1 / 54

Physical Properties of Solutions

Physical Properties of Solutions. Chapter 12 Dr. Ali Bumajdad. Chapter 12 Topics. Physical Properties of Solutions. Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization Solution Process Concentrations (%mass, Mole fraction, molality, molarity, ppm)

kermit-maddox
Download Presentation

Physical Properties of Solutions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Physical Properties of Solutions Chapter 12 Dr. Ali Bumajdad

  2. Chapter 12 Topics Physical Properties of Solutions • Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization • Solution Process • Concentrations (%mass, Mole fraction, molality, molarity, ppm) • Factors affecting Solubility (T and P) • Colligative properties of nonelectrolyte solution: • 1)Vapour pressure lowering • 2)B.P. Elevation • 3) F.P. Depression • 4) Osmotic Pressure () • Using Colligative properties to determine molar mass • Colligative properties of electrolyte solution Dr. Ali Bumajdad

  3. Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization • Solution:homogenous mixture of 2 or more substances • Solute: is(are) the substance(s) present in the smaller amount(s) • Solvent: is the substance present in the larger amount Solution not necessarily liquid, it could be gas or solid

  4. solvation Solute + solvent Solution crystallization • Saturated solution: contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. • Unsaturated solution: contains less solute than the solvent has the capacity to dissolve at a specific temperature. • Supersaturated solution: contains more solute than is present in a saturated solution at a specific temperature (not stable). Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate • Crystallization: dissolved solute comes out of solution and forms crystals.

  5. Solution Process • Three types of interactions in the solution process: • solvent-solvent interaction (endothermic) • solute-solute interaction (endothermic) • solvent-solute interaction (endothermic or exothermic) Hsoln>0, solution may form and maynot Hsoln<0, solution favourable DHsoln = DH1 + DH2 + DH3

  6. “like dissolves like” When solute-solute or solvent-solvent interaction are much more than solute-solvent interaction a solution will not form Two substances with similar intermolecular forces are likely to be soluble in each other. • non-polar molecules are soluble in non-polar solvents • CCl4 in C6H6 • polar molecules are soluble in polar solvents • C2H5OH in H2O • ionic compounds are more soluble in polar solvents • NaCl in H2O or NH3 (l)

  7. Miscible: two liquids completely soluble in each other in all • proportions • Solvation: ion or molecules surrounded by solvent molecules Q) Can we form a solution of octane (C8H18) in water? • No because: • - octane not polar while water is polar • In this case H3 is small while H1 and H2 large +ve • hence, Hsoln is large positive Sa Ex. :Predict whether each of the following substances is more likely to dissolve in CCl4 or in Water: C7H16, Na2SO4, HCl, I2.

  8. (a) Br2 in C6H6 (b) KClin NH3 (c) CH2O in H2O

  9. moles of A XA= (1) sum of moles of all components % by mass = x 100% mass of solute x 100% = mass of solution mass of solute (2) mass of solute + mass of solvent Concentration Units • Concentration of a solution is the amount of solute present in a given quantity of solvent or solution. 1) Percent by Mass 2) Mole Fraction(X)

  10. (3) moles of solute liters of solution (4) moles of solute m = mass of solvent (kg) M = Concentration Units Continued 3) Molarity(M) 4) Molality(m)

  11. n= mass 2) D= mass M.m Vol

  12. Sa. Ex.:A solution is made by dissolving 13.5 g glucose (C6H12O6) in 0.100 kg of water. What is the mass percentage of solute in this solution? 11.9% • Sa. Ex. • Calculate the mass percentage of NaCl in a solution containing 1.50 g of • NaCl in 50. 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? 2.91% 90.5g

  13. Sa. Ex. a) A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 ml of water. Calculate the molality of glucose in the solution. b) What is the molality of a solution made by dissolving 36.5 g of naphthalene (C10H8) in 425 g of toluene (C7H8) c) A solution of hydrochloric acid contains 36% HCl by mass (i) Calculate the mole fraction of HCl in the solution. (ii) Calculate the molality of HCl in the solution 0.966 m 0.670 m XHCl= 0.22 15 m

  14. moles of solute moles of solute m= m= moles of solute M = mass of solvent (kg) mass of solvent (kg) liters of solution 5.86 moles C2H5OH = 0.657 kg solvent Q) What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is 0.927 g/mL? • Assume 1 L of solution: • (1) mass of solute: (nMm=mass) • 5.86 moles ethanol contains 270 g ethanol • (2) mass of solution: solution density solution volume • 927 g of solution mass of solvent = mass of solution – mass of solute = 927 g – 270 g = 657 g = 0.657 kg = 8.92 m

  15. Sa. Ex. A solution contains 5.0 g of toluene (C7H8) and 225 g of benzene and has a density of 0.876 g/ml. Calculate the molarity of the solution. Sa. Ex. A solution containing equal masses of glycerol (C3H8O3, Mm=92gmol-1) and water has a density of 1.10 g/ml. Calculate: (a) The molality of glycerol (b) The mole fraction of glycerol (c) The molarity of glycerol in the solution. 0.21 M 10.9 m 0.163 5.97 M

  16. Solubility increases with increasing temperature Solubility decreases with increasing temperature • Factors affecting Solubility (T and P, and nature of solute and solvent) Like dissolve like (1) Temperature a) Solid solubility and temperature Usually s  T For solid

  17. Fractional crystallization: is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have 90 g KNO3 contaminated with 10 g NaCl. • Fractional crystallization: • Dissolve sample in 100 mL of water at 600C • Cool solution to 00C • All NaCl will stay in solution (s = 34.2g/100g) • 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

  18. Usually s  1 For gas T (1) Temperature (b)Gas solubility and temperature Solubility usually decreases with increasing temperature

  19. (5) (there are some exceptions when the gas react with water) c= kP Henry’s law Vapour Pressure: pressure of gas above the liquid low P high P low c high c (2) Pressure a) Gas solubility and Pressure The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law). c is the concentration (M) of the dissolved gas P is the pressure of the gas over the solution (vapor or partial pressure) kis a constant for each gas (mol/L•atm) that depends only on temperature

  20. (5) c= kP Sample Ex. :Calculate the concentration of CO2 in soft drink (partial pressure of CO2 = 4.0 atm, and the Henery's law constant for CO2 in water is 3.1  10-2 mol/L atm. =0.12 mol/l

  21. = vapor pressure of solutionafter adding solute = vapor pressure of pure solvent 0 P1 = X1 P 1 For nonvolatile solute (no measurable vapour pressure) (6) Raoult’s law 0 P 1 P 1 Colligative Properties of Nonelectrolyte Solutions • Colligative properties: properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. (e.g. adding salt and ethylene glycol reduce F.P.) 1) Vapor-Pressure Lowering Adding nonvolatile solute to a solvent always lower the vapour pressure Where: X1= mole fraction of the solvent

  22. (1 - X2) (7) 0 0 0 P 1 = P 1 P 1 P 1 - P1 = DP = X2 If the solution contains only one solute: X1 = 1 – X2 X2= mole fraction of the solute The decrease in vapour pressure (P) directly proportion to solute concentration (not to the nature of solute)

  23. 0 0 0 0 PA = XA P A PB = XB P B PT = XA P A +XB P B Ideal Solution When both components are volatile:1) Ideal behaviour PT = PA + PB

  24. Force A-A Force A-A Force B-B Force B-B Force A-B Force A-B < > & & When both components are volatile: 2) Nonideal behaviour PT is greater than predicted by Raoults’s law PT is less than predicted by Raoults’s law

  25. Fractional Distillation Apparatus Used to separate two liquids from each other based on their boiling point

  26. (7) 0 P1 = X1 P 1 (2) (6) X1 = n1 n1+n2 0 0 P 1 P 1 - P1 = DP = X2 Vapour pressure of water at 30 °C = 31.82mmHg Raoult’s Law X2=1-X1

  27. Sa. Ex. Glycerin (C3H8O3) is a nonvolatile nanelectrolyte with a density of 1.26 g/ml at 25 °C. Calculate the vapor pressure at 25 °C of a solution made by adding 50.0 ml of glycerin to 500.0 ml of water. The vapor pressure of pure water at 25 °C is 23.8 torr. Sa. Ex. The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution? 23.2 torr 0.290

  28. 0 DTb = Tb – T b 0 T b is the boiling point of the pure solvent 0 Tb > T b (8) DTb = Kbm Tb0 Tb 2) Boiling-Point Elevation By the addition of nanvolatile solute T b is the boiling point of the solution DTb > 0 mis the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) for a given solvent (depend only on solvent)

  29. 0 DTf = T f – Tf 0 T f is the freezing point of the pure solvent 0 T f > Tf (9) DTf = Kfm Tf Tf0 3) Freezing-Point Depression By the addition solute T f is the freezing point of the solution DTf > 0 m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) for a given solvent (depend only on solvent)

  30. The decrease in Tfupon addition of solute is due to the more energy needs to be removed to form order solution than to form order solvent (solution is more disorder) The solid that separate when a solution freeze is the pure solvent only that why the V.P. of solid is the same as that for pure liquid (see the phase diagram)

  31. De-icing of airplanes

  32. 0 DTf = T f – Tf moles of solute m= mass of solvent (kg) 478 g 62.01 g 3.202 kg solvent = = 2.41 m (9) DTf = Kfm 0 Tf = T f – DTf Q) What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g/mol. (Kf water = 1.86 0C/m) DTf = Kfm = 1.86 0C/m x 2.41 m = 4.48 0C = 0.00 0C – 4.48 0C = -4.48 0C

  33. Sa. Ex. Automotive antifreeze consists of ethylene glycol (C2H6O2), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass% solution of ethylene glycol in water. Sa. Ex. Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus trees. (9) (8) DTb = Kbm DTf = Kfm (Kb water=0.52 and Kf water = 1.86 °C/m) Tb = Tb0 +Tb Tf = Tf0 -Tb 102.8 °C -10.0 °C

  34. Cucumber in concentrated brine lose water Pickle 4) Osmotic Pressure (p) • Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one. • Semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules. • Osmotic pressure (p) is the pressure required to stop osmosis. more concentrated dilute

  35. Osmotic pressure in unit atm (10) p = MRT Colligative property = depend on concentration only High P Low P M is the molarity of the solution R is the gas constant (0.0821 L atm/mol K) T is the temperature (in K)

  36. When the concentrations of the two solution are equal and, hence, have the • same osmotic pressure, the solutions called isotonic • Isotonic: the two solutions are of equal concentration and, hence, • they have the same osmotic pressure • When concentrations of the two solution are different, the lower concentration • called hypotonic and the higher concentration called hypertonic • Hypotonic: the solution of lower concentration • Hypertonic: the solution of higher concentration

  37. A cell in an: A cell in hypertonic solution A cell in isotonic solution A cell in hypotonic solution the outer solution is less concentrated than the solution in the cell the outer solution is more concentrated than the solution in the cell the two solution are equal in concentration

  38. (10) p = MRT 1.23 M

  39. Sa. Ex.:The average osmotic pressure of blood is 7.7 atm at 25 °C. what concentration of glucose (C6H12O6) will be isotonic with blood? Sa. Ex.:What is the osmotic pressure at 20 °C of a 0.0020 M sucrose (C12H22O11) solution? 0.31 M 0.048 atm

  40. (9) (8) Kb Kf m = n m = n mass of solvent in Kg mass of solvent in Kg DTb = Kbm DTf = Kfm m= DTf m= DTb (11) (11) M.m = mass in g M.m = mass in g m = mass in g m = mass in g M.m. (mass of solvent in Kg) M.m. (mass of solvent in Kg) m(mass of solvent in Kg) m(mass of solvent in Kg) • Using Colligative Properties to determine molar mass B.P. Elevation F.P. Depression

  41. Osmotic Pressure (10) p = MRT M = p RT M = n VL M = mass in g M.m. (VL) (12) M.m = mass in g M(VL) (13) M.m = (mass in g) R T (VL) 

  42. Kf m= DTf (11) M.m = mass in g m(mass of solvent in Kg) Kf of benzene = 5.12 °C/m 127 g/mol C10H8

  43. 1 atm = 760 mmHg 6.51 104 g/mol

  44. Sa. Ex.: Coamphor (C10H16O) melts 1t 179.8 °C, and it has a particularly large freezing-point-depression constant Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. what is the molar mass of the solute?

  45. (8) (9) (7) Vapor-Pressure Lowering Boiling-Point Elevation DTb = Kbm 0 P1 = X1 P 1 Freezing-Point Depression DTf = Kfm p = MRT (10) Osmotic Pressure (p) Colligative Properties of Nonelectrolyte Solutions Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

More Related