II III I IV. Colligative Properties of Solutions Ch. 16 – Mixtures & Solutions.
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Transcript of II III I IV. Colligative Properties of Solutions Ch. 16 – Mixtures & Solutions.
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II
III
I IV. Colligative Properties of Solutions
Ch. 16 – Mixtures & SolutionsCh. 16 – Mixtures & Solutions
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A. DefinitionA. Definition
Colligative PropertyColligative Property
• property that depends on the number
of solute particles, not their identity in
an ideal solution
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B. TypesB. Types
Freezing Point DepressionFreezing Point Depression (Tf)
• f.p. of a solution is lower than f.p. of the pure solvent
Boiling Point ElevationBoiling Point Elevation (Tb)
• b.p. of a solution is higher than b.p. of the pure solvent
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B. TypesB. Types
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Freezing Point Depression
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B. TypesB. Types
Solute particles weaken IMF in the solvent
Boiling Point Elevation
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B. TypesB. Types
Applications• salting icy roads• making ice cream• antifreeze
• cars (-64°C to 136°C)• fish & insects
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C. CalculationsC. Calculations
T: change in temperature (°C)
i: Van’t Hoff Factor (VHF), the number of particles into which the solute dissociates
m: molality (m)
K: constant based on the solvent (°C·kg/mol) or (°C/m)
T = i · m · K
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C. CalculationsC. Calculations
T
• Change in temperature• Not actual freezing point or boiling point• Change from FP or BP of pure solvent
• Freezing Point (FP) TF i is always subtracted from FP of pure
solvent
• Boiling Point (BP) TB i is always added to BP of pure
solvent
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C. CalculationsC. Calculations
ii – VHF – VHF
• Nonelectrolytes (covalent)• remain intact when dissolved • 1 particle
• Electrolytes (ionic)• dissociate into ions when dissolved• number of ions per formula unit• 2 or more particles
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C. CalculationsC. Calculations
ii – VHF – VHF
• Examples
• CaCl2
• Ethanol C2H5OH
• Al2(SO4)3
• Methane CH4
•i =
• 3
• 1
• 5
• 1
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C. CalculationsC. Calculations
KK – molal constant – molal constant
•KKFF – molal freezing point constant• Changes for every solvent • 1.86 °C·kg/mol (or °C/m) for water
•KKBB – molal boiling point constant• Changes for every solvent • 0.512 °C·kg/mol (or °C/m) for water
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C. Calculations: Recap!C. Calculations: Recap!
T : subtract from F.P. : subtract from F.P.
add to B.P.add to B.P. ii – VHF : covalent = 1 – VHF : covalent = 1
ionic ionic >> 2 2K : K : KKF waterF water = = 1.86 °C·kg/mol
KKB water B water = = 0.512 °C·kg/mol
T = i · m · K
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At what temperature will a solution that is composed of 0.730 moles of glucose in 225 g of water boil?
C. CalculationsC. Calculations
m = 3.24mKB = 0.512°C/m
TB = i · m · KB
WORK:
m = 0.730 mol ÷ 0.225 kg
GIVEN:b.p. = ?TB = ?
i = 1 TB = (1)(3.24m)(0.512°C/m)
TB = 1.66°C
b.p. = 100.00°C + 1.66°C
b.p. = 101.66°C
100 + Tb
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C. CalculationsC. Calculations
Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.
i = 2
m = 4.8m
KF = 1.86°C/m
TF = i · m · KF
WORK:
m = 0.48mol ÷ 0.100kg
GIVEN:
f.p. = ?
TF = ? TF = (2)(4.8m)(1.86°C/m)
TF = 18°C
f.p. = 0.00°C – 18°C
f.p. = -18°C
0 – TF
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D. Osmotic PressureD. Osmotic Pressure
Osmosis: The flow of solvent into a solution through a semipermeable membrane
Semipermeable Membrane: membrane that allows solvent to pass through but not solute
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D. Osmotic Pressure D. Osmotic Pressure
Net transfer of solvent molecules into thesolution until the hydrostatic pressureequalizes the solvent flowin both directions
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Because the liquid level for the solution is higher, there is greater hydrostatic pressure on the solution than on the pure solvent
Osmotic Pressure:
The excess hydrostatic pressure on the solution compared to the pure solvent
D. Osmotic PressureD. Osmotic Pressure
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Osmotic Pressure:
Minimum Pressurerequired to stop flowof solvent into the solution
D. Osmotic PressureD. Osmotic Pressure
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D. Osmotic PressureD. Osmotic Pressure
Osmosis at Equilibrium
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= i M R T
where:
π = osmotic pressure (atm)osmotic pressure (atm)
i = VHFVHF
M = Molarity (moles/L)
R = Gas Law Constant
T = Temperature (Kelvin)
E. Osmotic Pressure CalculationsE. Osmotic Pressure Calculations
0.08206 L atm/mol K
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E. Osmotic Pressure CalculationsE. Osmotic Pressure Calculations
Calculate the osmotic pressure (in torr) at 25oC of aqueous solution containing 1.0g/L of a protein with a molar mass of 9.0 x 104 g/mol.
i = 1
M = 1.11 x 10-5 M
R = 0.08206 L atm/mol K
T = 25oC = 298 K
WORK:
M = 1.0 g prot.
GIVEN:
= ?
1.11 x 10-5 M
= (1)(1.11x10-5)(.08206)(298)
= 2.714 x 10-4 atm
= 0.21 torr
1 mol prot. 1 L sol’n 9.0 x 104 g
=
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If the external pressure is larger than the osmotic pressure, reverse osmosis occurs
One application is desalination of seawater
F. Reverse OsmosisF. Reverse Osmosis
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F. Reverse Osmosis F. Reverse Osmosis
•Net flow of solventfrom the solution to the solvent