1 / 70

Colligative Properties

Colligative Properties. of Solutions. Review. Parts of a Solution. SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) Solute + Solvent = Solution. Definitions.

rhian
Télécharger la présentation

Colligative Properties

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. Colligative Properties of Solutions

  2. Review Parts of a Solution • SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) • SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) • Solute + Solvent = Solution

  3. Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

  4. Aqueous Solutions If ions are present in aqueous solutions? They are called ELECTROLYTES HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

  5. Aqueous Solutions Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugar ethanol ethylene glycol

  6. Definition • Colligative Property • property that depends on the concentration of solute particles, not their identity • The presence of solutes affects the properties of the solutions

  7. Types A. Vapor Pressure Lowering: v.p. is lower in a solution than that of its pure solvent B. Freezing Point Depression (tf) f.p. of a solution is lower than f.p. of the pure solvent C. Boiling Point Elevation (tb) b.p. of a solution is higher than b.p. of the pure solvent D. Osmotic Pressure: The external pressure that must be applied to stop osmosis.

  8. Common Units % by mass – g solute/ g solution x 100% % by volume – Liters solute/ L solution x 100% Mole % - moles solute/moles solution x 100% Molarity (M) – moles solute/ L solution Molality (m) – moles solute/ kg solvent

  9. Molarity Molarity is probably the most common unit of concentration in chemistry. Why is that?

  10. Molarity Molarity (M) = moles solute L solution This is both a chemically relevant unit and a practically relevant unit. L solution is very easy to measure in the lab! MOLES are easy to calculate. Reaction occur based on the relative number of moles.

  11. % by mass This is one of the easiest units to use. Why?

  12. % by mass This is one of the easiest units to use. Why? Mass is easy to measure in the lab!

  13. % by mass % by mass = g solute x 100% g solution

  14. Colligative Property A colligative property is a property that depends ONLY on the amount of the substance present NOT on the identity of the substance. In other words, it doesn’t matter if it is salt, sugar, gasoline, or tennis balls – it will behave the same way!

  15. Types • Vapor Pressure Reduction • Related to boiling point • Freezing Point Depression • Salt on the road • Anti-freeze in your radiator • Boiling Point Elevation • Anti-freeze in your radiator • Osmotic Pressure • Membrane diffusion

  16. Vapor Pressure Lowering • Vapor pressure is the pressure caused by molecules that have escaped from the liquid phase to the gaseous phase. • The vapor pressure of the solution is lower than the vapor pressure of the pure solvent. • The solute thus reduces the freezing point and elevates the boiling point.

  17. Vapor Pressure At equilibrium, the rate of evaporation (liquid to gas) equals the rate of condensation (gas to liquid). The amount of gas is the “vapor pressure” Surface of liquid

  18. Vapor Pressure • Liquids exert vapor pressure (VP). As the temperature increases, VP increases! • Raoult found that when solute was added to pure solvent, it lowered the VP of the solvent--the more solute added the more depressed the VP became. This has became known as Raoult's Law. • For example, sugar lowers the VP of pure water by blocking water from vaporizing at surface

  19. Raoult's Law:  P = XP° P = VP of solution  P° = VP of pure solvent X = mole fraction of solvent • Vapor Pressure is the pressure caused by the molecules that have escaped from the liquid phase to the gas phase. • Adding nonvolatile solute, lowers the concentration of water molecules at the surface of the liquids. • This lowers the tendency of water molecules at the surface of the liquid.

  20. Problem 1.00g of nonvolatile sulfanilamide, C6H8O2N2S, is dissolved in 10.0g of acetone, C3H6O. The vapor pressure of pure acetone at the same temperature is 400 mmHg. Calculate the vapor pressure of the solution.

  21. Solution 1) Calculate moles of solute (sulfanilamide): (C6H8O2N2S) = 1.00 ÷ 172.1 = 0.0058 mol 2) Calculate moles of solvent (acetone): (C3H6O) = 10.0 ÷ 58 = 0.172 mol 3) Calculate the mole fraction of the solvent: Xsolvent = nsolvent ÷ (nsolute + nsolvent) Xa = 0.172 ÷ [0.172 + 0.0058] = 0.967 4) Calculate the vapor pressure: Pa = XaPao Pa = 0.967 x 400 mmHg = 386.8 mmHg = 387 mmHg

  22. ProblemWhat is the change in vapor pressure when 164 g of glycerin (C3H8O3) is added to 338 mL of H2O at 39.8 °C. The vapor pressure of pure H2O at 39.8 °C is 54.74 torr. The density of H2O at 39.8 °C is 0.992 g/mL.SolutionRaoult's Law can be used to express the vapor pressure relationships of solutions containing both volatile and nonvolatile solvents. Raoult's Law is expressed byPsolution = ΧsolventP0solvent wherePsolution is the vapor pressure of the solutionΧsolvent is mole fraction of the solventP0solvent is the vapor pressure of the pure solvent

  23. Types: Freezing Point Depression

  24. Change in Freezing Point Ethylene glycol/water solution Pure water The freezing point of a solution is LOWERthan that of the pure solvent

  25. Change in Freezing Point Common Applications of Freezing Point Depression Ethylene glycol – deadly to small animals Propylene glycol

  26. Change in Freezing Point Common Applications of Freezing Point Depression • Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? • sand, SiO2 • Rock salt, NaCl • Ice Melt, CaCl2

  27. Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -) Compound Theoretical Value of i glycol 1 NaCl 2 CaCl2 3 Ca3(PO4)2 5

  28. Boiling Point Elevation and Freezing Point Depression ∆T = K•m•i m = molality K = molal freezing point/boiling point constant

  29. Freezing Point Depression • Molality is the concentration of a solution expressed in moles of solute per kilogram of solvent • m = molality (moles of solute / mass of solvent) • A solution that contains 1 mole of solute, sodium hydroxide, NaOH, for example, dissolved in exactly 1 kg of solvent is a “one-molal” solution.

  30. One mole of NaOH has a molar mass of 40.0 g, so 40.0 g of NaOH dissolved into 1 kg of water results in a one-molal NaOH solution. If 20.0 g of NaOH, which is 0.50 moles of NaOH, is dissolved in exactly 1 kg of water, the concentration of the solution is 0.50 m NaOH. If 80.0 g of NaOH, which is 2.0 moles of NaOH, is dissolved in exactly 1 kg of water, the concentration of the solution is 2.00 m NaOH.

  31. Freezing Point Depression Calculate the Freezing Point of a 4.00 molal glycol/water solution. Kf = 1.86 oC/molal (See Kf table) Solution ∆TFP = Kf • m • i = (1.86 oC/molal)(4.00 m)(1) ∆TFP = 7.44 FP = 0 – 7.44 = -7.44 oC(because water normally freezes at 0)

  32. The freezing point of a 1 molal solution of anynonelectrolyte solute in water is found to be 1.86oC lower than the freezing point of pure water. When 1 mole of a nonelectrolyte solute is dissolved into 1 kg of water, the freezing point of the solution is -1.86oC instead of 0oC. When 2 mole of a solute is dissolved into 1 kg of water, the freezing point of the solution is -3.72oC. This is 2 x -1.86oC Molal freezing point constant for water: kf = -1.86 °C/m Each solvent has its own characteristic molal freezing-point constant

  33. Freezing Point Depression At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆TFP = Kf • m • i ∆TFP = (1.86 oC/molal) • 5.4 m • 2 ∆TFP = 20.1oC FP = 0 – 20.1 = -20.1 oC

  34. Molal freezing point constant: kf = -1.86 °C/m • m= molality (moles of solute / mass of solvent) • Kf= was determined experimentally. Does not depend on the identity of the substance! • tf = Kfm • tf = freezing pt of solution – freezing point of pure solvent

  35. Determine the f.p. of a solution of fructose, C6H12O6, made by dissolving 58.0g of fructose in 185g H2O.

  36. Determine the f.p. of a solution of fructose, C6H12O6, made by dissolving 58.0g of fructose in 185g H2O. -3.24 oC

  37. Determine the molal concentration of a solution of ethylene glycol, HOCH2CH2OH, if the solutions’sf.p. is -6.40C. • What mass of ethylene glycol would you dissolve in 500.0g of water to prepare such a solution?

  38. Determine the molal concentration of a solution of ethylene glycol, HOCH2CH2OH, if the solutions’sf.p. is -6.40C. • 3.44m • What mass of ethylene glycol would you dissolve in 500.0g of water to prepare such a solution? • 107g

  39. Boiling Point Elevation Boiling point elevation is related to freezing point depression. They both use a similar equation.

  40. Boiling Pt. Elevation Freezing Pt. Depression ΔTb = T boiling, solution – T boiling, pure solvent = Kb m ΔTf = T freezing, pure solvent – T freezing, solution = Kf m m = molality of the solution Kb = boiling constant Kf = freezing constant

  41. Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution? Kb = 0.52 oC/molal for water (see Kb table). Solution ∆TBP = Kb • m • i 1. Calculate solution molality = 4.00 m 2. ∆TBP = Kb • m • i ∆TBP = 0.52 oC/molal (4.00 molal) (1) ∆TBP = 2.08 oC BP = 100 + 2.08 = 102.08 oC (water normally boils at 100)

  42. NOTE ABOUT THE SIGN CONVENTION ΔTb = T boiling, solution – T boiling, pure solvent = Kb m ΔTf = T freezing, pure solvent – T freezing, solution = Kf m Δ = change

  43. K is a SOLVENT property Colligative properties don’t care what the solute is – they only care how much solute there is. The boiling point constant and freezing constant are SOLVENT properties only.

  44. A Review Problem How many grams of sucrose (C11H22O11) are needed to lower the freezing point of 100 g of water by 3° C?

  45. The Answer ΔTf = Kf m i We want to decrease the freezing point by 3°C 3° C = (1.86 °C/molal) m (1) m=1.61 molal = 1.61 moles solute/kg solvent

  46. 1.61 moles solute = 0.161 moles solute 1 kg solvent 0.100 kg water 0.161 moles sucrose * 330 g sucrose = 53.1 g 1 mole sucrose Sucrose = C11H22O11 330 g/mol = 11*12.01 g/mol + 22* 1.008 g/mol + 11* 16 g/mol

  47. Slight Variation on the problem How many grams of NaCl are needed to lower the freezing point of 100 g of water by 3 °C?

  48. The Answer ΔTf = Kf m i We want to decrease the freezing point by 3°C 3° C = (1.86 °C/molal) m (2) m=0.81 molal = 0.81 moles solute/kg solvent

  49. 0.81 moles solute = 0.081 moles solute 1 kg solvent 0.100 kg water

  50. 0.81 moles solute = 0.081 moles solute 1 kg solvent 0.100 kg water NaCl is an electrolyte: NaCl Na+ + Cl- You get 2 moles of solute per mole NaCl so i = 2 0.081 moles solute * 58.45 g NaCl = 4.7 g NaCl 1 mole NaCl

More Related