Solutions & Colligative Properties Lecture Notes

(HGPC-101) Dr. Hesham Moustafa Dr. Ahmed Salah Colligative Properties of Solutions  Colligative Properties Colligative Property: A property that depends only upon the number of solute particles (concentration ( ), and NOT upon thei...

(HGPC-101) Dr. Hesham Moustafa Dr. Ahmed Salah Colligative Properties of Solutions  Colligative Properties Colligative Property: A property that depends only upon the number of solute particles (concentration ( ), and NOT upon their identity. (For example, addition of Ethylene glycol or Urea to water will have the same effect.) Three Important Colligative Properties of Solutions. Vapor-pressure lowering Boiling-point elevation Freezing-point depression  Vapor-Pressure Lowering Vapor pressure: is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid (molecules are moving back and forth between liquid and vapor phases) A solution that contains a This is true because solute that is not easily in a solution, solute vaporized always has a particles reduce the lower vapor pressure number of free than the pure solvent. solvent particles able to escape the liquid.  Vapor-Pressure Lowering The decrease in a solution’s vapor pressure is proportional to the number of particles the solute makes in solution.  The above colligative properties depend on the mole fraction of the solvent. Mole fraction - The mole fraction of substance “A” is represented as XA # of moles of “A” XA = Total moles of solution (solute + solv Example #1: A solution is 1 mol Ethylene glycol and 9 mol water. Calculate the mole fraction of both the solute and the solvent.  1 mol Ethylene glycol XEthylene glycol = x 100 10 mol solution XEthylene glycol 10 = %  9 mol water XWater = x 100 10 mol solution XWater = 90% Example #2: What are the mole fractions of glucose and water in a solution containing 5.67g of glucose(C6H12O6) dissolved in 25.2g of water ?  GlucoseC6= H12O6180.1572 = g/mol 5.67g glucose 0.0315 = mol glucose Water =H2O =18.0152 g/mol 25.2g water1.40 = mol water  There are three formulas associated with Vapor Pressure Lowering. Plowering = Vapor pressuresolvent - VPsol 2. Plowering = PÅ x XB PÅ = vapor pressure of pure solvent XB = mole fraction of solute 3. Vpsolvent in a solution = PÅ - PA PA = change in VP of solvent (cal’c in. lowering example #1: V.P.water = 17.54 mmHg V.P.0.0100m antifreeze sol’n = 17.36 mmHg V.P.lowering = 0.18 mmHg Lowering vapor pressure, increases boiling po V.P. lowering example #2: How much will the vapor pressure of water drop when 5.67g of glucose are dissolved in 25.2g of water at 25C and what will the new vapor pressure be ? V.P. water 25 = 23.8 mmHg Clue: Use: Plowering = PÅ x XB  5.67g glucose = 0.0315 mol 25.2g water = 1.40 mol Total mol = 1.4315 mol Mol fraction glucose = 0.0220 Plowering = PÅ x XB = 23.8 mmHg x 0.0220 = 0.524 mmHg New V.P. = V.P. pure solvent - drop in V.P. = 23.8 mmHg - 0.524 mmHg = 23.3 mmHg  Boiling-Point Elevation Boiling Point: The temperature at which the vapor pressure of the liquid phase equals atmospheric pressure. Because of the decrease in vapor pressure, additional kinetic energy must be added to raise the vapor pressure of the liquid phase of the solution to atmospheric pressure to initiate boiling. Boiling-Point Elevation Boiling-Point Elevation: The difference in temperature between the boiling point of a solution and the boiling point of the pure solvent. The boiling point of a solution is higher than the boiling point of the pure solvent.  Boiling-Point Elevation The magnitude of the boiling-point elevation is proportional to the number of solute particles dissolved in the solvent. It takes additional kinetic energy for the solvent particles to overcome the attractive forces that keep them in the liquid.  What happened? Conc. dillute Initial Final What conditions exist when the equilibrium point is reached and no further change in the volumes occurs? What is the “Normal Boiling Point” of a substance? The temperature at which the vapor pressure equals 1 atmosphere. Therefore if: Vapor Pressure increases, Boiling Point decreases. Vapor Pressure decreases, Boiling Point increases. Boiling point elevation - Tb equals the boiling point of the solution minus the boiling point = pure Tbthe of i KbCmsolvent. where i = The number of ions if solvent is polar. Kb = boiling point elevation constant (depends on the solvent) Cm = molal concentration B.P. Elevation example #1: Naphthalene, C10H8 , is dissolved in Benzene until a 0.100m solution is attained. Benzene’s normal boiling point is 80.2C. It’s boiling point elevation constant is 2.61C/m. What is the new boiling point ?  Tb = iKbCm = 1 x Kb x Cm = 1 x 2.61C/m x 0.100m = 0.261C increase New B.P. = 80.2C + 0.261C = 80.461C B.P. Elevation example #2: 89.75g of Sodium chloride were added to 250mL water. Kb H20 = 0.512C/m. What is the new boiling point ?  Tb = iKbCm First calculate the molality. 89.75g NaCl 1.54 = “X” mol 250mL H2O = 250“X”g H2O 0.250 = “X” Kg H2O 1.54 mol Therefore: = 6.16m 0.250 Kg  Tb = iKbCm = 2 x Kb x Cm = 2 x 0.512C/m x 6.16m = 6.31C increase New B.P. = 100C + 6.31C = 106.31C  Freezing-Point Depression Freezing-Point Depression: The difference in temperature between the freezing point of a solution and the freezing point of the pure solvent (water). The presence of a solute in water disrupts the formation of the orderly pattern of ice. Therefore more kinetic energy must be withdrawn from a solution than from the pure solvent to cause the solution to solidify. Freezing-Point Depression  Freezing-Point Depression The magnitude of the freezing-point depression is proportional to the number of solute particles dissolved in the solvent and does not depend upon their identity. Which would be a better salt for putting on icy roads, NaCl or CaCl2? Freezing point depression - Tf equals the freezing point of the pure solvent minus the freezing point of the Tf = iKfCm solution. where i = The number of ions if solute is ionic. Kf = freezing point depression constant (depends on the solvent) Cm = molal concentration (m = #mol solute/Kg solvent) Freezing point depression example #1: How many grams of Ethylene glycol, C2H6O2 , must be added to 37.8g of water to give a freezing point of -0.150C ? Kf H2O = 1.858C/m Determination of molar mass from F.P. depression. Example #2: Safrole is contained in oil of sassafras and was once used to flavor root beer. A 2.39mg sample of safrole was dissolved in 103.0mg of Diphenyl ether. The solution had a melting point of 25.70C. Calculate the molecular weight F.P. pureofDiphenyl Safrole. ether = 26.84C Kf = 8.00C/m Clues: 1. Write the formula that you will use. 2. Plug the given values into the formula; or, set up the proportion needed to calculate the increase in boiling point. 3. Add the calculated increase in temperature to the Normal Boiling Point to determine the new boiling point.  Tf = iKfCm You must calculate the following to solve: Tf = 1.14C Cm = 0.1425m Final Answer = 162.83 g/mol Quiz time An aqueous solution was prepared by-1 dissolving 0.131g of a substance in 25.5g of water. The molality of the solution was determined by Freezing Point Depression to be.0.056m What is the molecular weight of the solute ? An aqueous solution is 0.0222m Glucose. -2 Water’s boiling point elevation constant is.0.512C/m What is the new boiling point ?  What is the boiling point of a solution of 0.152g of Glycerol, C3H8O3 , in 20g of water ? A solution was prepared by dissolving 0.915g of Sulfur, S8 , in 100.0g of Acetic acid. What is the new boiling point ? B.P. of Acetic acid = 118.5C Kb of Acetic acid = 3.08C/m Determination of molar mass from B.P. elevation. Example #1: A solution of 9.75g of a solute in 29.38g of water boils at 100.78C at 1 atm of pressure. What is the molar mass of the solute ?