Drill (pd 3)
5/11/2015
• What are the 3 rules for determining
solvation (what dissolves what)?
• Determine the number of grams of
solute needed to make a 0.182 m
solution of sucrose (C12H22O11) in
275 g of water.
Answers
• Like dissolves like:
– Polar solutes dissolve in polar solvents
– Nonpolar solutes dissolve in nonpolar solvents
– Ionic compounds are soluble in polar solvents
• molality = amount of solute (mol)
mass of solvent (kg)
0.182 m = X moles
0.275 kg
X = 0.05 moles x 342.3 g sucrose = 17.1 g sucrose
I mole sucrose
Announcement
• HW – Solubility Graph
• Test – Thursday or Friday this week
• Lab – Tomorrow or Wednesday this week
– wear closed toe shoes.
• Lab Notebooks – project grade
Drill
5/11/2015
• Why do you think salt lowers the freezing
point of ice on the road?
Answer
• Addition of solute to form a solution
stabilizes the solvent in the liquid phase,
and lowers the solvent chemical potential
so that solvent molecules have less
tendency to move to the gas or solid
phases. Liquid solutions slightly below the
solvent freezing point become stable
meaning that the freezing point decreases.
Agenda
•
•
•
•
Pass forward Molality WS (all periods),
Making a Solubility Curve Graph and WS
Notes on Polarity (WS)
Notes on Colligative Properties (WS)
SWBAT
• Calculate the freezing point depression
and boiling point elevation of a solute both
in the classroom and the lab.
Homework DUE
• Polar vs Nonpolar WS
Announcement
• Test – Thursday or Friday next week
• Lab – Tuesday or Wednesday next week –
wear closed toe shoes.
Colligative Properties
What is a colligative property?
•
A property that depends on the
concentration* of solute particles – NOT
the identity of the particle.
1. Vapor Point Reduction
2. Boiling Point Elevation
3. Freezing Point Depression
4. Osmotic Pressure
* Concentration will be in terms of molality (m)
Vapor pressure - is pressure caused
by molecules in the gas phase that are
in equilibrium with the liquid phase.
• (in a closed system particles go back
and forth between phases)
Vapor Pressure
Reduction
• The pressure of the vapor
over a solvent is reduced
when a solute is
dissolved in it.
• Vapor pressure reduction
is directly related to
concentration of the
solution.
Why does this occur?
• Increasing solute particles reduces the
proportion of solute to solvent. Fewer
solvent particles will be available to
leave the solution and enter the gas
phase.
• The solution will remain a liquid over a
larger temperature range.
Phase Diagram
Shows us what
phase the solvent
will be in at a
given temperature
and pressure.
If we add a solute
that lowers the
vapor pressure,
then our phase
diagram adjusts…
Boiling Point Elevation
Boiling point – the temp. at which vapor
pressure of a liquid is equal to
atmospheric pressure.
The boiling point of a solvent is raised
when a solute is dissolved in it. It is
directly related to the concentration of
the solute.
The change in boiling point is proportional to
the molality of the solution:
Tb = Kb  m  # of particles
Kb is the molal boiling point elevation constant, a
property of the solvent. It has the units ºC/m.
Tb is the boiling point elevation. It is the difference
between the boiling points of the pure solvent and
that of the solution.
# of particles:
nonelectrolytes = 1
electrolytes = # of ions produced in solutions
Freezing Point Depression
• Freezing point – the temp. where vapor
pressure of the solid and liquid phases
are the same.
• The freezing point of a solvent is
lowered when a solute is dissolved in
it. Freezing point depression is directly
related to concentration of a solution.
The change in freezing point can be found
similarly:
Tf = Kf  m  # of particles
Here Kf is the molal freezing point depression constant
of the solvent.
Tf is the freezing point depression. It is the difference
between the freezing points of the pure solvent and
that of the solution.
# of particles:
nonelectrolytes = 1
electrolytes = # of ions produced in solutions
Practice Problem
• NaCl is used to prevent icy roads and to freeze
ice cream. What are the boiling point and
freezing point of a 0.029 m aqueous solution of
NaCl? Kb = 0.512 C/m
1. Calculate Tb based on number of particles in
the solution.
Tb = Kb  m  # of particles
Tb = (0.512) (0.029) (2) = 0.030 C
Boiling Point = 100 C + 0.030 C = 100.030 C
Practice Problem
• NaCl is used to prevent icy roads and to freeze
ice cream. What are the boiling point and
freezing point of a 0.029 m aqueous solution of
NaCl? Kf = 1.86 C/m
1. Calculate Tf based on number of particles in
the solution.
Tf = Kf  m  # of particles
Tf = (1.86) (0.029) (2) = 0.11 C
Freezing Point = 0.00 C - 0.11 C = -0.11 C
Drill #5
4/22 & 23/14
• Using prefix meanings as a clue, which
direction does water move in hypotonic,
hypertonic and isotonic solutions?
• Hint: the prefix refers to concentration of
solute outside cell
SWBAT
• Describe the properties of hypertonic,
hypotonic and isotonic solutions.
• Calculate the freezing point depression
and boiling point elevation of a solute both
in the classroom and the lab.
Agenda
• Finish Notes on Colligative Properties Osmosis
• Antifreeze Lab Calculations
Homework DUE
• Effect of a Solute on Freezing and Boiling
Points WS
Homework
• Colligative Properties Quiz – next class!
• 15-4 Review and Reinforcement - #1 - #11
• 15-4 Practice Problems - #1, 3, 4, 7, 9,
11,12,15,16, and 18.
Osmosis
• Water diffuses from an area of high
concentration to an area of lower
concentration
Osmosis
• 3 types of osmosis (based on amount of solute
outside the cell)
– Hypertonic solution (higher)
– Hypotonic solution (lower)
– Isotonic solution (equal)
Hypertonic Solution
• Higher water concentration inside of the
cell compared to the outside
– More solutes on the outside of cell
– Water diffuses out of cell
– Cell shrinks
Hypotonic Solution
• Higher water concentration on the outside
of cell than inside
– Less solutes on the outside of cell
– Water diffuses into cell
– Cell gets bigger
Isotonic Solution
• Same water concentration outside and
inside of the cell
– Same solute concentration on outside and
inside of cell
– Water moves in and out at the same time
– Cell stays the same size
Pre-Lab Questions
We will begin with pre-lab question 5
• 5. Why is calculation step 5 necessary?
– Hint: What are you trying to calculate?
6. Write the mathematical
expression for molality.
7. Write the mathematical
expression that relates the math,
number of moles and molar mass
of a substance
• Hint: What is the formula for molar mass?
8. Using these two expressions,
derive an equation for calculating
the molar mass of a sample if you
know its mass and the freezing
point depression it causes for
water.
• Hint – Combine the equations from step 6
and 7 to determine the equation.
Post Lab Calculations
• 1-2 Record the freezing point depression
of your solutions
• 3-4 Determine the molality (m) of each
solution
– Hint: What equations do you know for
colligative properties
Post Lab Calculations Cont.
• To find the molar mass of antifreeze, you
first need to calculate the number of grams
of antifreeze per 1000 grams of solvent for
the solutions
– Hint: What concentrations were solutions 1
and 2?
• What is the ratio between the solute and the
solvent?
Post Lab Calculations Cont.
• Find the molar mass of antifreeze
– Use the equation you derived in the pre-lab
Complete the Critical Thinking
Questions
Questions not completed in class
will become your homework
Assignment
• Effect of a Solute on Freezing & Boiling
Points Worksheet #1-4
• Pre-lab Questions
Chapter 14 Test: this Friday, April 27
Topics include:
• Types of Mixtures – notes
• Solution Concentrations – molarity, molality
and dilution equations – problems
• Factors affecting solvation - notes
• Colligative Properties – boiling & freezing pts notes
• Osmosis – hypertonic, hypotonic, & isotonic
solutions - notes