1 / 14

III. Colligative Properties

III. Colligative Properties. Solutions. A. Definition. Colligative Property property that depends on the concentration of solute particles, not their identity. B. Types. Freezing Point Depression (  t f ) f.p. of a solution is lower than f.p. of the pure solvent

ehren
Download Presentation

III. 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. III. Colligative Properties Solutions

  2. A. Definition • Colligative Property • property that depends on the concentration of solute particles, not their identity

  3. B. Types • Freezing Point Depression (tf) • f.p. of a solution is lower than f.p. of the pure solvent • Boiling Point Elevation (tb) • b.p. of a solution is higher than b.p. of the pure solvent

  4. B. Types Freezing Point Depression View Flash animation.

  5. B. Types Boiling Point Elevation Solute particles weaken IMF in the solvent.

  6. B. Types • Applications • salting icy roads • making ice cream • antifreeze • cars (-64°C to 136°C) • fish & insects

  7. C. Calculations t: change in temperature (°C) k: constant based on the solvent (°C·kg/mol) m: molality (m) n: # of particles t = k · m · n

  8. C. Calculations • # of Particles • Nonelectrolytes (covalent) • remain intact when dissolved • 1 particle • Electrolytes (ionic) • dissociate into ions when dissolved • 2 or more particles

  9. C. Calculations • At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? GIVEN: b.p. = ? tb = ? kb = 3.60°C·kg/mol WORK: m = 0.73mol ÷ 0.225kg tb = (3.60°C·kg/mol)(3.2m)(1) tb = 12°C b.p. = 181.8°C + 12°C b.p. = 194°C m = 3.2m n = 1 tb = kb · m · n

  10. C. Calculations • Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. GIVEN: f.p. = ? tf = ? kf = 1.86°C·kg/mol WORK: m = 0.48mol ÷ 0.100kg tf = (1.86°C·kg/mol)(4.8m)(2) tf = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C m = 4.8m n = 2 tf = kf · m · n

  11. D. Concentrations of Solutions Percent Solutions • If both solute & solvent are liquids Percent by volume (% v/v) = volume of solute × 100% solution volume • If a solid is dissolved in a liquid Percent (mass/volume) (%(m/v)) = mass of solute (g) × 100% solution volume (mL) Must be the same unit: mL or L Must be this unit

  12. % (v/v) = volume of solute × 100% volume of solution Example 1 What is the percent by volume of ethanol (C2H6O) or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Volume of solute = 85 mL Volume of solution = 250 mL % (v/v) = 85 mL ethanol × 100% 250 mL solution = 34% ethanol

  13. Example 2 How many grams of glucose (C6H12O6) would you need to prepare 2.0 L of 2.8% glucose (m/v) solution? Solution volume = 2.0 L → change to mL Percent by mass = 2.8% • Percent (mass/volume) (%(m/v) = mass of solute (g) × 100% • solution volume (mL) 2.0L 1000mL = 2,000 mL 1L 2.8% = mass of solute (g) × 100% 2,000 mL 0.028 = X 2,000 mL X = 56 g of solute 100% 100%

  14. Percent Solution ProblemsYou do not have to write the problem. You MUST show your work. 1. What is the concentration, in percent (m/v), of a solution with 75g K2SO4 in 1500mL of solution? 2. A bottle of hydrogen peroxide antiseptic is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solution? 3. Calculate the grams of solute required to make 250 mL of 0.10% MgSO4 (m/v).

More Related