Physical Properties Physical Properties of Solutionsof Solutions

Chapter 13Chapter 13

Colligative Properties of Solutions

Colligative properties - properties that depend only on the number of solute particles in solution and not on the nature of the solute particles

Vapor-Pressure Lowering P1 = X1 P 1o

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

Vapor-Pressure Lowering

Raoult’s law

If the solution contains only one solute:

X1 = 1 – X2

P 1o- P1 = ΔP = X2 P 1o

P 1o = vapor pressure of pure solvent

X1 = mole fraction of the solvent

X2 = mole fraction of the solute

P1 = X1 P 1o

Fig 13.20

Boiling-Point Elevation

ΔTb = Tb – T bo

T b ≡ boiling point of pure solvent

o

T b ≡ boiling point of solution

ΔTb = Kb m

Kb ≡ molal boiling-point elevation constant (°C/m)

Fig 13.22

Boiling-Point Elevation

ΔTb = Kb m

Tb is added to the normal boiling point of the solvent

Freezing-Point Depression

ΔTf = Kf m

Tf is subtracted from the normal freezing point of the solvent

Osmotic Pressure ()

Osmosis - selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.

Semipermeable membrane - allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure () - pressure required to stop osmosis.

dilutemore

concentrated

Osmotic Pressure ()

= MRT

M is the molarity of the solution

R is the gas constant = 0.08206 (L atm)/(mol K)

T is the temperature (in K)

Chemistry In Action: RO Water

A cell in an:

isotonicsolution

Fig 13.24

hypotonicsolution

hypertonicsolution

Red bloodcell

Colligative Properties of Electrolyte Solutions0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions

Colligative properties ≡ properties that depend only on the number of solute particles and not on their nature

0.1 m NaCl solution 0.2 m ions in solution

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

nonelectrolytesNaCl

CaCl2

i should be

12

3

Boiling-Point Elevation ΔTb = i Kb m

Freezing-Point Depression ΔTf = i Kf m

Osmotic Pressure π = iMRT

Colligative Properties of Electrolyte Solutions

Sample Exercise 13.13 Molar Mass from Osmotic Pressure

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 at 25 °C was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.

Solutions

Colloids

Table 13.6 Types of Colloids

Colloid versus solution

• Collodial particles are much larger than solute molecules

• Collodial suspension is not as homogeneous as a solution

Solutions

Tyndall EffectTyndall Effect

The scatter of rays of light by particles in a colloidal suspension

Light is NOT scattered by particles in a solution

Fig 13.26

Solutions

Colloids in Biological SystemsColloids in Biological Systems

E.g.: Molecules in enzymes and antibodies

Have a polar, hydrophilic (water-loving) end and

A non-polar, hydrophobic (water-hating) end

Fig 13.28

Fig 13.29 Hydrophobic colloids

Solutions

Colloids in Biological Systems

Sodium stearate

Solutions

Colloids in Biological Systems

Fig 13.31 Stabilization of an emulsion of oil in water bystearate ions

Exam #2

Chapters 11 and 13

• Thirty multiple choiceThirty multiple choice (60%) (60%)

• Terminology, concepts, properties, etc.Terminology, concepts, properties, etc.

• CalculationsCalculations (40%) (40%)

• Heating/cooling and phase transitionsHeating/cooling and phase transitions

• Calc. concentration (M, m, or %(w/w) from given dataCalc. concentration (M, m, or %(w/w) from given data

• Det. molar mass from colligative property dataDet. molar mass from colligative property data

How much heat to convert 180. g H2O from -10 °C to 150. °C?

sp. ht. (ice) = 2.03 J/(g °C) ΔHfus = 6.01 kJ/molsp. ht. (water) = 4.18 J/(g °C)sp. ht. (steam) = 1.99 J/(g °C) ΔHvap = 40.8 kJ/mol

Ans:

561 kJq = mcΔT

q = mcΔT

q = mcΔT

ΔHfus

ΔHvap

Sample Exercise 13.7 Calculation of Molality and Molarity Using the Density of a Solution

A solution with a density of 0.876 g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene. Calculate the molarity of the solution.

A 202-mL benzene solution containing 2.47 g of an organicpolymer has an osmotic pressure of 8.83 mm Hg at 21.0 °C.Calculate the molar mass of the polymer.

π = MRT

M = π /RT =

K 294Kmol

atmL0.0821

Hg mm 760atm 1.00

Hg) mm (8.83

(4.81 x 10-4 mol/L) (0.202 L) = 9.72 x 10-5 mol

9.72 x 10-5 mol

2.47 g= 2.54 x 104 g/mol

= 4.81 x 10-4 M