V. Colligative Properties Ch. 13 - Solutions V. Colligative Properties

A. Definition Colligative Property property that depends on the concentration (# of moles) of solute particles, not their chemical nature (identity). For example: NaCl and MgSO4

B. Types 1. Freezing point depression (lowering) 2. Boiling point elevation (raising) 3. Vapor-pressure reduction 4. Osmotic pressure

1. Freezing Point Depression B. Types 1. Freezing Point Depression View Flash animation.

B. Types Freezing Point Depression (tf) f.p. of a solution is lower than f.p. of the pure solvent tf -difference between the freezing point of the solution and the freezing point of the pure solvent

B. Types Applications salting icy roads melts ice making ice cream- lowers temp. Antifreeze (ethylene glycol) cars (-64°C to 136°C)

2. Boiling Point Elevation B. Types 2. Boiling Point Elevation Solute particles weaken IMF in the solvent.

B. Types 2. Boiling Point Elevation (tb) b.p. of a solution is higher than b.p. of the pure solvent tb - difference between the boiling point of the solution and the boiling point of the pure solvent

C. Calculations t: change in temperature (°C) tf(b) = kf(b) · m · n t: change in temperature (°C) k: constant based on the solvent (°C·kg/mol) m: molality (m) n: # of particles 0.51 oC/m Boiling point k 1.86oC/m freezing point k

C. Calculations NaCl Na+ + Cl- n=2 MgCl2 Mg+2 + 2Cl- n=3 # of Particles: Nonelectrolytes (covalent) remain intact when dissolved C6H12O6 C6H12O6 n=1 Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles NaCl Na+ + Cl- n=2 MgCl2 Mg+2 + 2Cl- n=3

C. Calculations At what temperature will a solution that is composed of 0.73 moles of glucose (C6H12O6) in 225 g of water boil? GIVEN: b.p. = ? tb = ? kb = 0.51°C/ m WORK: m = 0.73mol ÷ 0.225kg tb = (0.51°C/ m)(3.2 m)(1) tb = 1.6°C b.p. = 100. °C + 1.6°C b.p. = 101.6°C m = 3.2 m 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: f.p. = ? tf = ? kf = 1.86°C/ m WORK: m = 0.48mol ÷ 0.100kg tf = (1.86°C/ m)(4.8 m)(2) tf = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C m = 4.8 m n = 2 tf = kf · m · n

D. Molar Mass determination Any colligative property can be used to determine the molar mass of an unknown (Exp. #24) STRATEGY: Calculate number of moles required to produce observed change in colligative property

D. Molar Mass determination Use: tb = kb · m · n to find # of moles tb = kb · x moles/kgsolv · n Molar Mass ( M )= mass solute/x moles

D. Molar Mass determination Or use this formula: mass(g)solu · Kb(f) tb(f) · kgsolv Molar Mass (M) =

D. Molar Mass determination A 10.0 g sample of an unknown compound that does not dissociate is dissolved in 100.0 g of water. The boiling point of the solution is elevated 0.433 0C above the normal boiling point. What is the molar mass of the unknown sample?