Catalyst

End

Refining Gasoline

Lecture 4.8 – Raoult’s Law and Osmosis

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.

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

€

X1 =moles of compound 1

total moles=n1

nt

Class Example

•A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO2, 18.0 mol percent O2, and 80.5 mol percent Ar. Calculate the partial pressure of O2in the mixture if the total pressure of the atmosphere is to be 745 torr.

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 N2, 12 mol percent Ar, and and 6.0 mol percent CH4. Calculate the partial pressure of each gas.

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

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

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 (Psolution) is the product of the mole fraction of the solvent (Xsolvent) and the pure vapor pressure (Po

solvent)

€

Psolution = X solventPsolvento

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.

€

ΔP = X solutePsolvento

Vapor Pressure for the System

Vapor Pressure lowers as more solute is added

Class Example

•Glycerin (C3H8O3) is a nonvolatile electrolyte with a density of 1.26 g/mL at 25 oC. Calculate the vapor pressure of 25 oC 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 oC is 23.8 torr and its density is 1.00 g/mL

Table Talk

•Calculate the vapor pressure of water above a solution prepared by adding 22.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. The vapor pressure. PH2O at this temperature is 187.5 torr.

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

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.

Question 2•25 grams of cyclohexane (Po = 80.5 torr,

molar mass = 84.16 g/mol) and 30 grams of ethanol (Po = 52.3 torr, molar mass = 92.14 g/mol) are both volatile components present in a solution. What is the partial pressure of ethanol?

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.

Question 4

•The Tf of camphor is 179.80°C and it’s Kf 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?

Question 5•Determine the freezing point of a solution

of 60.0 g of glucose, C6H12O6, dissolved in 80.0 g of water.

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

Osmotic Pressure•Osmotic Pressure (Π) is the pressure

that will be required to stop osmosis

€

Π=n

V

⎛

⎝ ⎜

⎞

⎠ ⎟RT = MRT

Class Example•The average osmotic pressure of blood is

7.7 atm at 25 oC. What molarity of glucose (C6H12O6) will be within blood?

Table Talk•What is the osmotic pressure at 20 oC of a

0.0020 M sucrose (C12H22O11) solution?

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 oC. What is this protein’s molar mass?

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.

Closing TimeClosing 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