IV. Colligative Properties of Solutions (p. 498 – 504)

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IV. Colligative Properties of Solutions (p. 498 – 504) Ch. 14 – Mixtures & Solutions IV. Colligative Properties of Solutions (p. 498 – 504)

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

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

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

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

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

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

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

C. Calculations At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? GIVEN: WORK:

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 = 3.04°C·kg/mol WORK: m = 0.73mol ÷ 0.225kg tb = (3.04°C·kg/mol)(3.2m)(1) tb = 9.7°C b.p. = 181.8°C + 9.7°C b.p. = 192°C m = 3.2m n = 1 tb = kb · m · n

C. Calculations Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. GIVEN: WORK:

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