Chapter 19: Molality and Colligative Properties

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Chapter 19 Molality and Colligative Properties

• HW Ch. 19 Blue Book 1-17, 19 (on problems that
  are a-z, please do a and b only)

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Plan of the Day

• Begin Ch. 19 Notes Examples
• Freezing Point Depression Lab (aka Ice Cream Lab)
  5/31
• Each student needs to bring
• ½ cup milk
• ½ cup heavy whipping cream
• ¼ tsp vanilla
• 1 (1qt.) ziploc bag
• 1 (gallon) ziploc bag
• HW Assignment - 1-17 19
• TEST 6/4

Divide conquer is the best method
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Molality

• The volume of a solution changes with a change in
  temperature which alters the molarity. (ex what
  happens when you boil 3 cups of water do you
  still have 3 cups?) http//www.youtube.com/watch?v
  WNrSexmBDXUfeaturerelated
• Masses, however, do not change with temperature.
• So we use molality (m)- the number of moles of
  solute per kilogram of solvent (mol/Kg, another
  concentration ratio).

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Example 1

• If 60.0 g of NaOH are dissolved in 1500g of
  water, what is the concentration of this solution
  using Molality?
• Analyze the problem what info are we given?
• mass of solute, 60.0 g NaOH
• mass of solvent, 1500 g H2O
• Solve for the unknown
• Couple of ways to do this
• 1. Railroad-track the whole thing all at
  once
• OR
• 2. Break it up
• calculate number of moles of solute
• Convert mass of water from g ? kg
• Plug in and solve for molality
• 1.00m

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Example 2

• Calculate the mass of ethanol, C2H5OH, which
  must be dissolved in 750.0 g water to make a 2.00
  m solution.
• Multiply the mass of water by the
  concentration (2.00 m), then convert to grams of
  ethanol by using molecular mass of ethanol
• 750g H2O 2.00 mol C2H5OH 46.1g C2H5OH
• 1000 g H2O 1
  mol C2H5OH
• Answer 69.2g C2H5OH

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Example 3

• Determine the mass of H2SO4 which must be
  dissolved in 2500 g of H2O to make a 4.00m
  solution.
• Multiply the mass of H2O by the concentration
  ratio. Then convert from moles ? grams of H2SO4
  using the formula mass of the acid.
• 2500g H2O 4.00 mol H2SO4 98.1 g H2SO4
• 1000 g H2O
  1mol H2SO4

981 g H2SO4
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Example 4

• What is the percentage by mass of Mg(NO3)2 in
  a 2.50 m solution?
• The concentration involves 1000g H2O. Find
  the g Mg(NO3)2 add these two masses together
  for total mass, then calculate percentage
  Mg(NO3)2 .
• 2.50 mol Mg(NO3)2 148.0 Mg(NO3)2
• 1000 g H2O 1 mol Mg(NO3)2
• Mass of soln 370.0 g 1000 g 1370 g soln
• Mg(NO3)2 370.0
• 1370 g

370 g Mg(NO3)2 1000 g H2O
27.0
X 100
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Example 5

• How many molecules of ethanol must be dissolved
  in 500.0 g of water to make a 1.00 m solution?
• Multiply the mass of water by the concentration.
  Then multiply by Avogadros /mol.
• 500.0 g H2O 1.00 mol C2H5OH 6.02 x 1023
  molecules
• 1000 g H2O
  1 mol
• Answer 3.01 x 1023 molecules C2H5OH

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Colligative Properties

• What are they?
• The word Colligative means depending on the
  collection
• Change the physical properties of the solvent.
• Depends on the number of particles of the solute
  NOT which solute is used!

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Colligative Properties

• Lowers the vapor pressure!
• Raises the boiling point!
• Lowers or depresses the freezing point!
• Osmotic pressure
• Why?

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Colligative Properties

• When a solute is dissolved in a solvent, the
  vapor pressure of the solvent is reduced.
• The reduction depends on the number of solute
  particles in a given amount of solvent.
• The French chemist, Raoult, first discovered the
  vapor pressure lowering relationship
  experimentally in 1882 which lead to

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Colligative Properties

• Raoults Law Any nonvolatile solute at a
  specific concentration lowers the vapor pressure
  of the solvent by an amount that is
  characteristic of that solvent.

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Vapor Pressure Lowering

• The vapor pressure above a liquid is lowered due
  to the attractive forces of the solvent on the
  dissolved solute particles.
• Because of this, less solvent particles have the
  energy to transition to the gaseous state
  (evaporate), and therefore the vapor pressure is
  lower.
• So The greater the number of solute particles in
  a solvent, the lower the VP

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• 1 solvent has a large surface area to
  evaporate from
• 2 mixed with solute fewer solvent particles
  at surface

Which one has lower VP?
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Boiling Point Elevation

• Similar factors (as with the vapor pressure
  lowering), contribute to the increase of the
  boiling point of a solvent .
• The more solute particles the higher the BP (the
  lower the VP)
• Practical application adding salt to water to
  increase the BP of water to cook foods.

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Boiling Point Elevation
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Freezing Point Depression

• Freezing occurs when the particles no longer have
  the energy to overcome their interparticle
  attractive forces they organize and solidify
  (molecules slow way down, loss of kinetic
  energy).
• Adding solute to a pure solvent lowers the FP!
• WHY?
• Because the solute interferes with the solvents
  interparticle attractions, therefore the solid
  forms at cooler or lower temperature.
• So the FP of a solution is always lower than the
  FP of a pure solvent.

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Freezing Point Depression
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___ Pure Solvent ---- Solution
100oC
0oC
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Osmotic Pressure

• What is osmosis?
• The amount of additional pressure caused by the
  water molecules that move into a concentrated
  solution is called osmotic pressure. (The
  diffusion of water)
• This pressure depends on the number of solute
  particles in a given volume of solution.

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As water is moving ? the pressure exerted by the
additional water molecules, osmotic pressure, is
increasing on the left side of the semipermeable
membrane. Higher osmotic pressure on left, lower
osmotic pressure on right.
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Colligative Properties (now the math)

• The change in the freezing and boiling pts varies
  directly with the concentration of particles.
• Molal freezing pt constant 1.86C for water.
  Each mole of solute causes the freezing pt of
  water to drop by this much.
• Molal boiling pt constant 0.512C for water.
  Each mole of solute causes the boiling point to
  rise by this much.

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Colligative Properties

• These can be used to determine
• The freezing point of the water
• The boiling point of the water
• The molecular mass of the solute from the
  freezing point or the boiling point
• (see table 19-1 for other constants)

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Colligative Properties

• Ex. 3
• Calculate the freezing point of a solution
  containing 5.70 g of sugar, C12H22O11, in
• 50.0 g of water.(Molal freezing pt constant
  1.86C for water. )
• Convert grams of solute per gram of water to
  moles of solute per kg of water (molality). Then
  multiply by the conversion ratio to obtain the
  change in FP
• 5.70 g C12H22O11 103 g H2O 1 mol
  C12H22O11 1.86C
• 50.0 g H2O 1kg H2O 342 g
  C12H22O11 1 m
• 0.620C,
• To determine the FP, subtract this from the FP of
  water
• 0 oC 0.620 - 0.620 oC

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Calculating Molecular Mass Ex. 4

• When 72.0 g of dextrose were dissolved in 100.0 g
  of water, the boiling point of the solution was
  observed to be 102.05 C. What is the molecular
  mass of dextrose?
• Step 1 determine the molality of the solution
• 100 oC - 102.05C 2.05C determine the rTb
• 2.05 oC m 4.00 m
• 0.512 C molal boiling pt.
  constant for H2O
• Step 2 determine the grams per mole
• 72.0 g dextrose 1 kg H2O 180 g
• 0.100 kg H2O 4.00 mol mol

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One last thing Colloids

• Colloids are not true solutions, but special
  types of mixtures that behave like solutions.
• There are two parts, the dispersed phase and
  continuous phase.
• Dispersed phase has particles from 1 to 100 nm in
  size and remain dispersed by the random motion of
  the molecules (kinetic energy).
• Any particle larger than 100 nm will usually
  settle out over time.

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Mark Rosengarten videos

• Antifreeze, Electrolytes
• http//www.youtube.com/watch?vn0W7Y2Gwi2Efeature
  related
• BP elevation FP depression
• http//www.youtube.com/watch?vtjHaIDSzHsofeature
  related
• Molality
• http//www.youtube.com/watch?vWNrSexmBDXUfeature
  related

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Freezing Point Depression Lab

• Do the lab eat the ice cream
• Complete the problems on the back of lab sheet ?
• Hand out Ch. 19 Test Review
• HW time
• Ch. 19 Test next Class!!!

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Plan for the Day

• 30 min Review
• Turn in NBs
• Chapter 19 Test

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Test Addition

• On test
• 2 c. rate of diffusion (osmotic pressure)

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