1 / 16

Chapter 13 Sec. 2

Chapter 13 Sec. 2. Solutions and Their Colligative Properties. Solutions have them but pure substances do not have them…. Colligative Properties - Properties of solutions which depend on the number of solute particles in the solution and not the nature of the solute.

jabari
Download Presentation

Chapter 13 Sec. 2

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. Chapter 13Sec. 2 Solutions and Their Colligative Properties

  2. Solutions have them but pure substances do not have them…. Colligative Properties - Properties of solutions which depend on the number of solute particles in the solution and not the nature of the solute. •Three Colligative Properties 1) Vapor pressure lowering 2) Boiling point elevation 3) Freezing point depression

  3. II. Units of Concentration • Molarity and % mass are not useful in colligative properties because the exact amount of solvent is unknown. • Only molality concentration units reflect the number of solute particles per solvent molecules and are useful with colligative properties.

  4. III. Colligative Properties Vapor pressure – the pressure above a liquid created by the liquid releasing gas molecules as it evaporates. The pressure is directly related to temperature. A.Vapor Pressure Lowering • The vapor pressure of the solution is lowered because the solute particles at the liquid/vapor boundary block the solvent particle from jumping into the vapor state.

  5. Dynamic Equilibrium When the rate of freezing is the same as the rate of melting, the amount of ice and the amount of water won't change on average (although there are short-term fluctuations at the surface of the ice). The ice and water are said to be in dynamic equilibrium with each other. The balance between freezing and melting processes can easily be upset. If the ice/water mixture is cooled, the molecules move slower. The slower-moving molecules are more easily captured by the ice, and freezing occurs at a greater rate than melting. Molecules in Motion Flash plug-in

  6. IV. Colligative Properties B.Boiling Point Elevation • For nonvolatile, nonelectrolyte solvents the change in boiling point (DTbp) is: Kbp=boiling point elevation constant DTbp = Kbpmsolute msolute=molality of solute The Boiling point Curve is shifted to the right. Raising the boiling Pt.

  7. Kfp=freezing point depression constant DTfp = Kfpmsolute IV. Colligative Properties C.Freezing Point Depression • For nonvolatile, nonelectrolyte solvents the change in freezing point (DTfp) is: msolute=molality of solute The Freezing point curve shifts to the left. Lowering the F.P. of the solvent.

  8. Practice Problems 1. Determine the freezing point of a solution of 60.0 g of glucose, C6H12O6, dissolved in 80.0 g of water. m= mol/kg m = 60.0g / 180g/mol / 0.080kg = 4.167m ∆Tfp = (-1.86°C/m) (4.167 m) ans: -7.75°C

  9. Practice Problems 2. What is the boiling point of a solution of 645 g of urea, CON2H4, dissolved in 980. g of water? . m= mol/kg m = 645g / 60g/mol / .980 kg = 10.97m ∆Tbp = (0.512°C/m) (10.97 m) ∆Tbp = 5.62°C BP = 100° + 5.62°C = 105.6°C

  10. IV. Colligative Properties D.Colligative Properties of Solutions Containing Ions • The change in VP, BP or FP is greater than expected for electrolyte (ionic salt) solutions. Predicted BP elevation of an aqueous 0.100 m NaCl solution DTbp = Kbp • msolute DTbp, calculated = (0.512 °C/m)(0.100 m) = 0.0512 °C Actual BP elevation of an aqueous 0.100 m NaCl solution DTbp, measured = 0.09470 °C (Almost double the DTbp calculated) • Colligative properties depend on the total number of solute particles in solution. Ionic compounds form ions in solution so the total number of solute particles in solution is equal to the total ions in solution.

  11. IV. Colligative Properties D.Colligative Properties of Solutions Containing Ions NaCl(s)  Na+(aq) + Cl-(aq) 0.100 m 0.100 m 0.100 m 0.200 m total THE MULTIPLIER -van’t Hoff factor (i) We will assume 100% ionization So for electrolytes or ionic solutions:

  12. IV. Colligative Properties D.Colligative Properties of Solutions Containing Ions Predicting van’t Hoff factors NaCl(s)  Na+(aq) + Cl-(aq) 1 particle + 1 particle = 2 particles ipredicted= 2 Na2SO4(s)  2 Na+(aq) + SO42-(aq) 2 particles + 1 particle = 3 particles ipredicted= 3

  13. Practice Problems 3. What is the expected boiling point of a brine solution containing 30.00 g of KBr dissolved in 100.00 g of water? m= mol/kg KBr  two ions , i = 2 m = 30.00g / 119.0g/mol / .100 kg = 2.52 m ∆Tbp = 2 (0.512°C/m) (2.52 m) = 2.6°C B.P. of water = 100.0°C + 2.6°C ans: 102.6°C

  14. Practice Problems 4. What is the expected boiling point of a CaCl2 solution containing 385 g of CaCl2 dissolved in 1.230 kg of water? m= mol/kg CaCl2 three ions , i = 3 m = 385 g / 111.1 g/mol / 1.230 kg = 2.82 m ∆Tbp = 3 (0.512°C/m) (2.82 m) = 4.3°C B.P. of water = 100.0°C + 4.3°C ans: 104.3°C

  15. To Measure a change in VP, BP elevation or FP depression. Moles of Solute Use mass of solvent Solution Conc. Molar Mass g solute mol solute IV. Colligative Properties E.Colligative Properties and Molar Mass Determination • Colligative properties can be used to determine the molar mass of a solute when it is dissolved in a solvent of appreciable vapor pressure and a known Kbp or Kfp. Example: Butylatedhydroxyanisole (BHA) is used as an antioxidant in margarine and other fats and oils; it prevents oxidation and prolongs the shelf life of food. What is the molar mass of BHA if 0.640 g of the compound, dissolved in 25.0 g of chloroform, produces a solution whose boiling point is 62.22 °C. (Chloroform BP = 61.70 °C, Kbp = 3.63 °C/m)

  16. Practice Problems 5. A solution of 0.827 g of an unknown non-electrolyte compound in 2.500 g of water has a freezing point of -10.18°C. Calculate the molar mass of the compound. Molar mass = g/mol solve for moles thru m -10.18 = (-1.86°C/m) (m) ; m = 5.47 mol/kg m = 5.47 mol/kg = mol/0.0025kg mol =.01367 Molar mass = 0.827g /.01367 mol ans: 60.5 g/mol

More Related