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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
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
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
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