Colligative Properties

1. Colligative Properties Grading Molarity/molality wkst & colligative property notes

2. Grading Molarity/molality wkst • molality side • 2.0 m NaCl • .21 m NaCl • .12 m I2 • 3.8 g I2 • (CCl4 is solvent) • 5) 0.134 kg • Molarity Side • .99 M NaCl • .118 M AgNO3 • 101 grams KNO3 • 0.27 L KCl • 2.5 g CuSO4iH2O • Backside: • 0.445 L (or 445 mL) 4) 7.51 L (or 7,510 mL) • 0.816 L (or 816 mL) 5) .0611 L (or 61.1 mL) • 0.215 L (or 215 mL)

3. How you doin’?

4. animate Freezing…from a molecule’s viewpoint… • Decrease in energy slows molecules/atoms down • Intermolecular forces have more effect (atoms have less energy to fight them) What happens when something freezes (i.e. water)? On a molecular level, draw 5 water molecules in their liquid state and then in their solid (ice) state. Similarities Differences

5. Freezing of a solution (not just pure water) Frozen water (ice) molecules are in an orderly pattern. The addition of another substance (a solute) disrupts and prevents water molecules from forming the pattern. See what I’m talkin’ ‘bout If you add another substance to water the freezing point of that solution will be lower than the freezing point of pure water….this is Freezing Point Depression We can calculate the change in freezing point: DTf = m*kf*d.f. Dissociation factor: How many particles the solute will break into in solution. Change in freezing temp. molality of solution constant = 1.86

6. Example problem: DTf = m*kf*d.f. What is the freezing point of a 2.0 m solution of NaCl in water? Kf = 1.86 oC/m Na+ Cl- DTf = (2.0m)(1.86oC/m)(2) = 7.44oC is the change in freezing temp Normal Freezing temp of water = 0oC Freezing temp. of this solution is -7.44oC

7. Boiling Point Elevation • Solute particles also get in the way of a solvent’s ability to boil thereby increasing the boiling temperature. • Calculated with: DTb = m*kb*d.f. Where kb is always 0.52 oC/m Add the answer (DTb) to 100oC, the boiling temp. of pure water.