Colligative Properties of Solutions

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Presentation transcript:

Dissociation is separation of ions that occurs when an ionic compound dissolves.

Colligative Properties of Solutions Properties that depend on the concentration of solute particles but not on their identity Vapor-Pressure Lowering Freezing-Point Depression Boiling-Point Elevation Osmotic Pressure

Vapor-Pressure Lowering A nonvolatile substance : has little tendency to become a gas under existing conditions.

Freezing-point depression(∆tf ) : the difference between the freezing points of the pure solvent and that of a solution ∆tf = i Kf m where: i = # of dissociated moles of solute particles per mole of compound (van't Hoff factor ) Kf = freezing-point constant m = molality of the solution

Boiling-point elevation (∆tb ) : the difference between the boiling points of the pure solvent and that of the solution ∆tb = i Kb m where: i = # of dissociated moles of solute particles per mole of compound (van't Hoff factor ) Kb = boiling-point constant m = molality of the solution

Molal Freezing-Point and Boiling-Point Constants

Sample Problem What is the freezing-point depression of water in a solution of 17.1 g of sucrose, C12H22O11, in 200. g of water? Given: solute mass and chemical formula = 17.1 g C12H22O11 solvent mass and identity = 200 g water Solution: Calculate the molality of the solution

2. Solve for change is freezing point. ∆tf = iKfm Solution: 2. Solve for change is freezing point. ∆tf = iKfm Since sugar does not dissociate , i = 1 ∆tf = (1)(−1.86°C/m)(0.250 m) = − 0.465 °C

Sample Problem What is the boiling-point elevation of a solution made from 20.1 g of NaCl and 400.0 g of water? The molar mass of NaCl is 58.44 g. Given: solute mass = 20.1 g solute molar mass = 58.44 g solvent mass and identity = 400.0 g of water Solution: Calculate the number of moles of the solute. 20.1 g NaCl ÷ 58.44 g/mol = 0.344 mol 2. Calculate the molality of the solution. 0.344 mol = 0.860 m 0.400 kg 3. Calculate the difference in boiling points. Since salt dissociates into two difference ions , i = 2 ∆tb = (2)( 0.860 m)(0.51°C/m) = 0.877 °C