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Solution colligative properties 12th HSC Maharashtra state board

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Solution Colligative Properties
Presentedby:FreyaCardozo
1
Presented by:
Freya
Cardozo

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Solution
 It is a homogenous mixture of two or more pure substances
 Solution= Solute + Solvent
 Solvent is the one pre...

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Types of solutions
 The solute and solvent can be in any of the following state solids, liquids
or gases
 Combinations o...

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Solution colligative properties 12th HSC Maharashtra state board

  1. 1. Solution Colligative Properties Presentedby:FreyaCardozo 1 Presented by: Freya Cardozo
  2. 2. Solution  It is a homogenous mixture of two or more pure substances  Solution= Solute + Solvent  Solvent is the one present in larger quantity  Solute is the one present in smaller quantities  Solutions with only one solute are called Binary solutions Presented by: Freya Cardozo 2
  3. 3. Types of solutions  The solute and solvent can be in any of the following state solids, liquids or gases  Combinations of these gives mainly 9 types of solutions Presented by: Freya Cardozo 3
  4. 4. Presented by: Freya Cardozo 4
  5. 5. Solubility  The solubility of a solute is its amount per unit volume of saturated solution at a specific temperature  Solubility unit= mol L-1 Presented by: Freya Cardozo 5
  6. 6. Factors affecting solubility  Nature of solute and solvent  Effect of temperature on solubility  Effect of pressure on solubility Presented by: Freya Cardozo 6
  7. 7. Nature of solute and solvent  Like dissolves like  Similar chemical character are more readily soluble in each other than different chemical natures  Similar substances also have similar intermolecular forces of attraction  Polar solutes dissolve in polar solvents because , solute-solute, Solute- solvent and solvent-solvent interactions are of similar magnitude  Eg. NaCl in water, Cholesterol in Benzene, Sugar in water Presented by: Freya Cardozo 7
  8. 8. Effect of temperature on solubility  For endothermic processes – Eg. KCl in water ↑T ↑Solubility because there is an increase in stress on the solution  For exothermic process – CaCl2 in water ↑T ↓Solubility But there hasn’t been any direct relation between exo and endothermicity on T Presented by: Freya Cardozo 8
  9. 9. T Vs Solubility  NaBr, NaCl, KCl : S slightly changes with T  KNO3, NaNO3, KBr: ↑T ↑Solubility  Na2SO4 ↑T ↓Solubility Gas molecules are condensed in liquid phase, this process is exothermic. Thus, Solubility of gases in water should decreases with increase in temperature. Presented by: Freya Cardozo 9
  10. 10. REAL LIFE EXAMPLE Industries take billions of gallons of water from rivers and lakes to cool down the equipment's which get heated during the industrial production process. Once the cool water is used the hot water produced is the returned to the water bodies again. Due to the increase in the temperature of water the solubility of gas in water decrease and thus it is very less available for fishes Presented by: Freya Cardozo 10
  11. 11. Effect of pressure on solubility  Pressure has no effect on solubility of liquids and solids as they are incompressible  But for gases the solubility is greatly influenced by pressure  The relation between P and S for gases can be given by HENRY’S LAW Presented by: Freya Cardozo 11
  12. 12. HENRY’S LAW  The law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution   Where, S= Solubilty of gase in molL-1 P=Pressure of gas over solution in bar KH= Henry constant. Unit is mol L-1 bar-1  When, P=1, KH=S. Thus, KH is the solubility of the gas in a liquid when its pressure over solution is 1 bae Presented by: Freya Cardozo 12
  13. 13. Example and Exceptions  CO2 added in cold drinks under a pressure  As compared to normal conditions under pressure the solubility is much higher  Release relives this pressure giving the effervescence  Exceptions are NH3 and CO2 because they react with water  As they form these compounds they have higher solubilities than expected by Henrys law Presented by: Freya Cardozo 13
  14. 14. Type 1 numericals  Based on Henry’s law  S=KHP Presented by: Freya Cardozo 14
  15. 15. QUESTION  The henry’s law constant of CH3Br is 0.159 mol L-1 bar-1 at 250C. What is the solubility of CH3Br in water at 250C and at 130mm Hg? (1mmHg= 0.00133 bar) Presented by: Freya Cardozo 15
  16. 16. RAOULT’S LAW  The law states that, “ The partial vapour pressure of any volatile component of a solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution.”  P= x P0  P= partial pressure  p0= partial pressure of pure component  x= mole fraction Presented by: Freya Cardozo 16
  17. 17. Mole fraction  Mole fraction represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture. It's a way of expressing the concentration of a solution  Represented by x  The mole fraction of solute is x2 and solvent is x1; x1+x2=1 Presented by: Freya Cardozo 17
  18. 18. Derive Raoult’s law 1. Let A1 and A2 be two volatile liquids, P1 and P2 be the respective partial pressures and P1 0 , P2 0 be vapour pressures of pure liquids 2. Let x1 and x2 be the mole fractions of A1 and A2 respectively 3. Raoult’s law can be written as P1= x1 P1 0 and P2 = x2 P2 0 Presented by: Freya Cardozo 18
  19. 19. Since, P1 0 and P2 0 are constant P Vs x2 is a straight line P1 Vs X1 and P2 Vs x2 are also straight line passing through the origin Presented by: Freya Cardozo 19
  20. 20. IDEAL SOLTIONS NONIDEAL SOLUTIONS 1. Obey Raoult’s law over entire range of concentrations 2. V.P always lies between V.P of pure components 3. No heat is evolved/ absorbed when 2 components are mixed. Enthalpy of mixing is zero. ∆mix H=0 4. No volume change. Thus Vol of sol. Is equal to sums of volumes of the 2 components mixed. ∆mixV=0 5. Solvent-solute, solute-solute and solvent-solvent molecular interactions are comparable 1. Do not obey Raoult’s law over entire range of concentrations 2. V.P of these solutions can be higher or lower than those of pure components 3. It shows two types of deviation:  Positive deviation  Negative deviation Presented by: Freya Cardozo 20
  21. 21. POSITIVE DEVIATION NEGATIVE DEVIATION  Solute-solvent I.F.A (weaker)< solute-solute I.F.A & solvent-solvent I.F.A  V.P is higher than the pure components  Eg. Acetone+ Ethanol Acetone+ carbon disulphide  Solute-solvent I.F.A (stronger)> solute-solute I.F.A & solvent-solvent I.F.A  V.P is higher than the pure components  Eg. Acetone + chloroform Phenol + Aniline Presented by: Freya Cardozo 21 I.F.A= Intermolecular forces of attraction
  22. 22. Colligative properties  Definition of colligative properties The physical properties of solutions that depend on the number of solute particles in solutions and not on their nature are called colligative properties  There are 4 colligative properties 1. Vapour pressure lowering 2. Boiling point elevation 3. Freezing point depression 4. osmotic pressure  Non-electrolytic and dilute solutions are used Presented by: Freya Cardozo 22
  23. 23. Vapor pressure  The pressure exerted by the liquid vapor over the surface of a liquid in a closed container when both are in equilibrium is called as vapor pressure Presented by: Freya Cardozo 23
  24. 24. VOLATILE SUBSTANCE NON-VOLATILE SUBSTANCE Presented by: Freya Cardozo 24
  25. 25. Vapour pressure lowering  When a non volatile non-ionizable solid is dissolved in a liquid solvent, the vapour pressure of the solution is lower than that of the solvent.  If the solute is non-volatile it does not contribute to the VP above the solution  Thus VP of solution=VP of solvent above the solution  Mathematically, If P1 0 is the VP of pure solvent and P1 is the VP of solvent above the solution, P1< P1 0  Thus, vapour pressure lowering is ∆= P1 0 – P1 Presented by: Freya Cardozo 25
  26. 26. Reason for this?  The vapour pressure of a liquid depends on the ease with which the molecules escape from the surface of liquid  When non-volatile solute is added it replaces some of the volatile solvent molecules at the surface and these non volatile solutes do not vaporizes and thus do not contribute to the vapour pressure.  Therefore, the number of solvent molecules available for vaporization per unit surface area is less than number at surface of pure solvent Presented by: Freya Cardozo 26
  27. 27. Raoult’s law for solution of non volatile solutes OR Prove that ∆P= P1 0 x2 OR Prove lowering of vapour pressure is a colligative property  For a solution containing non volatile solute, the vapour pressure of solvent over the solution is equal to the vapour pressure multiplied by its mole fraction in the solution P1= P1 0 x1  For a binary solution containing 1 solute, x1=1-x2 (Since, x1+x2=1)  But, we know that lowering of vapour pressure is given by ∆= P1 0 – P1  Thus, ∆P= P1 0x2  From the above equation it is clear that the ∆P depends on x2 which is number of solute particles, thus lowering of vapour pressure is a colligative property Presented by: Freya Cardozo 27
  28. 28. Relative Lowering of Vapour pressure  The ratio of vapour pressure lowering of solvent divided by the vapour pressure of pure solvent is called relative lowering of vapour pressure.  Thus,  Relative lowering of VP is equal to the mole fraction of solute in the solution. Therefore, relative lowering of VP is also a colligative property Presented by: Freya Cardozo 28
  29. 29. Relationship between molar mass of solute and lowering of vapour pressure  We know that relative lowering of vapour pressure is equal to mole fraction of solute i.e.  The mole fraction of a component of solution is equal to its moles divided by the total moles in the solution. Thus,  n1= moles of solvent and n2= moles of solute  In dilute solutions, n1>>n2, thus n1+n2~n1.  Thus the mole fraction is Presented by: Freya Cardozo 29
  30. 30.  Suppose a solution is prepared by adding W2 g of solute in W1 g of solvent. The moles of solute and solvent in the solution are,  Where, M1 and M2 are molar masses of solvent and solute respectively. Substituting in the equation, Presented by: Freya Cardozo 30
  31. 31. WHAT DO YOU THINK? Presented by: Freya Cardozo 31
  32. 32. TYPE 2 numericals ∆P= P1 0X2 Presented by: Freya Cardozo 32
  33. 33. Presented by: Freya Cardozo 33In an experiment, 18.04 g of mannitol was dissolved in 100 g of water. The vapour pressure of water was lowered by 0.309 mm Hg from 17.535 mmHg. Calculate the molar mass of mannitol
  34. 34. Presented by: Freya Cardozo 34 A solution is prepared by dissolving 394g of a non volatile solute is 622g of water. The vapour pressure of solution is found be 30.74 mm Hg at 300C. If the vapour pressure at 300C is 31.8 mm Hg, what is the molar mass of solute?
  35. 35. Presented by: Freya Cardozo 35The vapour pressure of pure benzene (molar mass 78 g/mol) at a certain temperature is 640 mmHg. A non-volatile solute of mass 2.315g is added to 49g of benzene. Th vapour pressure of solution is 600 mmHg. What is the molar mass of the solute?
  36. 36. Presented by: Freya Cardozo 36
  37. 37. Boiling point elevation  B.P : The temperature at which the vapour pressure equals to the applied pressure/ atmospheric pressure(if open container)  Solutions with non volatile solute have higher B.P than pure solvent  If Tb 0 is B.P of pure solvent and Tb is of the solution then the difference between them ∆Tb  The difference between the B.P of solution and that of the pure solvent at any given point is called the boiling point elevation Presented by: Freya Cardozo 37
  38. 38. Lowering of V.P and Elevation in B.P 1. In the graph Vapour pressure of the solvent and solution are plotted as a function of temperature 2. From the previous discussions we know that the vapour pressure of solution with non volatile solute<vapour pressure of pure solvent 3. Thus from graph we can see that CD= vapour pressure of solution(lower curve) AB= vapour pressure of pure solvent(higher curve) 4. The V.P difference goes on increasing with temperature Presented by: Freya Cardozo 38
  39. 39. 5. The intersection of the curves AB and CD on the X axis shows that the boiling point of solution is more than the pure solvent 6. A liquid boils when its V.P is equal to 1 atm(atmospheric pressure). Therefore, in order for the solution to boil a higher temperature will be needed to reach 1 tm than the solvent. 7. That’s why the solution needs to be heated more to be boiled. Hence, there will be elevation in the B.P of solution even though there is a lowering of vapour pressure. [V.P is inversely related to B.P] Presented by: Freya Cardozo 39
  40. 40. Expressing concentration MOLARITY MOLALITY Presented by: Freya Cardozo 40
  41. 41. B.P elevation and conc. Of solute 1. The B.P elevation is directly proportional to molality of the solution. 2. Where, m= molality of solution Kb= boiling point elevation constant/ molal elevation constant/ ebullioscopic constant 3. If, m=1, ∆Tb=Kb Thus, ebullioscopic constant is the B.P elevation produced by 1 molal solution Presented by: Freya Cardozo 41
  42. 42. Batao.. Presented by: Freya Cardozo 42
  43. 43. Why molality over molarity? Because, we are studying systems where temperature is not constant thus we have to choose terms that do not depend on temperature molality is temperature independent whereas molarity depends on temperature This can be seen from the units also Presented by: Freya Cardozo 43 Molality= mol/kg Molarity= mol/L
  44. 44. Relationship between mass of solute and boiling point elevation  Suppose a solution is made by dissolving W2 g of solute in W1 g of solvent  Moles of solute in solvent= W2/ M2 (M2= molar mass) Mass of solvent= W1g = W1 g/ 1000 g/kg = W1/ 1000 kg  Molality can be given by Presented by: Freya Cardozo 44
  45. 45. TYPE 3: Elevation of B.P Presented by: Freya Cardozo 45
  46. 46. Presented by: Freya Cardozo 46
  47. 47. Presented by: Freya Cardozo 47
  48. 48. Presented by: Freya Cardozo 48
  49. 49. Presented by: Freya Cardozo 49 A solution containing 0.73 g of camphor (molar mass 152 gmol-1) in 36.8 g of acetone (boiling point 56.30C) boils at 56.55 0C. A solution of 0.564 g of unknown compound in the same weight of acetone boils at 56.46 0C. Calculate the molar mass of the unknown compound. [Oct 2014]
  50. 50. Freezing point  Freezing of point of a liquid is the temperature at which the liquid and solid are in equilibrium and the two phases have the same vapour pressure Presented by: Freya Cardozo 50
  51. 51. Depression in Freezing point  It is observed that addition of non volatile solute to solvent lowers the freezing point  Thus, The F.P of solution( with non-volatile solute)< F.P of pure solvent  Now, if Tf 0 is freezing point of pure solvent and Tf of the solution, Tf 0 > Tf  The difference between the two is called ∆ Tf Presented by: Freya Cardozo 51
  52. 52. Freezing point depression & Vapour pressure lowering  Consider the vapour pressure diagram 1. Curve AB= V.P of solid solvent 2. Curve CD= V.P of pure liquid 3. Curve EF = V.P of solution  The vapour pressure of solution is always lower than the solvent and thus is present at the lower part of the graph  The non-volatile solute does not dissolve in the solid solvent Presented by: Freya Cardozo 52
  53. 53.  Curve AB and CD  Intersect at B – This is where both the solid and liquid phases have the same V.P. Thus, The temperature corresponding to point B is freezing point of the pure solvent Tf 0  Curve EF and AB  Intersect at E – This is the point at which the solid solvent is in equilibrium with the solution Thus, The temperature corresponding to point E is freezing point of the solution Tf  It is clear from the figure that the freezing point of solution is lower than that of the pure solvent Presented by: Freya Cardozo 53
  54. 54. But why so? DUE TO THE ATTRACTIVE FORCES BETWEEN THE MOLECULES  In pure liquid the attractive forces among molecules are large enough to cause the change of phase from liquid to solid  In solution, there are solute molecules in between the solvent particles.. this causes more separation of solvent molecules than in the solvent  Therefore, there is a decrease in the attractive forces between the solvent molecules. So the temperature of the solution is lowered below the freezing point of solvent to cause phase channge Presented by: Freya Cardozo 54
  55. 55. Presented by: Freya Cardozo 55 Only solvent molecules Solute + solvent molecules Easier to change phase Solute molecules obstruct the phase change. More T needed(Lower Temps)
  56. 56. Real life application Presented by: Freya Cardozo 56
  57. 57. Freezing point depression and molality  Freezing point depression ∆Tf is directly proportional to molality of the solution  Kf= Freezing point depression constant / cryoscopic constant  If, m=1 ; ∆Tf = Kf  The cryoscopic constant thus is the depression in freezing point produced by 1 molal solution of a non volatile solute Presented by: Freya Cardozo 57
  58. 58. Presented by: Freya Cardozo 58
  59. 59. Relationship between Molar mass of solute and F.P depression Presented by: Freya Cardozo 59
  60. 60. Presented by: Freya Cardozo 60
  61. 61. Presented by: Freya Cardozo 61
  62. 62. Osmosis  The net flow of solvent molecules into the solution through a semipermeable membrane OR  The net flow of solvent molecules into the solution from a more dilute solution to more concentrated solution through a semipermeable membrane is called osmosis Presented by: Freya Cardozo 62 SOLVENT SOLUTION Lower Conc. Higher conc.
  63. 63. Semipermeable membrane  It is a film such as cellophane which has pores large enough to allow the solvent molecules to pass through them  The pores are small enough to not allow the flow of the larger solute molecules or ions of higher mass  Selectively allows passage of solvent molecules Presented by: Freya Cardozo 63 Solute molecules Semi- P.M Solvent molecules
  64. 64. Osmotic pressure  In the thistle tube solution of interest (sugar solution) is placed and it is immersed in beaker filled with pure water  A semipermeable membrane is placed at the mouth of the tube  Some solvent passes through the membrane into the solution  This causes rise in liquid level in the tube  Now, the hydrostatic pressure in the tube pushes the solvent back into the container Presented by: Freya Cardozo 64
  65. 65. Reverse osmosis  The liquid rises in tube and then stops, when the liquid stops rising that is the pressure which stops the flow and is called the osmotic pressure  This hydrostatic pressure that stops the osmosis is of the solution is the osmotic pressureΠ and is equal to 1. Height of the liquid column 2. Density of liquid column 3. acceleration due to gravity Presented by: Freya Cardozo 65
  66. 66. Presented by: Freya Cardozo 66 Type Definition Examples Isotonic solutions Two or more solutions having the same osmotic pressure are said to be isotonic No net flow of solvent in either direction 0.1 M urea= 0.1 M sucrose Both have equal osmotic pressure but Different conc in g/L Hypertonic solutions If two solutions have unequal osmotic pressure, the more concentrated solution with the higher osmotic pressure is said to be hypertonic In a sucrose AND urea solution, Sucrose Higher osmotic pressureHypertonic Hypotonic solutions If two solutions have unequal osmotic pressure, the more dilute solution with the lower osmotic pressure is said to be hypertonic In a sucrose AND urea solution, Urea Lower osmotic pressureHyptonic
  67. 67. Presented by: Freya Cardozo 67
  68. 68. Osmotic pressure and conc. Of solution  For dilute solution the osmotic pressure can be given by,  V= volume of the solution in dm3 n2= number of moles of non volatile solute R= real gas constant = 0.08206 dm3 atm K-1 mol-1 Π= Osmotic pressure in atm Presented by: Freya Cardozo 68  We know that Concentration= n2/V  Concentration can be written in terms of Molarity M  Thus, equation becomes  Here, we can use molarity instead of molality because the osmotic pressures are measured at a constant temperature.
  69. 69. Presented by: Freya Cardozo 69
  70. 70. Molar mass of solute from osmotic pressure Presented by: Freya Cardozo 70
  71. 71. Reverse osmosis  The direction of osmosis if from pure solvent to solution but this can be reversed by applying a pressure higher than the osmotic pressure  The pure solvent then flows from solution into pure solvent through semipermeable membrane. This is called reverse osmosis.  Eg. Fresh water and salty water separation using semipermeable membrane  On application of pressure higher than osmotic pressure, the salty water passes into fresh pure water  This leaves the salt behind Presented by: Freya Cardozo 71
  72. 72. Presented by: Freya Cardozo 72
  73. 73. Colligative properties of electrolytes 1. The solutions of electrolytes also exhibit colligative properties which do not obey the relations of non-electrolytes 2. The colligative properties of the solutions of electrolytes are greater than those to be expected for solutions of non electrolytes of the same concentrations 3. The molar masses of electrolytes in aqueous solutions determined by colligative properties are found to be considerably lower than the formula masses Presented by: Freya Cardozo 73
  74. 74. Why is colligative properties of electrolytes more than nonelectrolytes Electrolytes- Dissociates Increases the number of particles Increase in colligative particles Eg. NaCl(lesser C.P) Vs Sucrose(more C.P) Presented by: Freya Cardozo 74
  75. 75. Van’t Hoff factor (i)  In order to account for the dissociation/ association of electrolytes and to calculate their colligative properties Van’t Hoff suggested the factor i  It can be defined as the ratio of colligative property of a solution of electrolyte divided by the colligative property of nonelectrolyte solution of the same concentration.  No subscript  electrolyte solutions With subscript nonelectrolyte solutions Presented by: Freya Cardozo 75
  76. 76. Definitions  It is also defined as , Presented by: Freya Cardozo 76 i=2 KNO3 NaCl i=3 CaCl2 Na2SO4 i=1 for non electrolytic solutions
  77. 77. Why colligative properties of higher concentration solutions smaller than expected?  The electrostatic forces between the oppositely charged ions bring about the formation of ion pairs  Each ion pair consists of one or more cations and one or more anions held together by electrostatic attractive forces  This results in decrease in the number of particles in solution causing reduction in the expected I value and colligative properties Presented by: Freya Cardozo 77
  78. 78. Modifications of expressions of colligative properties Presented by: Freya Cardozo 78
  79. 79. Van’t Hoff factor i and degree of dissociation  For weak electrolytes Dissociation is related tαo Degree of dissociation α  Consider the equation  Initially 1 mol 0 0 At equilibrium (1- α)mol x α y α Total moles after dissociation = (1- α)+ x α + y α= 1+ α (x+y-1)= 1+ α(n-1) [Since, n=x+y] Vant Hoff factor can thus be given by, Presented by: Freya Cardozo 79
  80. 80. Presented by: Freya Cardozo 80
  81. 81. Presented by: Freya Cardozo 81
  82. 82. Presented by: Freya Cardozo 82

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