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

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

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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

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

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

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

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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

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

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

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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 = 3.04°C·kg/mol t b = (3.04°C·kg/mol)(3.2m)(1) t b = 9.7°C b.p. = 181.8°C + 9.7°C b.p. = 192°C

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

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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

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