Molality. Molality (m) is the number of moles of solute per kilogram of solvent. mol of solute kg solvent. Molality ( m) =

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1 Molality Molality (m) is the number of moles of solute per kilogram of solvent. Molality ( m) = mol of solute kg solvent

2 Sample Problem Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water. m = = = mol of solute kg solvent ( 13.5g) m 1 mol KF kg

3 Mole Fraction Mole fraction (χ) is the ratio of the number of moles of a substance over the total number of moles of substances in solution. χ = i = number of moles of i total number of moles n n i T

4 Sample Problem -Conversions between units- ex) What is the molality of a M aluminum nitrate solution (d = 1.012g/mL)? Work with 1 liter of solution. mass = 1012 g mass Al(NO 3 ) 3 = mol g/mol = 42.6 g ; mass water = 1012 g -43 g = 969 g 0.200mol Molality = = mol / kg 0.969kg

5 Sample Problem Calculate the mole fraction of 10.0g of NaCl dissolved in 100. g of water. χ = NaCl = mol of NaCl mol of NaCl + mol H O 1 mol NaCl ( 10.0g) 58.5g NaCl 1 mol NaCl 1 mol H + 2O g NaCl 18.0g H2O ( 10.0g) ( 100.g H O) =

6 Conc. Based on mass g solute mass % or weight % = 100 g solution gsolute ppm = 10 gsolution gsolute ppb = 10 g solution g solute ppt = 10 gsolution

7 ppm, ppb, ppt in dilute aqueous solution ppm ppb ppt 1gsolute 1mgsolute = 6 10 g of solution L solution 1g solute 1µ g solute = 9 10 g of solution L solution 1g solute 1ng solute = g of solution L solution

8 11.8 Effect of Temperature on Solubility The solubility of a gas decreases with temperature.

9 Principles of Solubility Nature of solute and solvent Most nonelectrolytes that are appreciably soluble in water are hydrogen bonded (methanol, hydrogen peroxide, sugars). Other types of nonelectrolytes are generally more soluble in nonpolar or slightly polar solvents such as benzene or toluene.

10 Principles of Solubility (cont.) Effect of temperature a. Increase in T favors endothermic process: solid + water solution H usually positive, so solubility increases with T gas + water solution H usually negative, so solubility decreases with T Effect of pressure Henry s Law Negligible, except for gases, where solubility is directly proportional to the partial pressure of the gas. Carbonated beverages.

11 emperature and pressure effects on gaseous solubility

12 Solubility of O2 Copyright Houghton Mifflin Company. All rights reserved

13 Effect of Temperature on Solubility The solubility of an ionic solid generally increases with temperature.

14 Effect of Pressure on Solubility Henry s Law P gas = k gas C gas P gas = pressure of the gas above the solution C gas = concentration of the gas k gas = Henry s law constant Henry s law holds best for gases O 2 and N 2, does not hold HCl

15 Sample Problem A liter of water dissolves g of oxygen at 25 o C at a pressure of 760. torr. What would be the concentration of oxygen (in g/l) if the pressure were increased to 1880 torr at the same temperature? P C P = C g/l C = torr 1880 torr C = g/l

16 11.9 Colligative Properties Vapor pressure lowering the vapor pressure of a solvent is lowered by the addition of a nonvolatile solute. cf) volatile solute

17 Colligative Properties of Nonelectrolytes The properties of a solution differ considerably from those of the pure solvent. The solution properties depend primarily upon concentration of solute particles rather than type.

18 Colligative Properties of Nonelectrolytes Vapor pressure lowering Boiling point elevation, freezing point lowering Osmosis, osmotic pressure

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20 Raoult s Law P = χ P o solution solvent solvent P o solvent = vapor pressure of the pure solvent It is independent of the nature of the solute but directly proportional to its concentration.

21 Sample Problem What will be the vapor pressure of a solution made by dissolving 6.25g of glucose, C 6 H 12 O 6, in 50.0g of water at 25 o C? How much was the vapor pressure of the pure water lowered? The vapor pressure of water at 25 o C is 23.8 torr 1 mol mol glucose = 6.25 g = mol glucose 180. g 1 mol mol water = 50.0 g = 2.78 mol water 18.0g 2.78 χ water = = o P soln = χ waterpwater = ( 0.988)( 23.8 torr) = 23.5 torr vapor pressure lowering = 23.8 torr torr = 0.2 torr

22 apor pressure lowering and freezing point depression

23 Colligative Properties Boiling point elevation the change in the boiling point is: T b = ik b m i = sum of the coefficients of the ions (i = 1 for molecular compounds) K b = boiling point elevation constant m = molality

24 Colligative Properties Freezing point depression the change in the freezing point is: T f = ik f m i = sum of the coefficients of the ions (i = 1 for molecular compounds) K f = freezing point depression constant m = molality

25 Colligative Properties

26 Boiling point elevation, freezing point lowering Results from vapor pressure lowering.

27 Antifreeze solution

28 Sample Problem Calculate the boiling point elevation and the freezing point depression of a solution made by dissolving 12.2g of KCl in 45.0g of water. K b = o C/m and K f = 1.86 o C/m i = 2 for KCl K + + Cl 1 mol mol KCl = 12.2g = mol 74.6g mol KCl mol mkcl = = = 3.64 m kg water kg o ( )( )( ) o Tb ikbm C/ m 3.64 m 3.73 C = = = o ( )( )( ) o Tf ik fm C/ m 3.64 m 13.5 C = = =

29 Osmosis, osmotic pressure Water moves through semi-permeable membrane from region of high vapor pressure (pure water) to region of low vapor pressure (solution) π = nrt/v = MRT 1 M solution at 25 C has osmotic pressure of 24.5 atm Pickle, Fresh water, Nutrient solution

30 Colligative Properties Osmotic pressure П = imrt i = sum of the coefficients of the ions (i = 1 for molecular compounds) M = molarity R = gas constant ( L atm/mol K) T = temperature in Kelvin

31 Osmosis

32 Osmosis

33 Osmosis

34 Electrolytes Colligative effects are greater because of increased number of particles. T f = 1.86 C m i a. where i is approximately equal to the number of moles of ions per mole of solute: (only in very dilute solution) NaCl(s) Na + (aq) + Cl - (aq) i = 2 CaCl(s) Ca 2+ (aq) + 2Cl - (aq) i = 3 a. Actually, i is usually less than predicted because of ionic atmosphere effects and ion pair.

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36 Sample Problem What is the osmotic pressure of a 100. ml solution containing 9.50 g of glucose, C 6 H 12 O 6, at 20.0 o C? 1 mol mol glucose = 9.50 g = mol 180. g mol M glucose= = mol/l L L atm π = (1)( mol/l) ( [ ] K ) = 12.7 atm K mol