Transcript Catalyst - MAEDA AP Chemistry

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
moles of compound 1
n1
X1 =

total moles
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 (Posolvent)
Psolution  X
o
solvent solvent
P
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:
P  X
o
solute solvent
P
• The vapor pressure lowering depends only on
the amount of solute added to a solution.

Vapor Pressure lowers
as more solute is added
Vapor Pressure
for the System
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
n 
   RT  MRT
V 

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 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