Based on the nature of the solvent, the solutions are classified into 1) Aqueous solutions - in which water is the solvent. 2) Nonaqueous solutions - in which the solvent is other than water E.g., CHCl3 , C6H6, CCl4 etc., 3) Alcoholic solutions - in which Ethyl alcohol is the solvent.

* Based on the relative quantity of solute, the solutions are classified into Concentrated Solutions : The relative amount of solute is very high. Dilute Solutions : The relative amount of solute is very low. Methods of expressing concentration : Concentration or strength : The amount of solute present in specific amount of solution is generally referred to as concentration or strength of solution. Following are some methods which are used to measure the concentration of a solution. 1) Weight percentage (i) %

w V

denotes weight of solute in 100 mL of solution.

(ii) %

w w

denotes weight of solute in 100 g of solution.

2) Molarity (M) 3) Normality (N) 4) Molality (m) 5) Mole fraction (  )

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Molarity (M) : The number of moles of a solute present in one litre of the solution. M=

M=

Volume of solution in litres  V  n V  in L 

Weight of the solute W  Gram molecular Weight of thesolute GMW M =

M=

w 1 x GMW V  in L 

W 1000 x GMW V  in mL 

Units : mole. Litre-1= Molar (M) Note : Molarity depends on temperature.

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n=

no.of moles of solute  n 

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* When one mole of the solute is present in 1litre of solution, then the molarity (M) = 1 M (one molar solution) * When 0.1 mole of the solute is present in 1 litre of solution, then the molarity (M) = 0.1 M ( decimolar solution)

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Problems 1) What is the molarity of 500 mL of solution made by dissolving 2 moles of glucose in water? 2) What is the molarity of 250 mL of an aqueous solution containing 10 grams of NaOH? 3) Calculate the weight of solute in 100 mL of glucose solution. 4) What is the molarity of 10.6% Na2CO3 solution? 5) Calculate the molarity of 9.8% H2SO4 solution. The specific gravity of the solution is 1.12. Normality : The number of gram equivalent weights of the solute present in one litre of solution. N=

number gram equivalent weights of the solute (n e ) volume of the solution (V) in lit

ne 

weight of the solute (w) Gram equivalent weight of the solute (GEW)

N=

W 1 x GEW V(in L) or

W 1000 x GEW V(in mL) Units : gram equivalent weights . Litre-1 = Normal (N) The one litre of solution containing 1 gram equivalent weight of the solute is called one normal solution. If 0.1 gram equivalent weights are present, then that solution is called decinormal solution. N=

Calculation of Gram equivalent weights (GEW) * Gram equivalent weight of a substance depends on it’s nature and the chemical reaction in which it is participating. It can be calculated by using following formula.

Solutions & Colligative properties

GEW 

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GMW Equivalent factor

Equivalent weights of acids : GEW =

Gram molecular weight  or  formula weight basicity

Basicity : Number of replaceable hydrogens present in an acid molecule is called basicity of acid. Acid HCl (Hydrochloric acid)

Basicity

M.wt

Eq.wt

1

36.5

36.5  36.5 1

H2SO4 ( Sulfuric acid)

HO

S

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O OH

2

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O N

O

HO

98  49 2

1

63

63  63 1

OH

3

98

98  32.9 3

OH

2

82

82 = 41 2

1

66

66  66 1

1

60

60  60 1

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HNO3 ( Nitric acid)

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O

98

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O

H3PO4( Orthophosphoric acid) HO

P

OH

O

H3PO3( orthophosphorous acid) H

P

OH

O

H3PO2 (Hypophosphorus acid) H

P

H

OH

O

CH3COOH (Acetic acid) H3C

C O

H

Equivalent weights of bases : GEW =

Molecular weight  or  formula weight of base Acidity

Acidity : Total number of OH ions furnished by a base is called acidity of the base. 1) NaOH -

GEW 

GFW 40   40 g Acidity 1

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2) KOH

GFW 56   56 g Acidity 1

GEW  3) Ca  OH2

GFW 74   37 g Acidity 1 Equivalent weight of Salts: GEW 

GEW=

Formula Weight of the salt total amount of charge present on either cations  or  Anions

 GEW 

FW 58.5   58.5 g 1 1

 GEW 

FW 106   53g 2 2

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(4) AlC l3    A l 3   3C l 

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(2) Na2CO3   2 Na   CO3 2

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(1) NaCl   Na   Cl 

FW 133.5   44.5 g 3 3 Equivalent weight of oxidants and Reductants:

GMW Number of electrons lost or gained by a molecule

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GEW =

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 GEW 

A + B   Reductant Oxidant A   A n+ + ne  : 

n

B+ne  B :

A n+ +

Bn-

oxidation

Reduction

(1)KMnO 4 In acidic medium +

H Mn 7+ + 5e-   Mn +2

GMW 158 = = 31.6g 5 5 In basic medium GEW =

-

OH Mn 7+ + 3e-   Mn 4+

GEW =

GMW 158 = =52.66g 3 3

(2) K 2 Cr2 O7 +

H Cr 6+ +3e-   Cr 3+

GEW =

GMW 294 = =49g 6 6

(since two chromium atoms gain 2 x 3 = 6 electrons)

Solutions & Colligative properties

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Relation between Molarity and Normality Molarity ( M ) 

n W 1  X V GMW V

Normality ( N ) 

But

GEW 

ne W 1  X V GEW V

GMW Equivalent factor

For acids N = M x basicity For bases N = M x acidity

N = M x total charge either on cations or anions

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For salts

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1  W N  X  x equivalent factor  GMW V  N = M x equivalent factor

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Law of Dilution : For a solution, let the initial Molarity Initial Volume No of moles in solute

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For oxidants or reductants N = M x loss or gain of electrons Problems 1) What is the normality of 100 mL of solution containing 5.3 g of Na2CO3 ? 2) Calculate the weight of solute in 250 mL of 0.8 N H2SO4? 3) What is the normality of 20% NaOH solution? 4) The specific gravity of 2.45% H2SO4 solution is 1.245. Calculate its normality.

= M1 =V1 = n1

n1  n1 = M1V1 V1 After adding small volume of solvent, Final molarity =M2 Final volume = V2 No of moles in solute = n1  M1 =

 M2 =

n1  n1 = M 2 V2 V2

 M 1V1  M 2V2

similarly N1V1  N 2V2

Concentration of mixed solution In the first solution Molarity = M1 Volume = V1

In the second solution Molarity = M2 Volume = V2

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no. of moles of solute = n1 no of moles of solute = n2 For the mixed solution formed by above solutions, M mix 

N Mix =

N1V1 +N 2 V2 V 1  V2

Calculations involved in acid-base titrations Formulae 1. NaVa = NbVb Where Na = Normality of acid solution Nb = Normality of base solution Va = Volume of acid solution Vb = Volume of base solution M a Va M b Vb  na nb

Where M a = Molarity of acid solution

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M b = Molarity of base solution

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2.

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similarly

n1  n2 M 1V1  M 2V2  V1  V2 V1  V2

Va = Volume of acid solution

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Vb = Volume of base solution

n a = Stoichiometric coefficient of acid in the balanced equation

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n b = Stoichiometric coefficient of base in the balanced equation Problems: 1) How much water is to be added to 100 mL of 2 M H2SO4 to get 0.5 M H2SO4 ? 2) What is the normality of solution formed by adding 750 mL of water to 250 mL of 0.6 N solution of acetic acid? 3) 100 mL of 2 M HCl and 200 mL of 3 M HCl are mixed. What is the molarity of the mixture? 4) Equal volumes of 0.8 N and 0.2 N Na2CO3 solutions are mixed. What is the final normality of the solution formed? 5) 20 mL of 0.5 M NaOH solution is required to completely neutralize 40 mL of HCl. Calculate the molarity of HCl solution. 6) Calculate the volume of 0.4 M NaOH solution required to completely neutralize 100 mL of 0.2 M H2SO4 solution. 7) 10 mL of Na2CO3 solution is required to neutralize 20 mL of 0.2 M HCl completely. What is the strength of Na2CO3 solution? Calculate the amount of Na2CO3 in a 500 mL of solution. 8) Calculate the volume of 1.5 N acid solution required to completely neutralize 300 mL of 0.5 N alkaline solution. Molality (m) The number of moles of solute present per kilogram of the solvent is called as Molality (m) of the solution. molality(m) =

i.e.,

m=

number of moles of the solute (n) weight of the solvent in kilograms (W1 )

n W1 (in Kg)

Solutions & Colligative properties

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But

n

w2 GMW2



m=

w2 1 x GM W 2 W1 (in Kg)

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or or

m=

w2 1000 x GM W 2 W1 (in g)

Units : mole.Kg -1 = Molal (m) Note: Molality does not change with temperature.

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Problems 1) 4.6 g of ethyl alcohol is dissolved in 500 g of water. Calculate the molality of solution formed. 2) Calculate the molality of 20% NaOH solution. 3) The specific gravity of 6.3% HNO3 solution is 1.063. What is the molality of solution?

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Mole Fraction   : The ratio between the number of moles of one component of the solution to the total number of moles of all components of that solution is called as mole fraction of that component.

χi =

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The mole fraction of ith component in a solution containing ‘n’ components is given by ni n1 +n 2 +............n i .........n n

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For binary solutions

W1 n1 GMW1 the mole fraction of solvent  1   W1 W2 n1  n2  GMW1 GMW2

similarly mole fraction of solute   2

* For dilute Solutions : n2<<<
χ 2 =

w2 MW1 x MW2 w1

W2 n2 GMW2   W1 W2 n1  n2  GMW1 GMW2

Solutions & Colligative properties

and

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χA  χB  1

Problems: 1) A solution containing 18 g of glucose and 178.2 g of water. What are the mole fractions of glucose and water in this solution? Vaporization : The spontaneous escaping of liquid molecules in to the space from the surface of the liquid is called Vaporization of the liquid.

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Rate of Vaporization : Number of molecules escaping in unit time from the surface of the liquid is called rate of vaporization. The factors affecting the rate of vaporization are 1. Area of the Surface 2. Nature of the liquid 3. Temperature 4. Velocity of gases flowing above the surface of the liquid

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Condensation : The spontaneous entry of vapour molecules into the liquid phase from vapor phase is called condensation.

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Rate of Condensation :The Number of molecules entering the liquid phase from the vapour phase in unit time is called rate of condensation. The state at which the rate of vaporization becomes equal to the rate of condensation is called equilibrium state.

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Vapour Pressure : The pressure exerted by vapour molecules over the surface of a liquid when both liquid and its vapour are in equilibrium with each other at a given temperature is called vapor pressure of the liquid. (at that temperature) * Vapour pressure depends on the nature of liquid and temperature. * The liquid with weak intermolecular forces has high vapor pressure. * The vapour pressure increases with increase in temperature as the average kinetic energy also increase. Boiling Point : The temperature at which the vapour pressure of liquid becomes equal to the atmospheric pressure is called boiling point. Volatile liquids : The liquids with high vapour pressure and low boiling points. Eg., Ethers, Ethyl alcohol, Acetone etc., Non - Volatile Liquids : The liquids with low vapour pressure and high boiling points. Ex : H2O, octane, Benzene, carbon tetrachloride etc.,. Similarly solutes are also classified in to volatile solutes and nonvolatile solutes. Problems 1. Calculate the vapour pressure of a solution containing 10 g of a nonvolatile solute in 80 g of ethanol at 298 K. Given the M.W. of solute as 120 the vapour pressure of alcohol is 22.45 mm of Hg at 298 K. 2.The vapour pressure of a solution containing 108.24 g of a nonvolatile solute in 1000 g at H2O at 20oC is 17.354 mm of Hg. The vapour pressure of H2O at 20oC is 17.54 mm Hg. Find the molecular weight of the substance. 3.The vapour pressure of a 4% solution of a nonvolatile solute in H2O at 373K is 74.5 cm Hg.

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Calculate the molecular weight of the solute. 4.What is the vapour pressure of 3 M of solution at 23oC? Vapour pressure of H2O at 23oC is 19.8 mm of Hg. COLLIGATIVE PROPERTIES The properties of solutions which only depends on the number of solute particles in it and not on the nature of the solute are called colligative properties. 1. Relative lowering of vapor pressure (R.L.V.P) 2. Elevation in Boiling point  ΔTb  3. Depression in freezing point  ΔTf  4. Osmotic pressure (  )

o

P P  P = Po Po

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R.L.V.P =

YA

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Relative Lowering of vapour pressure (R.L.V.P) and Raoult’s law : Lowering of vapour pressure  ΔP  (L.V.P) : The vapour pressure of a solution (p) containing a nonvolatile solute is always less than the vapour pressure of pure solvent (P0). This phenomenon is called Lowering of vapour pressure  ΔP  . vapour pressure of pure solvent = P0 vapour pressure of solution = P Lowering of vapour pressure (L.V.P) =  ΔP  = P0 - P Relative Lowering of vapour pressure (R.L.V.P): The ratio of lowering of vapour pressure  ΔP  to the vapour pressure of pure solvent (Po) is called Relative Lowering of vapour pressure (R.L.V.P)

RAOULT’S LAW : The relative lowering of vapour pressure of dilute solutions containing a nonvolatile solute is equal to the mole fraction of that solute ΔP = X2 Po

Where X 2 = mole fraction of the solute n2 P o -P = Po n1 +n 2

for dilute solution n1 >>>> n 2 

n1  n 2  n1

And

P o -P n 2 = Po n1



W2 MW1 P  x o P M W2 W1

Where Po = vapour pressure of the pure solvent P = vapour pressure of the dilute solution W2 = wt of the solute

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MW2 = M.wt of the solute W1 = wt of the solvent MW1 = M.wt of the solvent Applications : The unknown molecular weight of substances can determined by using this law. Limitations : 1. Only applicable to dilute solutions 2. Solute must be nonvolatile 3. Solute should not under goes any molecular association or dissociation in the solution. 4. There is should be no chemical interaction between solute and solvent. Ideal solutions:- The solutions which obey Raoult’s law are called ideal solutions.

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Problems: 1) What is the vapour pressure of solution formed by dissolving 6 g of urea in 92 g of ethanol at 298 K? (The vapour pressure of alcohol is 22.45 mm Hg at 298 K) 2) The vapour pressure of solution formed by dissolving 120 g of non volatile solute in 1800 g of water at 20oC is 17.24 mm Hg. What is the molecular weight of solute? ( The vapour pressure of water at 20oC is 17.54 mm Hg)

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Elevation in Boiling point (ΔTb ) As the vapour pressure of solution containing non volatile solute is less than that of pure solvent, the temperature at which the vapour pressure of solution becomes equal to the atmospheric pressure is always greater than the boiling point of solvent. i.e., the boiling point of solution (Tb) is always greater than the boiling point of pure solvent (Tb0)

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Tb  Tb o

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Tb  Tb o  Tb is called elevation in boiling point. The increase in the boiling point of solution containing non volatile solute when compared to

the boiling point of pure solvent is called elevation in boiling point ( Tb ). Relation between elevation in boiling point (ΔTb ) and molality (m) of the solution. In the following diagram, the curves ABC, DEF & GHK show the relations between boiling points and vapour pressure of pure solvent, solution-I and solution-II respectively.

C

t en

I II n n it o o i t lu lu So o S

B

P0 P1 Pressure

lv So

F

K

T1

T2

E P2

H A D G T0 Temperature

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In the above graph To, T1, T2 are boiling points of pure solvent, solution-I and solution-II respectively Po, P1, P2 are vapour pressure of pure solvent, solution-I and solution-II at To By considering above curves as almost straight lines, From BFE & BKH triangles BE BF  BH BK Po  P1 T1  To or P  P  T  T o 2 2 o

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P1 T1 or P  T 2 2

P  X2 Po

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where X2 = Mole fraction of solute

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P  Tb From Raoult’s law

or P  Po X 2

YA

so Tb  X 2

X2 

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or Tb  KX 2 ( where K= A proportionality constant) But for dilute solutions n2 W2 MW1  x n1 MW2 W1

W2 1000 Molality (m) = MW x W 2 1  X2 

m x MW1 1000

m x MW1 1000 Now from the law of thermodynamics Tb = K x

RTo 2 K v H ΔTb =

RT0 2 m x MW1 x ΔvH 1000

or Tb  Kb m Where K b 

RT0 2 x MW1 RT0 2 =  v H x 1000 1v x 1000

( since  v H  lv x MW1 )

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Where Kb= molal elevation constant R = Gas constant To = boiling point of pure solvent MW1 = molecular weight of solvent  v H = Molar heat of vaporization of solvent lv = latent heat of vaporization Molal Elevation of boiling point constant or ebbulioscopic constant (Kb) : The elevation in boiling point of one molal solution is called molal elevation constant. Units: K. Kg. mol-1

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Depression in freezing point of the solution ( T f ) The freezing point of solution (Tf) containing a non volatile solute is always less than the freezing point of pure solvent (Tfo). This is called depression in freezing point ( T f ) . Tf  Tf o

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T f o  T f  T f = depression in freezing point.

P

I II n nt n o e o i i lv ut ut so C Sol ol S F

Pressure

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YA

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* In the following diagram, the curves OP, ABC & DEF show the relations between freezing points and vapour pressure of pure solvent, solution-I and solution-II respectively.

O

Po

A

P1

B

P2 D

E

T2 T1 T0

Temperature

For dilute solutions, these curves are almost straight lines and hence from triangles ΔOAB & ΔODE OB AB  OE DE P0  P1 T0  T1  P0  P2 T0  T2

P1 T1  P2 T2



P  T

From Raoult’s law P  X2 P0

or

P  P0 X 2

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where X 2 = mole fraction of solute P  X 2 

Tf  X 2

or Tf  KX 2 For dilute solutions X2 



m x MW1 1000

Tf  K x

m x MW1 1000

From the laws of thermodynamics

 Tf 

or

Tf  K f m

where K f 

RT0 2 x MW1 RT0 2  f H x1000 l f x1000

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RT0 2 m x MW1 x f H 1000



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RT0 2 f H

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K

R = Gas constant T0 = Freezing point of solvent

MW1 = Molecular weight of solvent

YA

Kf = Molal depression constant or Cryoscopic constant

V. AD VA I AG T

 f H = Molar heat of freezing of solvent lf = latent heat of freezing

Definition of Kf: Depression in freezing point of one molal solution containing a nonvolatile solute is called molal depression in freezing point constant or cryoscopic constant. Units: K. Kg. mol-1 PROBLEMS 1. 0.1 moles of a nonvolatile and non-electrolytic solute is dissolved in 100 g of a solvent to elevate it’s boiling point by 0.12oC. Calculate Kb of the solvent. 2. The elevation in boiling point of a solution containing 0.20 g of solute in 20 g of ether is 0.17oC. Kb for ether is 2.16 K.kg/mol. Calculate the molar mass of solute. 3. 3 g of solute is dissolved in 22 g of water. The depression of freezing point shown by the solution is 1.45 K. kf for H2O = 1.86 K kg mol-1. Calculate molar mass. Osmotic pressure (  ) Osmosis : The flow of solvent from a dilute solution to a concentrated solution when both of these solutions are separated by a semi permeable membrane is called osmosis. Semipermeable membrane :The membrane which allows only solvent molecules to flow through it and not the solute particles is known as semipermeable membrane. Eg., Parchment paper,

Solutions & Colligative properties

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Cellophane paper, Some animal membranes, Membranes prepared by precipitating some inorganic salts like copper ferro cyanide. Osmotic pressure ( ) : The pressure required to stop the flow of solvent molecules from solvent to solution when both are separated by semi permeable membrane. Reverse Osmosis : The flow of solvent from solution to solvent which are separated by a semi permeable membrane by applying pressure greater than the osmotic pressure over the solution is called reverse osmosis. Isotonic Solution :The solutions with same osmotic pressure.

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e.g. Blood is Isotonic with saline [ 0.9 % w v NaCl solution]

YA

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Uses of Osmosis : 1. Plants take up water from the soil by the phenomenon of osmosis through roots hairs. 2. If the osmotic pressure of the contents of the living cell is not equal to that of the contents surrounding it outside, two phenomena may occur. a) If the out side concentration is higher, water may enter the cell. The cell bulges and finally bursts. This is called Hemolysis. b) If the concentration in the cell is lower, water may come out of the cell and the cell collapses. This is called Plasmolysis. Both are dangerous

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Van’t Hoff theory of dilute solution : According to Van’t Hoff, the solute in dilute solutions behaves as a gas and hence all the gas laws can be applied to dilute solutions. Boyle-Van’t Hoff’s law: The osmotic pressure (  ) of a dilute solution is directly proportional to the concentration of the solution at constant temperature. C But the concentration of solution is inversely proportional to the volume. 1 C V 1     (1) V or

 V = K  constant ---------> This is like Boyle’s law

Charle-van’t Hoff’s law: The osmotic pressure of a dilute solution at constant concentration is directly proportional to the absolute temperature.   T - (2)

  K'T From equations (1) & (2) T π  V

---------> This is like Charle’s law

Solutions & Colligative properties

T V

π  S

or

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15

( S is constant)

1 C   CST V 

But 

But the value of ‘S’ is almost equal to gas constant ‘R’ Hence

 = CRT

 =

or



( C =

 V  nRT W n  MW W.R.T V  MW

number of moles of solute(n) ) Volume of the solution(V)

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But

n RT V

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W.R.T V W.R.T 1 MW  XC ( V  ) or  C Where W = weight of solute & MW = Molecular weight of solute

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MW 

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Problems 1.An aqueous solution (1 litre) contained 6 g urea (mol.wt = 60). It’s osmotic pressure is found equal to solution of glucose ( mol. wt = 180). How much weight of glucose is present in one litre of the solution. 2.At 283 K, The osmotic pressure of 2% solution of X is 7.87 X 104 Nm-2. Calculate molar mass of X. 3.An aqueous solution glucose containing 20 X 10-3 kg of glucose in a litre. Calculate the osmatic pressure of the solution at the temperature 250C. Ostwald’s Dynamic Method :In this Experiment four bulbs which are connected to each other are taken. The fourth bulb is connected to a U-shaped tube containing anhydrous CaCl2 The first two bulbs (A) are filled with solution whereas the last two bulbs (B) contain pure solvent. Dry air

A

A

solution

B

B

solvent

(Anhydrous) CaCl2

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The weights of bulbs and U-tube are determined before the experiment. A current of dry air is passed through the bulbs. It carries away the vapour of solvent while passing through the bulbs. The loss of weight in first two bulbs (A) is proportional to vapour pressure of solution (P). wt. loss in ‘A’ Bulbs  x   P But the vapour pressure of solvent is greater than that of the solution, some more vapour enters into the air while passing through the last two bulbs. Hence the loss in last two bulbs (B) is proportional to (Po -P). wt. loss in ‘B’ Bulbs (y) α Po -P Gain in wt.of ‘U’ shaped tube  z  = total wt. loss in A and B bulbs ( x  y )

Po -P =X 2 Po y  X2 x+y

or

y W2 MW1  x x  y MW2 W1

i.e.,

MW2 

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i.e.,

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z =x+y = P + (Po -P) =Po According to Raoult’s law

W2 x y x MW1 x W1 y

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Where MW2=Molecular weight of solute W2 = weight of solute MW1 = Molecular weight of solvent W1 = weight of solvent PROBLEMS 1. At a certain temperature dry air is sent into a solution & solvent continuously. The solution is prepared by dissolving 3.50 g of solute in 100 g of solvent. The loss in weights of containers containing solution and solvent are 0.975 g, 0.025 g respectively. The molecular weight of solvent is 78. What is the molar mass of solute ? Cottrell’s Method In this method, Beckman’s thermometer is used to determine the small magnitudes of elevation in boiling point or depression in freezing point accurately. It consists of a mercury reservoir at one end and a bulb at the other end. These are connected by a capillary. The elevation or depression can be measured in a range of -6oC to 300oC by adjusting the amount of mercury in the bulb with help of reservoir. The apparatus consists of a main tube provided with a side tube. The main tube contains another inner tube through which the Beckman’s thermometer is inserted. A platinum wire is fused at the bottom of the outer tube and heated during the boiling process.

Solutions & Colligative properties

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1

Cottrells apparatus 1. beckmann Thermometer 2.Condenser 2 3

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3. Boiling tube 4.Solution 5. Inverted funnel (Cottrell's device)

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Beckmann Thermometer

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W2 1000 x MW2 W1

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ΔTb  Tb -Tb o =K b x

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An inverted funnel called cottrell’s device is placed in the solution. The stem of this funnel is split into three symmetrical tubes. This device pumps the bubbles of the solution along with the vapour on to the thermometer bulb so that the liquid and vapour will be in equilibrium on the surface of the bulb at the boiling point. Thus this device helps in reducing super heating. A known weight (w1) of solvent is taken in the Cottrell’s tube and the boiling point (Tbo)is determined. Then a known weight of solute (w2) is dissolved in the solvent and the boiling point (Tb) is measured. The molar mass of the solute is determined using the following equation.

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Where MW2=Molecular weight of solute W2 = weight of solute W1 = weight of solvent Kb= Molal elevation constant of solvent. PROBLEMS 1. The boiling point of CHCl3 was raised by 0.325 K when 5.141 X 10-4 kg of anthracene is dissolved in 3.5 X 10-3 kg of CHCl3. Calculate molar mass of anthracene.(kb = 3.9 K kg mol-1). Rast Method This method is used to determines the depression in freezing point of a solution. As camphor is used as solvent, this method is generally used for solid solutions. A known weight (w1) of finely powdered camphor is mixed with a known weight (w2) of solute and melted to form a homogeneous solution and then cooled and powdered.. This mixture is taken in a capillary tube and the melting point (Tf) is determined accurately. The melting point (Tfo) of pure camphor is determined separately. The molar mass of the solute is determined by using the following equation. ΔTf =K f x

W2 1000 x MW2 W1

Where MW2=Molecular weight of solute W2 = weight of solute W1 = weight of solvent Kf= Molal freezing point constant of solvent (camphor) Berkeley - Hartley Method This method is used to determine the osmotic pressure. The apparatus consists of a porous tube opened at both the ends. Copper ferrocyanide is precipitated in the pores as semi permeable

Solutions & Colligative properties

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membrane by electrolytic method. Then the two ends are closed with rubber corks. A thistle funnel is fixed in one rubber cork and a capillary tube is introduced in another cork. The porous tube is fixed in an outer cylindrical tube made up of gun metal. This outer tube is provided with a manometer. Water is added into the porous tube through the funnel and the level of water in the capillary is recorded. Then the space between the inner and outer tubes is filled with the solution of known concentration (C) which is formed by dissolving W g of solute. 1

5 6

Berkeley-Hartley method 1.Solvent reservoir 2.Semipermeable

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3.Solution 4. Capillary tube

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5.Piston

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2

7. Solvent

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3

6. Pressure guage

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As osmosis starts, the level of liquid in the capillary decreases. Now the pressure is applied externally to adjust the level of water in the capillary to original position. This external pressure applied is taken as osmotic pressure(  ) and the molar mass (MW) of the solute is calculated by using the following equation. W.R.T MW  xC 

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Abnormal molar mass and van’t Hoff factor (i): Solutions which obey Raoult’s law are called ideal solutions. But these solutions deviate from ideality when the concentration is increased or when the solute undergoes either dissociation or association or when there is interaction between solute and solvent. The colligative properties measured for these non-ideal solutions are termed as abnormal colligative properties. Van’t Hoff factor (i): It is the ratio of experimentally determined abnormal colligative property to that of theoretically calculated colligative property is called van’t Hoff factor. i=

experimental value of colligative propery Calculated value of colligative property

The colligative property is inversely proportional to the molar mass of the solute and hence i=

Calculated molar mass of solute experimental molar mass of solute

For ideal solutions i = 1 For non-ideal solutions i < 1 when solute undergoes dissociation. i > 1 when solute undergoes association. The equations for calculating different colligative properties can be modified to correct non ideality as given below. ΔP = i.X 2 Po

Solutions & Colligative properties

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Tb  i.K b m Tf  i.Kf m

 = i.CRT The relation between ‘i’ and degree of dissociation or association ‘ ’ can be calculated as follows.

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Relation between ‘i’ and ‘  ’ when solute undergoes dissociation: If one mole of ‘A’ ionize by giving ‘n’ moles of ions and the degree of dissociation is ‘ ’ , then A  nB initial 1 mole 0 At equilibrium (1-  ) n Total no.of particles in the solution = (1-  ) + n   = 1+n  -  = 1+  (n-1)

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Total no.of particles in the solution 1    n  1  no.of particles taken 1 i = 1+  (n-1) i-1=  (n-1)

dissociation



i-1 n-1

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

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van't Hoff factor (i) 

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Relation between ‘i’ and ‘  ’ when solute undergoes association: If ‘n’ moles of A combine to give A n molecules and the degree of association is ‘  ’, then,

nA 1 mole



A n initial 0  at equilibrium (1- ) n   1 Total no.of particles in the solution = (1- ) + = 1+   = 1+ ( -1) n n n no.of particles in the solution no.of particles taken 1 1+ ( -1) n  1 1 i.e., i = 1+ ( -1) n 1 i-1= ( -1) n i-1  = association 1 1 n

 van't Hoff factor (i) =

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TEST YOUR UNDERSTANDING State whether the following statements are true or false. 1) Molarity is the number of moles of solute dissolved in 1 litre of solvent. 2) 1 litre of 1M HCl solution is added with 1 litre of water. The molarity of final solution will be 2M. 3) The equivalent weight of H2SO4 in the following reaction is 49 g. H 2SO 4 + KOH   KHSO 4 + H 2O

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4) The molality of an aqueous solution is 2 m at 270C. The molality of same solution will be 2 m only, at 1270C. 5) Vapour pressure of H2O at 200C is 17.54 mm Hg. The vapour pressure is increased to 17.60 mm Hg when 0.1 moles of urea is added. 6) Depression in freezing point of one molal solution containing a nonvolatile solute is called molal depression in freezing point constant or cryoscopic constant. 7) The osmatic pressure of 1M NaCl is equal to the osmatic pressure of 1M sucrose solution. 8) The basicity of H3PO2 is equal to 3 9) An aqueous solution of 0.1M urea has same boiling point as that of aqueous solution of 0.1M sucrose solution. 10) The Van’t - Hoff factor (i) of aqueous solution of Acetic acid is greater than 1, whereas it’s value is less than 1 in benzene solution.