solution ppt

26
Molality and Mole Fraction In Chapter 3 we introduced two important concentration units. 1. % by mass of solute % 100 solution of mass solute of mass = w/w % 2. Molarity soluti of Liters solute of moles = M

Upload: manoj-kumar-sharma

Post on 22-Jul-2016

14 views

Category:

Documents


2 download

TRANSCRIPT

Page 1: Solution PPT

Molality and Mole Fraction

• In Chapter 3 we introduced two important concentration units.

1. % by mass of solute

%100solution of masssolute of mass = w/w%

2. Molarity

solution of Literssolute of moles = M

Page 2: Solution PPT

Molality and Mole Fraction

m moles of solutekg of solvent

in dilute aqueous solutions molarity and molality are nearly equal

• Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

Page 3: Solution PPT

Molality and Mole Fraction• Calculate the molarity and the molality of an aqueous solution that is 10.0%

glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g

Page 4: Solution PPT

Molality and Mole Fraction• Calculate the molality of a solution that contains 7.25 g of benzoic acid

C6H5COOH, in 2.00 x 102 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g

Page 5: Solution PPT

Molality and Mole Fraction• Mole fraction is the number of moles of one component divided by

the moles of all the components of the solution– Mole fraction is literally a fraction using moles of one

component as the numerator and moles of all the components as the denominator.

• In a two component solution, the mole fraction of one component, A, has the symbol XA.

B of moles ofnumber +A of moles ofnumber A of moles ofnumber

AX

Page 6: Solution PPT

Molality and Mole Fraction

• The mole fraction of component B - XB

1.00. equalmust fractions mole theall of sum The1 that Note

B of moles ofnumber +A of moles ofnumber B of moles ofnumber

A

B

B

XX

X

Page 7: Solution PPT

Molality and Mole Fraction• What are the mole fractions of glucose and water in a

10.0% glucose solution?

Page 8: Solution PPT

Colligative Properties of Solutions• Colligative properties are properties of solutions that depend solely

on the number of particles dissolved in the solution.– Colligative properties do not depend on the kinds of particles

dissolved.• Colligative properties are a physical property of solutions.

There are four common types of colligative properties:1. Vapor pressure lowering2. Freezing point depression3. Boiling point elevation4. Osmotic pressure

Vapor pressure lowering is the key to all four of the colligative properties.

Page 9: Solution PPT

Lowering of Vapor Pressure and Raoult’s Law• Addition of a nonvolatile solute to a solution lowers

the vapor pressure of the solution.– The effect is simply due to fewer solvent molecules at

the solution’s surface.– The solute molecules occupy some of the spaces that

would normally be occupied by solvent.• Raoult’s Law models this effect in ideal solutions.

Page 10: Solution PPT

Lowering of Vapor Pressure and Raoult’s Law

• This graph shows how the solution’s vapor pressure is changed by the mole fraction of the solute, which is Raoult’s law.

Page 11: Solution PPT

Fractional Distillation• Distillation is a technique used to separate solutions that have two or more volatile components with differing boiling points.

• A simple distillation has a single distilling column.– Simple distillations give reasonable

separations.• A fractional distillation gives increased

separations because of the increased surface area.– Commonly, glass beads or steel wool are

inserted into the distilling column.

Page 12: Solution PPT

Boiling Point Elevation• Addition of a nonvolatile solute to a solution raises the

boiling point of the solution above that of the pure solvent.– This effect is because the solution’s vapor pressure is

lowered as described by Raoult’s law.– The solution’s temperature must be raised to make the

solution’s vapor pressure equal to the atmospheric pressure.

• The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.

Page 13: Solution PPT

Boiling Point Elevation

solvent for the constantelevation point boiling molal K

solution ofion concentrat molal elevationpoint boiling T :where

KT

b

b

bb

m

m• Boiling point elevation relationship is:

Page 14: Solution PPT

Freezing Point DepressionRelationship for freezing point depression

is:

T Kwhere: T freezing point depression of solvent

molal concentration of soltuionK freezing point depression constant for solvent

f f

f

f

m

m

Page 15: Solution PPT

Freezing Point Depression• Notice the similarity of the two relationships

for freezing point depression and boiling point elevation.

• Fundamentally, freezing point depression and boiling point elevation are the same phenomenon.– The only differences are the size of the effect which is

reflected in the sizes of the constants, Kf & Kb.• This is easily seen on a phase diagram for a solution.

mm bbff K T vs.KT

Page 16: Solution PPT

Freezing Point Depression

Page 17: Solution PPT

Boiling Point Elevation• What is the normal boiling point of a 2.50 m glucose,

C6H12O6, solution?

Page 18: Solution PPT

Freezing Point Depression• Calculate the freezing point of a solution that contains 8.50 g

of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6.

Page 19: Solution PPT

Determination of Molecular Weight by Freezing Point Depression

• The size of the freezing point depression depends on two things:

1. The size of the Kf for a given solvent, which are well known.

2. And the molal concentration of the solution which depends on the number of moles of solute and the kg of solvent.

• If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.

Page 20: Solution PPT

Determination of Molecular Weight by Freezing Point Depression

• A 37.0 g sample of a new covalent compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound?

Page 21: Solution PPT

Colligative Properties and Dissociation of Electrolytes

• Electrolytes have larger effects on boiling point elevation and freezing point depression than nonelectrolytes.– This is because the number of particles released in solution

is greater for electrolytes• One mole of sugar dissolves in water to produce one mole of

aqueous sugar molecules.• One mole of NaCl dissolves in water to produce two moles

of aqueous ions:– 1 mole of Na+ and 1 mole of Cl- ions

Page 22: Solution PPT

Osmotic Pressure• Osmosis is the net flow of a solvent between

two solutions separated by a semipermeable membrane.

– The solvent passes from the lower concentration solution into the higher concentration solution.

• Examples of semipermeable membranes include:

1. cellophane and saran wrap2. skin3. cell membranes

Page 23: Solution PPT

Osmotic Pressure• Osmosis is a rate controlled phenomenon.

– The solvent is passing from the dilute solution into the concentrated solution at a faster rate than in opposite direction, i.e. establishing an equilibrium.

• The osmotic pressure is the pressure exerted by a column of the solvent in an osmosis experiment.

M

M

RTwhere: = osmotic pressure in atm

= molar concentration of solution

R = 0.0821 L atmmol K

T = absolute temperature

Page 24: Solution PPT

Osmotic Pressure

For very dilute aqueous solutions, molarity and molality are nearly equal. M m

mfor dilute aqueous solutions only

RT

Page 25: Solution PPT

Osmotic Pressure• Osmotic pressures can be very large.

– For example, a 1 M sugar solution has an osmotic pressure of 22.4 atm or 330 p.s.i.

• Since this is a large effect, the osmotic pressure measurements can be used to determine the molar masses of very large molecules such as:

1. Polymers2. Biomolecules like

• proteins• ribonucleotides

Page 26: Solution PPT

Osmotic Pressure• A 1.00 g sample of a biological material was dissolved in enough

water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.