1. II III I I. The Nature of Solutions (p. 401 - 410, 425 - 433) Ch. 13 & 14 - Solutions

2. A. Definitions  Solution -  Solution - homogeneous mixture Solvent Solvent - present in greater amount Solute Solute - substance being dissolved

3. A. Definitions Solute Solute - KMnO 4 Solvent Solvent - H 2 O

4. B. Solvation  Solvation –  Solvation – the process of dissolving solute particles are separated and pulled into solution solute particles are surrounded by solvent particles

5. B. Solvation Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules DISSOCIATIONIONIZATION View animation online.animation

6. B. Solvation  Dissociation separation of an ionic solid into aqueous ions NaCl(s)  Na + (aq) + Cl – (aq)

7. B. Solvation  Ionization breaking apart of some polar molecules into aqueous ions HNO 3 (aq) + H 2 O(l)  H 3 O + (aq) + NO 3 – (aq)

8. B. Solvation  Molecular Solvation molecules stay intact C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq)

9. B. Solvation NONPOLAR POLAR “Like Dissolves Like”

10. B. Solvation  Soap/Detergent polar “head” with long nonpolar “tail” dissolves nonpolar grease in polar water

11. C. Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

12. C. Solubility  Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln

13. C. Solubility  Solubility Curve shows the dependence of solubility on temperature

14. C. Solubility  Solids are more soluble at... high temperatures.  Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: nitrogen narcosis, the “bends,” soda

15. II III I II. Concentration (p. 412 - 418) Ch. 13 & 14 - Solutions

16. A. Concentration  The amount of solute in a solution.  Describing Concentration % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

17. A. Concentration SAWS Water Quality Report - June 2000

18. B. Molality mass of solvent only 1 kg water = 1 L water

19. B. Molality  Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.21 g MgCl 2 = 3.2 m MgCl 2 0.25 kg water

20. B. Molality  How many grams of NaCl are req’d to make a 1.54m solution using 0.500 kg of water? 0.500 kg water1.54 mol NaCl 1 kg water = 45.0 g NaCl 58.44 g NaCl 1 mol NaCl

21. C. Dilution  Preparation of a desired solution by adding water to a concentrate.  Moles of solute remain the same.

22. C. Dilution  What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

23. D. Preparing Solutions  500 mL of 1.54M NaCl 500 mL water 45.0 g NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL mark 500 mL volumetric flask  1.54m NaCl in 0.500 kg of water

24. D. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Filling the volumetric flask”)

25. D. Preparing Solutions Copyright © 1995-1996 NT Curriculum Project, UW-Madison (above: “Using your hand as a stopper”)

26. D. Preparing Solutions  250 mL of 6.0M HNO 3 by dilution measure 95 mL of 15.8M HNO 3 95 mL of 15.8M HNO 3 water for safety 250 mL mark combine with water until total volume is 250 mL Safety: “Do as you oughtta, add the acid to the watta!”

27. Solution Preparation Lab  Turn in one paper per team.  Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution.  For each of the following solutions: 1) 100.0 mL of 0.50M NaCl 2) 0.25m NaCl in 100.0 mL of water 3) 100.0 mL of 3.0M HCl from 12.1M concentrate.

28. II III I III. Colligative Properties (p. 436 - 446) Ch. 13 & 14 - Solutions

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

30. B. Types  Freezing Point Depression  Freezing Point Depression (  t f ) f.p. of a solution is lower than f.p. of the pure solvent  Boiling Point Elevation  Boiling Point Elevation (  t b ) b.p. of a solution is higher than b.p. of the pure solvent

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

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

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

34. 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

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

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

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