Chapter 19: Molality and Colligative Properties

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)

HW Ch. 19 Blue Book: #1-17, 19 (on problems that are a-z, please

do a and b only)

Plan of the DayPlan 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

MolalityMolality

• 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=WNrSexmBDXU&feature=related

• 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).

Example #1Example #1If 60.0 g of NaOH are dissolved in 1500g of water, what

is the concentration of this solution using Molality?1. Analyze the problem – what info are we given?

mass of solute, 60.0 g NaOHmass of solvent, 1500 g H2O

2. Solve for the unknownCouple of ways to do this: 1. “Railroad-track” the whole thing all at onceOR 2. Break it up…

calculate number of moles of soluteConvert mass of water from g kgPlug in and solve for molality

=1.00m

Example #2Example #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 ethanol750g H2O 2.00 mol C2H5OH 46.1g C2H5OH

1000 g H2O 1 mol C2H5OH

Answer: 69.2g C2H5OH

Example #3Example #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

Example # 4Example # 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

X 100 = 27.0 %

Example # 5Example # 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 Avogadro’s # /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

Colligative PropertiesColligative Properties

• What are they?What are they?– The word Colligative means “depending The word Colligative means “depending

on the collection”on the collection”– Change the Change the physical propertiesphysical properties of the of the

solvent.solvent.– Depends on the number of particles of Depends on the number of particles of

the solute NOT which solute is used!the solute NOT which solute is used!

Colligative PropertiesColligative Properties

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

point!• Osmotic pressure• Why?

Colligative PropertiesColligative 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…

Colligative PropertiesColligative Properties

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

Vapor Pressure LoweringVapor 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

• #1 – solvent has a large surface area to evaporate from

• #2 – mixed with solute = fewer solvent particles at surface

Beaker #1 Beaker #2

Pure Solvent Solution

Which one has lower VP?

Boiling Point ElevationBoiling 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.

Boiling Point ElevationBoiling Point Elevation

Freezing Point DepressionFreezing 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.

Freezing Point DepressionFreezing Point Depression

0oC

100oC

___ =

Pure Solvent

---- = Solution

Osmotic PressureOsmotic 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.

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.

Colligative Properties (now the math)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.

Colligative PropertiesColligative PropertiesThese 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)

Colligative PropertiesColligative Properties

Ex. 3Calculate 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

water0 oC – 0.620 = - 0.620 oC

Calculating Molecular Mass Ex. 4Calculating Molecular Mass Ex. 4When 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 - 102.05˚C = 2.05˚C determine the Tb

2.05 oC m = 4.00 m 0.512 ˚C molal boiling pt. constant for H2O

Step 2: determine the grams per mole72.0 g dextrose 1 kg H2O = 180 g

0.100 kg H2O 4.00 mol mol

One last thing: ColloidsOne 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.

Mark Rosengarten videosMark Rosengarten videos

• Antifreeze, Electrolytes…:– http://www.youtube.com/watch?v=n0W

7Y2Gwi2E&feature=related• BP elevation & FP depression

– http://www.youtube.com/watch?v=tjHaIDSzHso&feature=related

• Molality– http://www.youtube.com/watch?v=WNrS

exmBDXU&feature=related

Freezing Point Depression Freezing Point Depression LabLab

Freezing Point Depression Freezing Point Depression LabLab

• Do the lab… eat the ice creamDo the lab… eat the ice cream–Complete the problems on Complete the problems on the back of lab sheet the back of lab sheet

• Hand out Ch. 19 Test ReviewHand out Ch. 19 Test Review• HW timeHW time• Ch. 19 Test next Class!!!Ch. 19 Test next Class!!!

Plan for the DayPlan for the DayPlan for the DayPlan for the Day

•~30 min Review~30 min Review•Turn in NB’sTurn in NB’s•Chapter 19 TestChapter 19 Test

Test AdditionTest AdditionTest AdditionTest Addition

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