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II III I IV. Colligative Properties of Solutions Ch. 16 – Mixtures & Solutions

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A. Definition Colligative Property Colligative Property property that depends on the number of solute particles, not their identity in an ideal solution

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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 animationFlash 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 T : change in temperature (° C ) i : Vant Hoff Factor (VHF), the number of particles into which the solute dissociates m : molality ( m ) K : constant based on the solvent (° C·kg/mol ) or (°C/ m ) T = i · m · K

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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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C. Calculations T Change in temperature Not actual freezing point or boiling point Change from FP or BP of pure solvent Freezing Point (FP) i T F is always subtracted from FP of pure solvent Boiling Point (BP) i T B is always added to BP of pure solvent

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C. Calculations i – VHF i – VHF Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved number of ions per formula unit 2 or more particles

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C. Calculations i – VHF i – VHF Examples CaCl 2 Ethanol C 2 H 5 OH Al 2 (SO 4 ) 3 Methane CH 4 i = 3 1 5 1

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C. Calculations K – molal constant K – molal constant K F K F – molal freezing point constant Changes for every solvent 1.86 °C·kg/mol (or °C/ m ) for water K B K B – molal boiling point constant Changes for every solvent 0.512 °C·kg/mol (or °C/ m ) for water

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C. Calculations: Recap! : subtract from F.P. T : subtract from F.P. add to B.P. add to B.P. i – VHF : covalent = 1 i – VHF : covalent = 1 ionic > 2 ionic > 2 K : K F water = K : K F water = 1.86 °C·kg/mol K B water = K B water = 0.512 °C·kg/mol T = i · m · K

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At what temperature will a solution that is composed of 0.730 moles of glucose in 225 g of water boil? C. Calculations m = 3.24 m K B = 0.512°C/ m T B = i · m · K B WORK: m = 0.730 mol ÷ 0.225 kg GIVEN: b.p. = ? T B = ? i = 1 T B = (1)(3.24 m )(0.512°C/ m ) T B = 1.66°C b.p. = 100.00°C + 1.66°C b.p. = 101.66°C 100 + T b

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C. Calculations Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. i = 2 m = 4.8 m K F = 1.86°C/ m T F = i · m · K F WORK: m = 0.48mol ÷ 0.100kg GIVEN: f.p. = ? T F = ? T F = (2)(4.8 m )(1.86°C/ m ) T F = 18°C f.p. = 0.00°C – 18°C f.p. = -18°C 0 – T F

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D. Osmotic Pressure Osmosis: The flow of solvent into a solution through a semipermeable membrane Semipermeable Membrane: membrane that allows solvent to pass through but not solute

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D. Osmotic Pressure Net transfer of solvent molecules into the solution until the hydrostatic pressure equalizes the solvent flow in both directions

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Because the liquid level for the solution is higher, there is greater hydrostatic pressure on the solution than on the pure solvent Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent D. Osmotic Pressure

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Osmotic Pressure: Minimum Pressure required to stop flow of solvent into the solution D. Osmotic Pressure

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Osmosis at Equilibrium

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= i M R T where: osmotic pressure (atm) π = osmotic pressure (atm) VHF i = VHF M = Molarity (moles/L) R = Gas Law Constant T = Temperature (Kelvin) E. Osmotic Pressure Calculations 0.08206 L atm/mol K

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E. Osmotic Pressure Calculations Calculate the osmotic pressure (in torr) at 25 o C of aqueous solution containing 1.0g/L of a protein with a molar mass of 9.0 x 10 4 g/mol. i = 1 M = 1.11 x 10 -5 M R = 0.08206 L atm/mol K T = 25 o C = 298 K WORK: M = 1.0 g prot. GIVEN: = ? 1.11 x 10 -5 M = (1)(1.11x10 -5 )(.08206)(298) = 2.714 x 10 -4 atm = 0.21 torr 1 mol prot. 1 L soln 9.0 x 10 4 g =

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Colligative Properties useful for: characterizing the nature of a solute after it is dissolved in a solvent determining molar masses of substances D. Osmotic Pressure

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If the external pressure is larger than the osmotic pressure, reverse osmosis occurs One application is desalination of seawater F. Reverse Osmosis

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Net flow of solvent from the solution to the solvent

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Carli Elizabeth Oster and Ayzsa Fulani Tannis are the most fantastic people in the whole wide world.

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