3. Solutions. detail notes on solutions and explanation

3. Solutions. detail notes on solutions

1. S O L U T I O N S

2. Definitions A solution is a homogeneous mixture of two or more substances OR A solution is a homogenous mixture of solute and solvent. OR A solution is a homogenous mixture of two substances consisting of one phase.

3. A solute is dissolved in a solvent . - solute is the substance being dissolved - solvent is the liquid in which the solute is dissolved an aqueous solution has water as a solvent Binary solution: A homogenous mixture consisting of one phase and containing only two components i.e. one solute and one solvent e.g. Solution of NaCl in water.

4. Dilute Solutions: A solution containing relatively small quantity of solute as compared with the amount of solvent. Concentrated Solution: A solution containing large amount of solute in the solution than that in dilute solution. Un-saturated solution: a solution in which more solute can be dissolved at a given temperature is called as an unsaturated solution.

5. A saturated solution is one where the concentration is at a maximum - no more solute is able to dissolve at a given temperature. A saturated solution represents an equilibrium: the rate of dissolving is equal to the rate of crystallization. The salt continues to dissolve but crystallizes at the same rate or under this condition, the number of molecules leaving the solute is equal to the number of molecules returning to the solid phase i.e. solute. Super Saturated Solution: A solution that contains a relatively larger amount of solute than that required for saturation it is prepared by heating and adding more and more solute.

6. SO L U T I O N S Materials may be mixed together to form three types of solutions ; On the basis of dispersed particles or solute molecules, solutions are classified as 1. True Solutions It is defined as a mixture of 2 or more components that form a homogenous molecular dispersions i.e., a phase system, the composition of which can vary over a wide range . In solutions , we will be mainly concerned with this type of solutions

7. 2. Coarse dispersions For example emulsions and dispersions. The diameter of the particles for the most part being larger than 0.1 µm. 3. Colloidal dispersions These represent a system with a particle size intermediate between a true solution and coarse dispersions, i.e., between 0.001 to 0.5µm. A colloidal dispersion may be considered a two-phase system (heterogeneous) or a single-phase system (homogeneous)

8. System A system is a bounded space or a definite quantity of substance that is under observation and experimentation . Phase A phase is a distinct homogenous part of a system separated by definite boundaries from other parts of the system. It may be considered as a single piece of ice floating in water or it may be distributed as small particles throughout the system like oil droplets in an emulsion or solid particles in a suspension.

9. Binary Solution A solution composed of only two substances is known as binary solution, and the components or constituents are referred to as Solvent and Solute. In a solution, the solute is dispersed uniformly throughout the solvent . The constituent in greater amount in a binary solution is designated as solvent and one in lesser amount as solute. Solutions

10. The physical properties of the system are classified as; Extensive properties Depending on the quantity of the matter in the system (e.g., mass and volume). Intensive properties Independent of the amount of the substance in the system (e.g ., temp ., pressure, density, surface tension. PROPERTIES OF SOLUTIONS

11. PROPERTIES OF SOLUTIONS Solutions The physical properties of substances may be classified as; Colligative properties Additive properties Constitutive properties

12. COLLIGATIVE PROPERTIES Solutions Depend upon the number of particles in a solution. The colligative properties are; Vapor pressure lowering Freezing pressure depression Boiling point elevation Osmotic pressure

13. Solutions ADDITIVE PROPERTIES These properties depend on the total constituents of the atoms in the molecules or on the sum of the properties of the constituents in a solution e.g ., molecular weight i.e., the sum of the masses of the constituent atoms present in the solution .

14. CONSTITUTIVE PROPERTIES These properties depend on the arrangements and to a lesser extent on the number and kind of atoms within a molecule. These properties give clues to the constitution of individual compounds and groups of molecules in a system. Many physical properties may be partly additive , or partly constitutive . For example, refraction of light, electric properties, surface , and interfacial tension.

15. TYPES OF SOLUTIONS Solutions Matter exists in 3 states i.e., solid liquid and gas, so 9 types of solutions are possible. Types of Solutions Example Solid in solid Brass (Alloys) Solid in liquid Sugar solution Solid in gas Aerosols Liquid in solid Dental amalgam Liquid in liquid Ethanol in water Liquid in gas Water in air Gas in solid Hydrogen in palladium Gas in liquid Water in air, Soda water Gas in gas Atmosphere

16. Types of Solutions: Based on the physi cal states of solute and solvent:

17. One way to distinguish between solutions that contain ions and those that contain molecules is an electrical conductivity test . ELECTROLYTES AND NON-ELECTROLYTES Solutions

18. Dissolution vs reaction Dissolution is a physical change—you can get back the original solute by evaporating the solvent. If you can’t, the substance didn’t dissolve, it reacted. 2 Ni(s) + HCl(aq) NiCl 2 (aq) + H 2 (g) NiCl (s) d r y Solutions

19. Degree of saturation Saturated solution Solvent holds as much solute as is possible at that temperature. Undissolved solid remains in flask. Dissolved solute is in dynamic equilibrium with solid solute particles. Solutions

20. Degree of saturation Unsaturated Solution Less than the maximum amount of solute for that temperature is dissolved in the solvent. No solid remains in flask. Solutions

21. Solutions Degree of saturation Supersaturated Solvent holds more solute than is normally possible at that temperature. These solutions are unstable; crystallization can often be stimulated by adding a “seed crystal” or scratching the side of the flask.

22. Ways of Expressing Concentrations of Solutions

23. 5 ways of expressing concentration Mass percent: (mass solute / mass of solution) * 100 Molarity(M): moles solute / Liter solution Molality* (m) - moles solute / Kg solvent Normality (N)- gram equivalent of solute/ liter solution Mole Fraction - moles solute / total moles solution * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.

24. MASS PERCENTAGE Solutions %age by weight (%w/w) = Grams of solute in 100 g of solution %age by volume (%v/v) = Milliliter of solute in 100ml of solution %age weight by vol(%w/v) = Grams of solute in 100ml of solution Milligram percent = Milligram of solute in 100ml of solution

25. Mass Percentage or Percentage Expression % expression is an expression of parts of solute per 100 parts Of the solution mass of A in solution Mass % of A = x 100 total mass of solution

26. % w/w: It expresses the no. of grams of the solute per 100 gram of the solution. e.g. a 10 % w/w aqueous glycerine solution means 10 g of glycerine dissolved (in sufficient water to make overall) 100 gram of the solution.

27. % v/v: It expresses the no. of milliliters of the solute per 100 milliliters of the solution. e.g. a 10 % v/v aqueous ethanolic solution means 10 ml of ethanol dissolved in sufficient water to make overall 100 mL of the solution.

28. % w/v It expresses the no. of grams of the solute per 100 mL of the solution. e.g. a 10 % w/v aqueous NaCl solution means 10 g of Nacl dissolved in sufficient water to make overall 100 mL of the solution.

29. % v/w It expresses the no. of mL of the solute per 100 gram of the solution. e.g. a 10 % v/w aqueous glycerine solution means 10 ml of glycerine dissolved in sufficient water to make overall 100 gram of the solution.

30. Parts per Million and Parts per Billion ppm = total mass of solution mass of A in solution  10 6 Parts per Million (ppm) Parts per Billion (ppb) ppb = mass of A in solution  10 9 total mass of solution Solutions

31. moles of A total moles in solution X A = Mole Fraction ( X ) In some applications, one needs the mole fraction of solvent , not solute —make sure you find the quantity you need!  Mole percent ?

32. M = mol of solute Molarity ( M ) L of solution Moles (gram molecular weight) of solute in 1 liter solution . All solutions of the same molarity contain the same number of solute molecules in a definite volume of solution . When a solution contains more than one solute, it may have different molar concentrations with respect to various solutes. Because volume is temperature dependent, molarity can change with temperature.

33. Molality ( m ) m = mol of solute kg of solvent Moles of solute in 1000g of solvent Because neither moles nor mass changes with temperature, molality (unlike molarity) is not te mperature dependent .

34. Normality (N) L of Solution Gram equivalent weight of solute in 1 liter of solution. Gram equivalent ? N = Gram eq. wt. of solute Solutions

35. M a ss / M a s s M o l e s / M o l e s M o le s/ M a ss M ol es / L

36. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

37. Molarity and Normality Difficulties encountered in expressing the molarity of ion or radicals in sol. A molar solution of NaCl is 1M with respect to both Na and Cl ions, where as a molar solution of Na 2 CO 3 is 1M with respect to CO 3 ions and 2M with respect to Na ions. A molar solution of NaCl is also 1N with respect to both Na and Cl ions, however a molar solution of Na 2 CO 3 is 2N with respect to both the Na and CO 3 ions. Both molar and normal solution have a disadvantage of changing values with temperature because of expansion or contraction of liquids. Another problem is with unknown volume of the solvent in final solution especially measuring vapor pressure or osmotic pressure.

38. Solutions Molality Molal solutions are expressed in terms of weight and does not have disadvantages as in case of molar or normal solutions. In aq. solutions more dilute than 0.1M, may be assumed for practical purposes having equivalent in terms of molality and molarity. Addition of another solute to a 1 molal solution does not affect the molality of previous solute although total volume and the weight of solutions has changed.

39. SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn 2+ What is the concentration of Zn 2+ in parts per million? PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? PRACTICE EXERCISE A commercial bleach solution contains 3.62 mass % NaOCl in water. Calculate (a) the molality and (b) the mole fraction of NaOCl in the solution.

40. SAMPLE EXERCISE 13.4 Calculation of Mass-Related Concentrations PRACTICE EXERCISE (a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution? Answers: (a) 2.91%, (b) 90.5 g of NaOCl (a) A solution is made by dissolving 13.5 g of glucose (C 6 H 12 O 6 ) in 0.100 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of groundwater was found to contain 5.4 g of Zn 2+ What is the concentration of Zn 2+ in parts per million? Solution (a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percentage of solute. Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water). Solve: g is 1  10 –6 g, Comment: The mass percentage of water in this solution is (100 – 11.9)% = 88.1%. (b) Analyze: In this case we are given the number of micrograms of solute. Because 1 5.4 g = 5.4  10 –6 g. Plan: We calculate the parts per million using Equation 13.6. Solve: Answers: (a) 0.505 m , (b) 9.00  10 –3

41. Colligative Properties Solutions Colligative properties depend only on the number of solute particles present, not on the identity of the solute articles Among colligative properties are Vapor pressure lowering Boiling point elevation Freezing point depression Osmotic pressure

42. Lowering of Vapor Pressure Solutions As solute molecules are added to a solution, the solvent becomes less volatile (=decreased vapor pressure ). Solute-solvent interactions contribute to this effect. Therefore, the vapor pressure of a solution is lower than that of a pure solvent .

43. According to Roult’s law, the VP of the solvent over a dilute sol is equal to the VP of pure solvent times the mole fraction of solvent in sol. P = P 1 O X 1 Since solute is considered to be nonvolatile, it is more convenient to express the VP of sol in terms of the conc. of the solute, rather than the mole fraction of solvent and it is done as; The sum of mole fraction is unity X 1 + X 2 = 1 X 1 = 1 – X 2 X 1 = mole fraction of solvent X 2 = mole fraction of solute Roult’s law is P = P 1 O X 1 replacing the value of X 1 P = P 1 O (1- X 2 ) P 1 O - P = P 1 O X 2 P 1 O – P = P 1 O P = X 2 = n 2 / n 1 + n 2 P/P o is relative lowering of VP hence it is colligative property

44. Solutions Determination of Vapor Pressure The VP of a solution may be determined by means of a manometer, and the VP lowering can be obtained by subtracting the VP of solution from the VP of pure solvent. Another method is using the isopiestic method for precise determination. Hill and Blades developed a thermoelectric method for the determination of VP.

45. S o l u t i o n s Boiling Point Elevation The normal boiling point is the temp at which the VP of liquid becomes equal to an external pressure of 760 mm of Hg. The boiling point of a solution of a nonvolatile solute is higher than that of the pure solvent, owing to the fact that the solute lowers the VP of the solvent, as shown in fig. The VP curve of the solution lies below that of pure solvent and the temp of the solution must be elevated to a value above that of the solvent in order to reach the normal boiling point. The elevation of the boiling point is shown in the fig. T o T

46. Solutions Boiling Point Elevation T – To = ΔT b The ratio of elevation of boiling point ΔTb to the VP lowering ΔP=P o -P at 100 o C is approximately a constant at this temp and is written as; ΔT b = k or ΔT b = kΔP ΔP More over since P o is a constant the boiling point elevation may be considered proportional to ΔP/ P o , the relative lowering of VP. According to Roult’s law the relative lowering of VP is equal to mole fraction of solute, therefore ΔT b = kX 2 Since boiling point elevation depends upon the mole

47. Boiling Point Elevation ΔT b = kX 2 ΔT b = k ΔT b = kM1 m 1000 ΔT b = k b m In which ΔT b is known as boiling elevation and k b is called the molal elevation constant or the ebullioscopic constant. k b has a characteristic val S u ol e utions for each solvent. In dilute solutions, X2 is approximately equal to, m (1000 / M1) m (1000 / M1) The relative VP lowering can be expressed in terms of molal concentration of the solute by setting the weight of solvent w1 equal to 1000 g for a aq. sol. ΔP n 2 w 2 /M 2 X2 = ---- ͇͂ --- = ------- Po n 1 1000/M 1

48. Solutions Determination of boiling point elevation Boiling point elevation is determined experimentally by placing a weighed amount of the solute and the solvent in a glass vessel provided with a thermometer and a reflux condenser. In the Cortell boiling point apparatus , the vapors and the boiling solvent are pumped by the force of ebullition through a glass tube and sprayed over the thermometer bulb to obtain an invariant equilibrium temp. The boiling point of the solvent is also measured in the same instrument.

49. Solutions Freezing Point Depression The normal freezing point or melting point of a pure compound is the temp at which the solid and the liquid phases are in equilibrium under a pressure of 1 atm. Equilibrium here means that the tendency for the solid to pass into the liquid state is the same as the tendency for the reverse process to occur.

50. S o l u t i o n s Freezing Point Depression The value To as shown in fig for water saturated with vapors at this pressure is arbitrarily assigned a temp of o C. The triple point of air free water at which ice, liquid and vapors are in equilibrium lies at a temp of 0.0098oC and pressure of 4.58 mm of Hg. Now if a solute is added in the liquid, the triple point, the escaping tendency or VP of solvent is lowered below that of pure solid solvent. The temperature must drop in order to reestablish equilibrium between the liquid and solid. Because of this fact the freezing point of solution is always less than pure solvent. T T o

51. Freezing Point Depression The change in freezing point can be found similarly:  T f = K f m Here K f is the molal freezing point depression constant of the solvent also called Cryoscopic constant. S o l u t i o n s  T f is subtracted from the normal freezing point of the solvent.

52. Determination of freezing point depression Solutions a) b ) Several methods are available for the determination of freezing point depression. They include; The Beckmann method The equilibrium method The Beckmann apparatus is shown in fig. It consist of jacketed tube with a sidearm through which the test material may be introduced. A Beckmann thermometer is supported in the tube and extend into the solution. A glass stirrer is also attached. The tube and jacket are supported in a vessel containing a cooling mixture of ice and salt. Freezing point of pure solvent and solution is read from the thermometer.

53. Boiling Point Elevation and Freezing Point Depression Solutions In both equations,  T does not D epend on what the solute is , but only How many p a r t i c l e s a r e dissolved.  T b = K b m  T f = K f m

54. Osmotic pressure Diffusion : Is the movement of particles along the concentration gradient i.e the movement of particles from higher solute concentration to lower solute concentration. Osmosis Is the spontaneous movement of water across a semipermeable membrane from an area of low solute concentration to an area of high solute concentration or osmosis is the movement of solvent molecule from dilute solution to concentrated solution through semi permeable membrane. ( semi-permeable membrane is a type of membrane which permits only solvent molecules to pass through it) Osmotic Pressure The hydrostatic pressure builds up on the semi-permeable membrane which just stops the osmosis of pure solvent into the solution through semi permeable membrane.

55. O s m o si s Solutions Semipermeable membranes allow some particles to pass through while blocking others. In biological systems, most semipermeable membranes (such as cell membrane) allow water to pass through, but block solutes.

56. Osmosis In osmosis, there is net movement of solvent from the area of higher solvent concentration ( low er solute concentration ) to the area of lower solvent concentration h ighe r ( solute c o n c e n t r a t i o n ) . Water tries to equalize the concentration on both sides until the pressure is too high.

57. 1 . 2 . 3 . The addition of non-volatile solute to solvent forms a solution in which the VP of the solvent is reduced (Raoult’s law). If pure solvent is now placed adjacent to the solution but separated from it by a semipermeable membrane, solvent molecules will pass through the membrane into the solution in an attempt to dilute out solute and raise the VP back to its original value (that of original solvent) The osmotic pressure that is set up as a result of the passage of solvent molecules may be determined either by measuring the hydrostatic heed appearing in the solution or by applying a known pressure that just balances the osmotic pressure and prevents any S n o e lut t ions movement of solvent molecules into solution . Osmosis cont.. It is advantageous to consider osmosis in terms of following steps;

58. Solutions Osmosis cont…. The osmotic pressure thus obtained is proportional to the reduction in VP brought about by the concentration of solute particles. Since this is a function of molecular weight of the solute, osmotic pressure is a colligative property.

59. Solutions The measurement of Osmotic pressure The osmotic pressure of solution, is not measured conveniently by observing the height that the solution attains in the tube at equilibrium. The concentration of the final solution is not known. A more exact measure of the osmotic pressure of the undiluted solution is obtained by determining the excess pressure on the solution side that just prevents the passage of solvent through the membrane. Osmotic pressure is defined as the excess pressure, pressure just greater than that above the solvent, that must be applied to the solution to prevent the passage of the solvent through semipermeable membrane. Osmotic pressure is the colligative property best suited to the determination of molecular weight of polymers such as proteins.

60. Solutions Osmotic Pressure The pressure required to stop osmosis, known as osmotic pressure  , , is calculated using Van’t Hoff equation;  = ( n ) RT = MRT V where M is the molarity of the solution If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic .

61. Osmosis and Blood Cells A cell placed in an isotonic solution. The net movement of water in and out of the cell is zero because the concentration of solutes inside and outside the cell is the same. In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die. In a hypotonic solution, the concentration of solutes outside of the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to burst.

62. Osmosis in Blood Cells If the solute concentration outside the cell is greater than that inside the cell, the solution is hypertonic . Water will flow out of the cell, and crenation results. Solutions

63. Osmosis in Cells If the solute concentration outside the cell is less than that inside the cell, the solution is hypotonic . Water will flow into the cell, and hemolysis results. Solutions

64. Molecular weight determination Solutions The 4 colligative properties may be used to calculate the molecular weight of non-electrolytes present as solutes. Thus lowering of VP of a solution containing non- volatile solute depends only on the mole fraction of the solute. This allows the molecular weight of the solute to be determined in the following manner.

65. Solutions Molecular weight determination cont…..  Since the mole fraction of solvent, n 1 = w 1 /M 1 And the mole fraction of solute n 2 = w 2 /M 2 We know X 1 + X 2 = 1 X 1 = 1 – X 2 According to Raoult’s law P = P O X 1 1 r e p l a c i n g the value of X 1 P = P O (1- X ) 1 2 P O - P = P O X 1 1 2 1 = P O – P ΔP = X 2 = n 2 = w /M 2 2 P O P O 1 1 n 1 + n 2 1 1 + w M w /M 2 2

66. In dilute solution in which w 2 /M 2 is negligible compared to w 1 /M 1 , the former term may be omitted for the denominator, and the equation simplifies to; ΔP w 2 /M 2 = ------- Po w 1 /M 1 The molecular weight of solute M 2 is obtained by rearranging w 2 M 1 P o M 2 = ---------- w 1 ΔP S o l u t i o n s Molecular weight determination cont…..

67. Solutions  Similarly molecular weight can be calculated using boiling point elevation, knowing kb and Tb. Since 1000w2/w1 the weight of solute per kg of solvent, Molality (moles/ kg of solvent) can be expressed as; w 1 M 2 w 2 /M 2 1000w 2 w 1 / M 1 ΔT b = k b m 1000w 2 w 1 M 2 1000 w 2 w 1 ΔT b Molecular weight determination cont…..

68. Solutions Similarly by depression in freezing point method, molecular weight of non-volatile solute can be calculated; ΔT f = k f m 1000w 2 w 1 M 2 1000 w w 1 ΔT f Molecular weight determination cont…..

69. IDEAL SOLUTIONS

70. Cohesive forces: forces that exist between two similar types of molecules i.e. between solute-solute or between solvent-solvent. Adhesive forces: forces that exist between different types of molecules i.e. between solute and solvent. “Solutions in which the adhesive and cohesive forces are same (equal) are known as ideal solutions” Or “Solutions that obey Raoult's law”

71. Raoult's Law: At a definite temperature, the partial pressure (PA) of component (A) in a liquid mixture is equal to the vapour pressure of that component in the pure state (P°A ) multiplied by the mole fraction (X A) of that component in the solution. PA = XA P°A

72. Mixtures of Volatile Liquids Both liquids evaporate & contribute to the vapor pressure Solution A Solution B Solution of A and B

73. Raoult’s Law: Mixing Two Volatile Liquids Since BOTH liquids are volatile and contribute to the vapour , the total vapor pressure can be represented using Dalton’s Law: P t = P a +P b The vapor pressure from each component follows Raoult’s Law: P t = xa P° a + xb P° b Also, xa + xb = 1 (since there are 2 components)

74. 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, + x To = 1 Benzene: when PBz = P°Bz = 384 torr & x B z = 0.5 Toluene: when Pto1 = P°™ = 133 torr & x ™ = 0.5 P = x P° + x P° T Bz BZ Tol TOL

75. Characteristics of an ideal solution: Ideal behavior is expected to be exhibited by the systems which comprises of the chemical similar compounds, because it is only in such systems that the conditions of equal intermolecular forces between components are likely to be satisfied. Examples: solutions of ethyl alcohol- methyl alcohol, chloroform- bromoform , benzene- toluene. Ideal solutions have zero enthalpy change i.e. heat is neither absorbed nor evolved during solution formation. The volume of the solution is exactly equal to the sum of the individual volumes of the components.

76. Non- Ideal or Real Solutions

77. Solutions in which cohesive and adhesive forces are not equal are known as non-ideal or real solutions Solutions in which solute-solute, solvent-solvent and solute-solvent attractive forces are not equal. Solutions which don’t obey Raoults's law

78. Deviations from Raoult's Law

79. Negative deviation from Raoult's Law: Real solution which showed negative deviation from Raoult's Law are those solution in which adhesive forces (i.e. solute-solvent) are stronger than the cohesive forces (i.e. solute-solute or solvent-solvent) and vapour pressure of the solution is less than expected from Raoult's law. i.e. A-B > A-A, B-B

80. Reasoning of lowering of vapour pressure It attractive forces between solute and solvent (A-B) are stronger than those exerted between solute -solute (A-A) and solvent- solvent (B-B) molecules, then this strong mutual affinity between solute and solvent molecules results in the formation of complex or compound which results in strong holding of solvent molecules and results in lowering of escaping tendency of solvent molecules and ultimately lowering of vapour pressure. When this occur, there may be decrease in solution volume occur than the sum of volume of the components. E.g. Chloroform- Ethanol, Benzene - Ethanol

81. Positive deviation from Raoult's Law: Real solution which showed positive deviation from Raoult's Law are those solution in which cohesive forces (i.e. solute-solute or solvent-solvent) are stronger than the adhesive forces ( i.e. solute-solvent) and vapour pressure of the solution is greater than expected from Raoult's law. i.e. A-B < A-A, B-B

82. Reasoning of elevation of vapour pressure It attractive forces between solute and solvent (A-B) are less than those exerted between solute - solute (A-A) and solvent- solvent (B-B) molecules, then the presence of “A” reduces the (B-B) attraction and similarly presence of “B” molecules reduces (A-A) attraction. This results in greater escaping tendency of A and B and ultimately partial vapour pressure of the components are greater than expected from Raoult's law showed positive deviation. When this occur, there may be increase in solution volume occur than the sum of volume of the components. E.g. Chloroform- Acetone, Pyridine-Acetic Acid

83. Solutions Applications of Colligative Properties Each colligative properties seems to have certain advantages and disadvantages for the determination of molecular weights. The boiling point method can be used only when the solute is nonvolatile and when the substance is not decomposed at boiling temp. The freezing point method is satisfactory for solutions containing volatile solutes, such as alcohol, since the freezing point of solution depends on the VP of the solvent alone. The freezing point method is easily executed and yields results of high accuracy for solutions of small m o l e c u l e s .

84. It is sometimes inconvenient to use freezing point or boiling point method, however, since they must be carried out at definite temperatures. Osmotic pressure measurements do not have this disadvantage, and yet the difficulties inherent in this method preclude its wide use. In summary, it can be said that the cryoscopic and newer techniques of VP are methods of choice, except for very high polymers, in which instance the osmotic pressure method is used.. Since the colligative properties are interrelated, it should be possible to determine the value of one property from a knowledge of any other. Applications of Colligative Properties cont. ….