Unit 6: Solutions and Kinetics Colligative Properties

Colligative Properties Colligative property: a property that depends only on the number of solute particles, and not the type of particle Examples: 1. Freezing Point Depression (lowering) 2. Boiling Point Elevation (raising) 3. Vapor Pressure Lowering We will focus on Freezing Point and Boiling Point.

Molality In order to work with colligative properties, we need to use a different type of concentration… Molality (m): the number of moles of solute per kilogram of solvent Units: m ex: 3.0 m is “3 molal” Difference between molarity and molality: https://www.youtube.com/watch?v=96oNrVnTk50

Molality Example What is the molality of a solution of 16.8 grams of lithium sulfate dissolved in 0.500 kg of water?

Molality Practice What is the molality of a solution of 47.3 grams of potassium iodide dissolved in 0.275 kg of water?

Covalent compounds: will not dissociate → df = 1 Dissociation Factor Dissociation factor: the number of particles a solute will break in to when dissolved in a solution Dissociation Factor Covalent compounds: will not dissociate → df = 1 Ionic compounds: will dissociate into ions → df = # of ions per molecule

Dissociation Factor Examples What is the dissociation factor for each compound? AlPO4 N2O4 LiCl CaI2 PCl5 Pb(OH)4 XeF4 Cu2CO3 2 1 3 5

Freezing Point Depression What happens when something freezes (for example, water)? Decrease in energy slows molecules/atoms down so intermolecular forces have more effect (atoms have less energy to fight them) Frozen water (ice) molecules are in an orderly pattern What happens when you add a solute? Adding Solute The addition of a solute disrupts and prevents water molecules from forming an orderly pattern.

Freezing Point Depression Freezing Point Depression: adding a substance to a pure solvent lowers the freezing point To calculate the change in freezing point: ΔTf = (m)(kf)(df) Dissociation factor: the number of particles the solute will break into in solution molality constant Dissociation Factor Covalent compounds: will not dissociate → d.f. = 1 Ionic compounds: will dissociate into ions → d.f. = # of ions per molecule

Freezing Point Depression Example What is the freezing point of 10.2 grams of NaCl in 5.1 kg of water? kf = 1.86 oC/m for water ΔTf = (m)(kf)(df) To find the new freezing point, SUBTRACT your answer from the normal freezing point

Boiling Point Elevation Solution containing nonvolatile solute Pure solvent Solute particles also get in the way of a solvent’s ability to boil, thereby increasing the boiling temperature.

Boiling Point Elevation Boiling Point Elevation: adding a substance to a pure solvent increases the boiling point To calculate the change in boiling point: ΔTb = (m)(kb)(df) Dissociation factor: how many particles the solute will break into in solution molality constant Dissociation Factor Covalent compounds: will not dissociate → d.f. = 1 Ionic compounds: will dissociate into ions → d.f. = # of ions per molecule

Boiling Point Elevation Example What is the boiling point of a solution containing 100.0 g MgCl2 dissolved in 250.0 g of water? kb = 0.512 oC/m for water ΔTb = (m)(kb)(df) To find the new boiling point, ADD your answer to the normal boiling point

Summary Freezing point depression:ΔTf = (m)(kf)(df) kf = 1.86 oC/m for water To find the new freezing point, SUBTRACT your answer from the normal freezing point Boiling point elevation:ΔTb = (m)(kb)(df) kb = 0.512 oC/m for water To find the new boiling point, ADD your answer to the normal boiling point