Colligative Properties of Solutions Colligative properties = physical properties of solutions that depend on the # of particles dissolved, not the kind of particle.

Colligative Properties n Lowering vapor pressure n Raising boiling point n Lowering freezing point n Generating an osmotic pressure

Colligative Properties n Lowering vapor pressure n Raising boiling point n Lowering freezing point n Generating an osmotic pressure

Boiling Point Elevation n a solution that contains a nonvolatile solute has a higher boiling pt than the pure solvent; the boiling pt elevation is proportional to the # of moles of solute dissolved in a given mass of solvent.

Boiling Point Elevation n  T b = k b m where:  T b = elevation of boiling pt m = molality of solute k b = the molal boiling pt elevation constant n k b values are constants; see table 15-4, p. 472 (honors text) n k b for water = 0.52 °C/m

Ex: What is the normal boiling pt of a 2.50 m glucose, C 6 H 12 O 6, solution? n “normal” implies 1 atm of pressure n  T b = k b m n  T b = (0.52  C/m)(2.50 m) n  T b = 1.3  C n T b =  C  C =  C

Ex: How many grams of glucose, C 6 H 12 O 6, would need to be dissolved in g of water to produce a solution that boils at 101.5°C? n  T b = k b m n 1.5  C= (0.52  C/m)(m) n m = 2.885

Freezing/Melting Point Depression n The freezing point of a solution is always lower than that of the pure solvent.

Freezing/Melting Point Depression n  T f = k f m where:  T f = lowering of freezing point m = molality of solute k f = the freezing pt depression constant n k f for water = 1.86 °C/m n k f values are constants; see table 15-5, p. 474 (honors text)

Ex: Calculate the freezing pt of a 2.50 m glucose solution. n  T f = k f m n  T f = (1.86  C/m)(2.50 m) n  T f = 4.65  C n T f = 0.00  C  C =  C

Ex: When 15.0 g of ethyl alcohol, C 2 H 5 OH, is dissolved in 750 grams of formic acid, the freezing pt of the solution is 7.20°C. The freezing pt of pure formic acid is 8.40°C. Determine K f for formic acid.  T f = k f m 1.20  C= (k f )( m) k f = 2.8  C/m

Ex: An antifreeze solution is prepared containing 50.0 cm 3 of ethylene glycol, C 2 H 6 O 2, (d = 1.12 g/cm 3 ), in 50.0 g water. Calculate the freezing point of this mixture. Would this antifreeze protect a car in Chicago on a day when the temperature gets as low as –10° F? (-10 °F = -23.3° C)  T f = k f m  T f = (1.86  C/m)(18.06 m)  T f = 33.6  C T f = 0  C – 33.6  C =  C YES!

Electrolytes and Colligative Properties Colligative properties depend on the # of particles present in solution. Because ionic solutes dissociate into ions, they have a greater effect on freezing pt and boiling pt than molecular solids of the same molal conc.

Electrolytes and Colligative Properties n For example, the freezing pt of water is lowered by 1.86°C with the addition of any molecular solute at a concentration of 1 m. –Such as C 6 H 12 O 6, or any other covalent compound n However, a 1 m NaCl solution contains 2 molal conc. of IONS. Thus, the freezing pt depression for NaCl is 3.72°C…double that of a molecular solute. –NaCl  Na + + Cl - (2 particles)

Electrolytes - Boiling Point Elevation and Freezing Point Depression The relationships are given by the following equations: n  T f = k f ·m·n or  T b = k b ·m·n  T f/b = f.p. depression/elevation of b.p. m = molality of solute k f/b = b.p. elevation/f.p depression constant n = # particles formed from the dissociation of each formula unit of the solute

Ex: What is the freezing pt of: a) a 1.15 m sodium chloride solution? n NaCl  Na + + Cl - n=2 n  T f = k f ·m·n n  T f = (1.86  C/m)(1.15 m)(2) n  T f = 4.28  C n T f = 0.00  C  C =  C

Ex: What is the freezing pt of: b) a 1.15 m calcium chloride solution? n CaCl 2  Ca Cl - n=3 n  T f = k f ·m·n n  T f = (1.86  C/m)(1.15 m)(3) n  T f = 6.42  C n T f = 0.00  C – 6.42  C =  C

Ex: What is the freezing pt of: c) a 1.15 m calcium phosphate solution? n Ca 3 (PO 4 ) 2  3Ca PO 4 3- n n=5 n  T f = k f ·m·n n  T f = (1.86  C/m)(1.15 m)(5) n  T f = 10.7  C n T f = 0.0  C – 10.7  C =  C

Determining Molecular Weights by Freezing Point Depression

n  T f = 0.56°C n  T f = k f ·m n 0.56°C = (5.12°C/m)(m) n m = Ex: A 1.20 g sample of an unknown molecular compound is dissolved in 50.0 g of benzene. The solution freezes at 4.92°C. Determine the molecular weight of the compound. The freezing pt of pure benzene is 5.48°C and the K f for benzene is 5.12°C/m.

Ex: A 37.0 g sample of a new covalent compound was dissolved in g of water. The resulting solution froze at –5.58°C. What is the molecular weight of the compound? n  T f = 5.58°C n  T f = k f ·m n 5.58°C = (1.86°C/m)(m) n m = 3.00 m