2. Catalyst End

3. Refining Gasoline

4. Lecture 4.8 – Raoult’s Law and Osmosis

5. Today’s Learning Targets LT. 4.11 – I can express the concentration of a solution in terms of mass percentage, mole fraction, molarity, and molality. LT 4.12 – I can explain Raoult’s Law and relate this law to mole fractions and vapor pressures. Furthermore, I can calculate vapor pressures using Raoult’s Law and describe the impact on a particular solution.

6. Mole Fractions Because each gas acts independently even when mixed, we can relate the amount of moles of gas in a mixture. The mole fraction (X) expresses the ratio of moles of gas to total moles of gas in the system

7. Class Example A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO 2, 18.0 mol percent O 2, and 80.5 mol percent Ar. Calculate the partial pressure of O 2 in the mixture if the total pressure of the atmosphere is to be 745 torr.

8. Table Talk The composition of the atmosphere of Saturn’s moon Titan has been estimated. The pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol percent N 2, 12 mol percent Ar, and and 6.0 mol percent CH 4. Calculate the partial pressure of each gas.

9. Colligative Properties Colligative Properties are properties that depend only on the quantity of a substance, not the identity of the substance. The following are all colligative properties: ▫Vapor Pressure Lowering ▫Osmotic Pressure ▫Boiling – Point Elevation ▫Freezing Point Depression

10. Vapor Pressure The liquid state of a substance is in equilibrium with the gas state Vapor Pressure is the pressure when the solution is at this equilibrium Solutions that have vapor pressure are said to be volatile

11. Raoult’s Law We often dissolve a non-volatile substance in a volatile liquid Raoult’s Law states that the partial pressure of a solvent vapor above a solution (P solution ) is the product of the mole fraction of the solvent (X solvent ) and the pure vapor pressure (P o solvent )

13. Raoult’s Law and Vapor Pressure Lowering We can also calculate the overall change in the vapor pressure when a solute is added by using the equation: The vapor pressure lowering depends only on the amount of solute added to a solution.

14. Vapor Pressure for the System Vapor Pressure lowers as more solute is added

15. Class Example Glycerin (C 3 H 8 O 3 ) is a nonvolatile electrolyte with a density of 1.26 g/mL at 25 o C. Calculate the vapor pressure of 25 o C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 o C is 23.8 torr and its density is 1.00 g/mL

16. Table Talk Calculate the vapor pressure of water above a solution prepared by adding 22.5 g of lactose (C 12 H 22 O 11 ) to 200.0 g of water at 338 K. The vapor pressure. P H2O at this temperature is 187.5 torr.

17. Battle Royale For this activity, you will need a white board and a marker. You will be pairing up to go head – to – head with various classmates Depending on whether you get the question correct it will be a win, lose, or tie

18. Question 1 What is the vapor pressure of a solution at 25 oC containing 78.0 grams of glucose (molar mass = 180.16 g/mol) in 500 grams of water? The vapor pressure of pure water at this temperature is 23.8 mm Hg.

19. Question 2 25 grams of cyclohexane (P o = 80.5 torr, molar mass = 84.16 g/mol) and 30 grams of ethanol (P o = 52.3 torr, molar mass = 92.14 g/mol) are both volatile components present in a solution. What is the partial pressure of ethanol?

20. Question 3 The equilibrium vapor pressures of pure benzene and toluene are 95.1 and 28.4 torr, respectively at 25°C. Calculate the total pressure, partial pressures and mole fractions of benzene and toluene over a solution of 1.0 mol of benzene in 3.0 mol of toluene at 25°C.

21. Question 4 The T f of camphor is 179.80°C and it’s K f is 39.7°C/m. When 200.0 mg of a compound (X) are added to 100.0 g of camphor, it’s freezing point drops to 179.29°C. What is the molar mass of X?

22. Question 5 Determine the freezing point of a solution of 60.0 g of glucose, C 6 H 12 O 6, dissolved in 80.0 g of water.

24. Osmosis Many substances contain semipermeable membranes that allow solvent molecules to pass through Osmosis is the movement of solvent through this membrane to the side of higher solute concentration

26. Osmotic Pressure Osmotic Pressure (Π) is the pressure that will be required to stop osmosis

27. Class Example The average osmotic pressure of blood is 7.7 atm at 25 o C. What molarity of glucose (C 6 H 12 O 6 ) will be within blood?

28. Table Talk What is the osmotic pressure at 20 o C of a 0.0020 M sucrose (C 12 H 22 O 11 ) solution?

29. Challenge Problem! The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution is found to be 1.54 torr at 25 o C. What is this protein’s molar mass?

30. Mind Maps Make a mind map of all Unit 4 topics At the center write “Unit 4 – Solids, Liquids, Gases, and Solutions” You will have 12 branches extending out; 1 per each learning target.

33. Closing Time Test Thursday/Friday Stations uploaded to the website Do book problems: 11.3, 11.5, 11.7, 11.17, 11.20, 11.21, 11.25, 11.30, 11.35, 11.61, 11.62, 11.63, 13.63, 13.65, 13.66, 13.77, 13.78, and 13.81