II III I IV. Colligative Properties of Solutions Ch. 16 – Mixtures & Solutions.

II

III

I IV. Colligative Properties of Solutions

Ch. 16 – Mixtures & SolutionsCh. 16 – Mixtures & Solutions

A. DefinitionA. Definition

Colligative PropertyColligative Property

• property that depends on the number

of solute particles, not their identity in

an ideal solution

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

B. TypesB. Types

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Freezing Point Depression

B. TypesB. Types

Solute particles weaken IMF in the solvent

Boiling Point Elevation

B. TypesB. Types

Applications• salting icy roads• making ice cream• antifreeze

• cars (-64°C to 136°C)• fish & insects

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

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

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

C. CalculationsC. Calculations

ii – VHF – VHF

• Examples

• CaCl2

• Ethanol C2H5OH

• Al2(SO4)3

• Methane CH4

•i =

• 3

• 1

• 5

• 1

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

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

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

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

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

D. Osmotic Pressure D. Osmotic Pressure

Net transfer of solvent molecules into thesolution until the hydrostatic pressureequalizes the solvent flowin both directions

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

Osmotic Pressure:

Minimum Pressurerequired to stop flowof solvent into the solution

D. Osmotic PressureD. Osmotic Pressure

D. Osmotic PressureD. Osmotic Pressure

Osmosis at Equilibrium

= 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

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

=

If the external pressure is larger than the osmotic pressure, reverse osmosis occurs

One application is desalination of seawater

F. Reverse OsmosisF. Reverse Osmosis

F. Reverse Osmosis F. Reverse Osmosis

•Net flow of solventfrom the solution to the solvent