Sample Problem. (b) Mass % H 2 SO 4 = kg H 2 SO 4 /1.046 kg total = 7.04%

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1 A Sample M solution Problem of H 2 SO 4 in water has a density of g/ml at 20ºC. What is the concentration in (a) mole fraction, (b) mass percent, (c) molality (MM = g/mol)? (a) Since the solution is mol/l and has a density of g/ml (or kg/l) density, 1.0 L solution contains mol (or 73.6 g) H 2 SO 4 and has a mass of kg: Mass of H 2 O in 1 L solution = kg kg = kg kg H 2 O = 972 g 1 mol/18.0 g = 54.0 mol H 2 O For H 2 SO 4, X = mol H 2 SO 4 /(0.750 mol H 2 SO mol H 2 0) =

2 Sample Problem A M solution of H 2 SO 4 in water has a density of g/ml at 20ºC. What is the concentration in (a) mole fraction, (b) mass percent, (c) molality (MM = g/mol)? (b) Mass % H 2 SO 4 = kg H 2 SO 4 /1.046 kg total = 7.04% (c) Since kg water has mol H 2 SO 4 in it, 1 kg water would have mol H 2 SO 4 dissolved in it: 1.00 kg H 2 O mol H 2 SO 4 /0.972 kg H 2 O = mol H 2 SO 4 Thus, molality of sulfuric acid is m

3 Colligative properties are physical properties of solutions that arise because of the number of solute molecules dissolved in solution and not on the kind of solute particles dissolved in solution. Pure Liquid Pure Liquid with solute

4 Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. The Four-Colligative Properties Vapor-Pressure Lowering Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure (π) P solution = X solvent P solvent ΔT b = K b msolution ΔT f = -K f msolution π = MRT

5 Vapor-Pressure Lowering: A non-volatile, non-ionic solute added to a pure solvent will lower the vapor pressure of the pure solvent according to Raoult s Law. Nonvolatile solute molecules reduce the number of volatile solvent molecules at the surface of the liquid reducing the # of solvent molecules escaping to the vapor phase. P solvt vapor p A = X A p A º pure solvent A solvent A + nonvolatile solute B P solution = χ solvent P solvent Vapor pressure of solution Mole Fraction of Solvent Vapor Pressure of Pure Solvent

6 At 100ºC what is the vapor pressure of a 50/50 % (v/v) solution of ethylene glycol, C 2 H 6 O 2, in water at 1 atm? (MM C 2 H 6 O 2 = g/mol d(c 2 H 6 O 2 ) = g/ml, d(h 2 O) = g/ml, d(50/50) = g/ml P solution = χ H2O P H2O Moles C 2 H 6 O 2 = 500. ml X g C 2 H 6 O 2 X 1 mol = 8.99 mol ml g Moles H 2 O = 500. ml X g H 2 O X 1 mol = 27.8 mol ml g Mole Fraction H 2 O = 27.8 mol/ (27.8 mol mol) = P H2O = χ H2O P H2O =.7556 X 760. torr = 574. torr

7 The vapor-pressure lowering can be recast in terms of the mole fraction of solute. P Soln = χ solvent P solvent (1) substituting P Soln = (1 - χ Solute ) P solvent (2) expanding 2 χ Solute + χ Solvent = 1 χ Solvent = 1 - χ Solute P Soln = P solvent - P solvent (χ Solute ) (3) ΔP = (P solvent - Psoln) = P solvent (χ Solute )

8 Calculate the vapor pressure lowering, ΔP, when 10.0 ml of glycerol (C 3 H 8 O 3 ) is dissolved in 500. ml of water at 50. o C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is g/ml. The density of glycerol is 1.26 g/ml. PLAN: Find the mol fraction, χ, of glycerol in solution and multiply by the vapor pressure of water. ΔP = (P solvent - Psolvent) = P solvent (χ Solute ) SOLUTION: 10.0 ml C 3 H 8 O 1.26 g C 3 H 8 O 3 mol C 3 H 8 O 3 3 x ml C 3 H 8 O x = mol C g C 3 H 8 O 3 H 8 O g H ml H 2 O 2 O mol H 2 O x x = 27.4 mol H ml H 2 O g H 2 O 2 O χ = ΔP = mol C 3 H 8 O 3 x 92.5 torr = torr mol C 3 H 8 O mol H 2 O

9 2. Boiling Point Elevation The addition of a nonvolatile non-ionic solute dissolved in a pure solvent increases the boiling the point of a solution. Solution boiling point Pure solvent boiling point ΔTb = Tbp - T bp = Kb m Boiling point elevation (+) Molal boiling point elevation constant ( C/m) Molality (mol/kg)

10 3. Freezing Point Depression The addition of a non-volatile non-ionic solute dissolved in a pure solvent decreases the freezing point of the solution. Pure solvent freezing point ΔTf = Tf - T fp = -Kf m Freezing point depression (-) Molal freezing point depression constant ( C/m) Solution Molality (mol/kg)

11 Molal Boiling-Point Eleveation and Freezing Point Depression Constants of Common Liquids

12 You add 1.00 kg of ethylene glycol antifreeze (C 2 H 6 O 2 ) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the resulting solution?

13 You add 1.00 kg of ethylene glycol antifreeze (C 2 H 6 O 2 ) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the resulting solution? PLAN: Find the number of mols of ethylene glycol and m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water. SOLUTION: m C 2 H 6 O 2 = 1.00 x 10 3 g C 2 H 6 O x mol C 2 H 6 O 2 2 = 16.1 mol C 2 H 6 O g C 2 H 6 O 2 m C 2 H 6 O 2 = 16.1 mol C 2 H 6 O kg H 2 O = 3.62 m C 2 H 6 O 2 ΔT b = o C/m x 3.62 m = 1.85 o C ΔT f = 1.86 o C/m x 3.62 m BP = o C FP = o C

14 Osmosis is the selective passage of solvent molecules through a semipermeable membrane from a dilute solution to a more concentrated one. dilute more concentrated

15 Osmosis is the diffusion of a solvent (frequently water) through a semi-permeable membrane, from a solution of low solute concentration (high water potential) to a solution with high solute concentration (low water potential), up a solute concentration gradient. pure solvent semipermeable membrane solution osmotic pressure An applied pressure is needed to prevent volume increase; this pressure is the osmotic pressure!

16 An pressure difference results from the net movement of solvent from a less-concentrated (hypotonic) to the more-concentrated (hypertonic) solution. For dilute solutions of electrolytes the osmotic pressure is given by: π = n V R T π = M R T Δπ = ΔM R T M is the molarity of the solution R is the gas constant T is the temperature (in Kelvin) Remember: The driving force is due to the difference in concentration of the solutions on each side of the membrane.

17 Calculate molarity of a aqueous solution at 300K which is found to have an osmostic pressure of 3.00 atm. M = π R T = π = M R T 3.00 atm L atm mol 1 K K = M A solution prepared by dissolving 20.0 mg of insulin in water and diluting to a volume of 5.00 ml gives an osmotic pressure of 12.5 torr at 300K. What is the molecular mass of the insulin? M = 12.5 torr 1 atm 760 torr L atm mol 1 K K = mol insulin L 4 mol insulin moles = L 0.005L = mol MolarMass = grams/mole = g/ mol = 5988 g/mol =

18 Suppose we have a molar solution of table sugar (sucrose) and a semi-permeable membrane not permeable to sucrose. What osmotic pressure in mm Hg and to what height could this pressure support a column of water (density Hg =13.6 g/ml and water = 1g/mL? π = M R T π = 0.02 M x L atm/mol K x 298K π =.49 atm x 760 torr/1 atm π = 371 mm Hg x 13.6 = 5.0 meters!

19 Cell membranes are semi-permeable membranes that are susceptable to diffusion of water and a some ions. No Osmotic Pressure Concentrations Are the Same Movement of solvent (water) from dilute to concentrated side! isotonic solution hypotonic solution hypertonic solution

20 Osmosis in an Onion Cell Plasmolyzed cell (cell membrane has shrunk from the cell wall

21 Now this is wild: Anology to Osmosis In a closed container the solution with the highest vapor pressure will completely transfer to the container of lower vapor pressure until the mole fractions of solvent are equal in both! Cool... Pure High Vapor Pressure Low Vapor Pressure

22 Dialysis and Osmosis Pres Memb Pure Water Water With High concentration of dissolved solute

23 Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles. The Four-Colligative Properties Vapor-Pressure Lowering P 1 = X 1 P 1 Boiling-Point Elevation Freezing-Point Depression Osmotic Pressure (π) ΔT b = K b m ΔT f = -K f m π = MRT

24 Biochemists have discovered more than 400 mutant varieties of hemoglobin (Hb), the blood protein that carries oxygen throughout the body. A physician studying a form of Hb associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.0 o C to make 1.50 ml of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this Hb mutant? PLAN: We know Π as well as R and T. Convert Π to atm and T to Kelvin. Use the Π equation to find the molarity M and then the amount and volume of the sample to calculate M. SOLUTION: # mol = g/m M = Π RT 2.08 x 10-4 mol L 21.5 mg x x g 10 3 mg = 3.61 torr x atm 760 torr ( L. atm/mol. K)( K) L 1.50 ml x = 3.12 x 10-7 mol 10 3 ml x x 10-7 mol = 2.08 x 10-4 M = 6.89 x 10 4 g/mol

25 Calculation of Molar mass We can calculate the Molar Mass of a substance using any four of the colligative properties solutions. We use the freezing point depression and osmotic pressure normally as both have much larger changes (easier to measure). If you measure the change in colligative properties for a 1.25 molal sucrose osmotic pressure and freezing point show the largest change and are easiest to measure (especially osmotic pressure).

26 Ionic solutes affect colligative properties differently than non-ionic solutes. We modify the non-ionic colligative equations by multiplying by the van t Hoff factor, i 0.1 m nonelectrolytes solution 0.1 m in solution 0.1 m NaCl solution 0.2 m ions in solution 0.1 m CaCl2 solution 0.3 m ions in solution