Colligative Properties Consider three beakers: 50.0 g of ice 50.0 g of ice + 0.15 moles NaCl 50.0 g of ice + 0.15 moles sugar (sucrose) What will the temperature of each beaker be? Beaker 1: Beaker 2: Beaker 3:

Colligative Properties The reduction of the freezing point of a substance is an example of a colligative property: A property of a solvent that depends on the total number of solute particles present There are four colligative properties to consider: Vapor pressure lowering (Raoult’s Law) Freezing point depression Boiling point elevation Osmotic pressure

Colligative Properties – Vapor Pressure A solvent in a closed container reaches a state of dynamic equilibrium. The pressure exerted by the vapor in the headspace is referred to as the vapor pressure of the solvent. The addition of any nonvolatile solute (one with no measurable vapor pressure) to any solvent reduces the vapor pressure of the solvent.

Colligative Properties – Vapor Pressure Nonvolatile solutes reduce the ability of the surface solvent molecules to escape the liquid. Vapor pressure is lowered. The extent of vapor pressure lowering depends on the amount of solute. Raoult’s Law quantifies the amount of vapor pressure lowering observed.

Colligative Properties – Vapor Pressure Raoult’s Law: PA = XAPOA where PA = partial pressure of the solvent vapor above the solution (ie with the solute) XA = mole fraction of the solvent PoA = vapor pressure of the pure solvent

Colligative Properties – Vapor Pressure Example: The vapor pressure of water at 20oC is 17.5 torr. Calculate the vapor pressure of an aqueous solution prepared by adding 36.0 g of glucose (C6H12O6) to 14.4 g of water. Given: PoH2O= 17.5 torr mass solute = 36.0 g of glucose mass solvent = 14.4 g of water Find: PH2O Raoult’s Law: PA = XAPOA

Colligative Properties – Vapor Pressure Solution: Answer: 14.0 torr

Colligative Properties – Vapor Pressure Example: The vapor pressure of pure water at 110oC is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.10 atm at the same temperature. What is the mole fraction of ethylene glycol in the solution? Both ethylene glycol and water are liquids. How do you know which one is the solvent and which one is the solute?

Colligative Properties – Vapor Pressure Given: PoH2O = 1070 torr Psoln = 1.10 atm Find: XEG Solution: Answer: XEG = 0.219 Raoult’s Law: PA = XAPOA

Colligative Properties – Vapor Pressure Ideal solutions are those that obey Raoult’s Law. Real solutions show approximately ideal behavior when: The solution concentration is low The solute and solvent have similarly sized molecules The solute and solvent have similar types of intermolecular forces.

Colligative Properties – Vapor Pressure Raoult’s Law breaks down when solvent-solvent and solute-solute intermolecular forces of attraction are much stronger or weaker than solute-solvent intermolecular forces.

Colligative Properties – BP Elevation The addition of a nonvolatile solute causes solutions to have higher boiling points than the pure solvent. Vapor pressure decreases with addition of non-volatile solute. Higher temperature is needed in order for vapor pressure to equal 1 atm.

Colligative Properties- BP Elevation The change in boiling point is proportional to the number of solute particles present and can be related to the molality of the solution: DTb = Kb.m where DTb = boiling point elevation Kb = molal boiling point elevation constant m = molality of solution The value of Kb depends only on the identity of the solvent (see Table 13.4).

Colligative Properties - BP Elevation Example: Calculate the boiling point of an aqueous solution that contains 20.0 mass % ethylene glycol (C2H6O2, a nonvolatile liquid). Solute = Solvent = Kb (solvent) = Tb = Kb  m

Colligative Properties - BP Elevation Molality of solute: DTb = BP = 102.1oC

Colligative Properties - BP Elevation Example: The boiling point of an aqueous solution that is 1.0 m in NaCl is 101.02oC whereas the boiling point of an aqueous solution that is 1.0 m in glucose (C6H12O6) is 100.51oC. Explain why.

Colligative Properties - BP Elevation Example: A solution containing 4.5 g of glycerol, a nonvolatile nonelectrolyte, in 100.0 g of ethanol has a boiling point of 79.0oC. If the normal BP of ethanol is 78.4oC, calculate the molar mass of glycerol. Given: DTb = 79.0oC - 78.4oC = 0.6oC mass solute = 4.5 g mass solvent = 100.0 g = 0.100 kg Kb = 1.22oC/m (Table 13.4) Find: molar mass (g/mol) Tb = Kb  m

Colligative Properties - BP Elevation Step 1: Calculate molality of solution Step 2: Calculate moles of solute present Step 3: Calculate molar mass

Colligative Properties - Freezing Pt Depression The addition of a nonvolatile solute causes solutions to have lower freezing points than the pure solvent. Solid-liquid equilibrium line rises ~ vertically from the triple point, which is lower than that of pure solvent. Freezing point of the solution is lower than that of the pure solvent.

Colligative Properties - Freezing Pt Depression The magnitude of the freezing point depression is proportional to the number of solute particles and can be related to the molality of the solution. Tf = Kf  m where DTf = freezing point depression Kb = molal freezing point depression constant m = molality of solution The value of Kf depends only on the identity of the solvent (see Table 13.4).

Colligative Properties - Freezing Pt Depression Example: Calculate the freezing point depression of a solution that contains 5.15 g of benzene (C6H6) dissolved in 50.0 g of CCl4. Given: mass solute = mass solvent = Kf solvent = Find: DTf Tf = Kf  m

Colligative Properties - Freezing Pt Depression Molality of solution: DTf =

Colligative Properties - Freezing Pt Depression Example: Which of the following will give the lowest freezing point when added to 1 kg of water: 1 mol of Co(C2H3O2)2, 2 mol KCl, or 3 mol of ethylene glycol (C2H6O2)? Explain why.

Colligative Properties - Freezing Pt Depression Reminder: You should be able to do the following as well: Calculate the freezing point of any solution given enough information to calculate the molality of the solution and the value of Kf Calculate the molar mass of a solution given the value of Kf and the freezing point depression (or the freezing points of the solution and the pure solvent).

Colligative Properties - Osmosis Some substances form semipermeable membranes, allowing some smaller particles to pass through, but blocking other larger particles. In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so. If two solutions with identical concentration (isotonic solutions) are separated by a semipermeable membrane, no net movement of solvent occurs.

Colligative Properties - Osmosis Osmosis: the net movement of a solvent through a semipermeable membrane toward the solution with greater solute concentration. In osmosis, there is net movement of solvent from the area of lower solute concentration to the area of higher solute concentration. Movement of solvent from high solvent concentration to low solvent concentration

Colligative Properties - Osmosis Osmosis plays an important role in living systems: Membranes of red blood cells are semipermeable. Placing a red blood cell in a hypertonic solution (solute concentration outside the cell is greater than inside the cell) causes water to flow out of the cell in a process called crenation.

Colligative Properties Placing a red blood cell in a hypotonic solution (solute concentration outside the cell is less than that inside the cell) causes water to flow into the cell. The cell ruptures in a process called hemolysis.

Colligative Properties - Osmosis Other everyday examples of osmosis: A cucumber placed in brine solution loses water and becomes a pickle. A limp carrot placed in water becomes firm because water enters by osmosis. Eating large quantities of salty food causes retention of water and swelling of tissues (edema).