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Properties of Solutions

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Presentation on theme: "Properties of Solutions"— Presentation transcript:

1 Properties of Solutions
Osmotic pressure and Examples Part IV in a series

2 Example: Aerosol Can When you use and aerosol can and spray, you may notice it gets colder. Why? What are the contents?

3 Answer: Aerosol Can Most propellants are butane or propane (they use to be CFC’s). The propellant is a liquid under pressure, due to its IMF’s. (London forces) As the pressure is released, the propellant changes phase from liquid to a gas. This phase change is endothermic, energy is absorbed and the can cools. If the phase change is endothermic, why does it occur?

4 Example Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na2SO4 (MW= 142g/mol) with 175 g water at 25 °C. The vapor pressure of pure water is torr at this temperature.

5 Solution Mole fraction of Water:
n of H2O = 175 g X (1mol/18g)= 9.72 mol n of Na2SO4= 35g X (1mol/142g) = 0.246mol For every one mole of sodium sulfate, 3 moles of particles are formed. Thus, n solute = 3(0.246 mol)= 0.738mol Mole fraction water= 9.72/( ) Mole fraction water = 0.929 Raoult’s: Psoln= (0.929)(23.76torr)=22.1torr

6 Osmotic Pressure A colligative property when a pure solvent and a solution are separated by a semipermeable membrane. This membrane allows solvent but not solute through.

7 Osmotic Pressure The flow of solvent though this membrane is called osmosis. When the liquid levels are different, there is greater hydrostatic pressure on the solution than the solvent. This excess pressure is osmotic pressure.

8 Osmotic pressure Can be quantified by:  = MRT
 is the osmotic pressure in atmospheres M = molarity of the solution R is the gas constant T= temp in Kelvin

9 Example: 1.00 x 10-3 g of a protein was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25 °C. What was the molar mass of the protein?

10 Solution:  = MRT = 1.12 torr X 1 atm/ 760 torr= 1.47 x10-3 atm
R= l atm/K mol T= = 298 K M=1.47 x 10-3 atm/[( L atm/K mol)(298k)] M=6.01 x 10-5 mol/L 0.001 g in one mL 1 g in 1 L thus: 1.00 g/ 6.01 x 10-5 mol X=1.66x 104 g

11 Colloids A suspension of tiny particles that do not settle is a colloid. Milk and other proteins may form colloids. The scattering of light is called the Tyndall effect.

12 Colloids What stabilizes a colloid?
The main factor appears to be Electrostatic Repulsion. A colloid is electrically neutral, but the molecule attracts a layer of ions of the same charge.

13 Colloids This group of ions in turn attracts another layer of oppositely charged ions. The particles now repel one another and do not easily form aggregates that are large enough to fall out of solution.

14 Coagulation The red colloidal hydrous iron (III) oxide, Fe2O3xH2O (left) is produced by adding concentrated FeCl3(aq) to boiling water. When a few drops of Al2(SO4)3(aq) are added, the suspension particles rapidly coagulate into a precipitate of Fe2O3xH2O


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