2A. Definition Colligative Property property that depends on the concentration of solute particles, not their identity
3B. Types Freezing Point Depression (tf) f.p. of a solution is lower than f.p. of the pure solventBoiling Point Elevation (tb)b.p. of a solution is higher than b.p. of the pure solvent
4Freezing Point Depression B. TypesFreezing Point DepressionView Flash animation.
5Boiling Point Elevation B. TypesBoiling Point ElevationSolute particles weaken IMF in the solvent.
6B. Types Applications salting icy roads making ice cream antifreeze cars (-64°C to 136°C)fish & insectsAntifreeze proteins are found in some fish, insects and plants. They bind to ice crystals and prevent them from growing to where they would damage the host.
7C. Calculations t: change in temperature (°C) t = k · m · nt: change in temperature (°C)k: constant based on the solvent (°C·kg/mol)m: molality (m)n: # of particles
8C. Calculations # of Particles Nonelectrolytes (covalent) remain intact when dissolved1 particleElectrolytes (ionic)dissociate into ions when dissolved2 or more particles
9C. CalculationsAt what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?GIVEN:WORK:
10C. CalculationsAt 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 = 3.04°C·kg/molWORK:m = 0.73mol ÷ 0.225kgtb = (3.04°C·kg/mol)(3.2m)(1)tb = 9.7°Cb.p. = 181.8°C + 9.7°Cb.p. = 192°Cm = 3.2mn = 1tb = kb · m · n
11C. CalculationsFind the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.GIVEN:WORK:
12C. CalculationsFind 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/molWORK:m = 0.48mol ÷ 0.100kgtf = (1.86°C·kg/mol)(4.8m)(2)tf = 18°Cf.p. = 0.00°C - 18°Cf.p. = -18°Cm = 4.8mn = 2tf = kf · m · n