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

A. Definition  Colligative Property property that depends on the concentration of solute particles, not their identity Examples: vapor pressure, freezing point, boiling point

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

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

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

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

C. Calculations  t : change in temperature (° C ) i : Van’t 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

C. Calculations  t 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

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

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

C. Calculations  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 °C·kg/mol (or °C/ m ) for water

C. Calculations: Recap! : subtract from F.P.   t : subtract from F.P. add to B.P. add to B.P.  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 = °C·kg/mol  t = i · m · K

 At what temperature will a solution that is composed of 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 = mol ÷ 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. = °C °C b.p. = °C  t b

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