Freezing Point Depression And Boiling Point

COLLIGATIVE
PROPERTIES:
FREEZING POINT
DEPRESSION AND
BOILING POINT
ELEVATION
DAY 1 – 04 FEBRUARY 2015

Colligative Properties
Depends on the NUMBER of solute, not on the nature of solute particles
Freezing Point Depression
Boiling Point Elevation
Vapor Pressure Lowering
Osmotic Pressure

Electrolyte and Nonelectrolytes
Electrolytes
•Separates in water forming a solution that conducts electric current •IONIC COMPOUNDS

Non- electrolytes
• does not allow the flow of an electric current
• COVALENT COMPOUNDS

Freezing
Point
Depressio n Freezing Point
Depression
Adding a solute to a solvent decreases freezing point of the solvent
Amount of solute Freezing Point of solvent Tf° (pure solvent) > Tf (solution) Freezing point of pure solvent – 0o
Tf = Tf° - Tf

Freezing Point
Depression
Formula (non-electrolytes):

Temperature

Freezing Point Depression constant
(1.860 C /m)
Molality (mol/kg)

Freezing Point
Depression
Formula (electrolytes):

ΔTf = i Kf m

ΔTf i freezing point temperature

van ‘t Hoff factor (sum of subscripts) Kf m Freezing Point Depression constant (1.860 C /m)
Molality (mol/kg)

Freezing Point
Depression
EXAMPLES

CaCl2

NaCl
Na
1

Cl
+

i =2

Ca
1

ΔTf
Cl2

+

1

i =3

2

Kf

i

Na PO

3
4
freezing point temperature

Na

(PO )

van ‘t Hoff
3 factor (sum of4 subscripts) 3

+

1

Freezing Point Depression constant (1.860 C /m)

m

i =4

Molality (mol/kg)

PRACTICE
What is the new freezing point of 200 g of water (Kf = -1.86 oC) if195 g of sucrose (C12H22O11) are added to it?

ANSWER
∆Tf = -5.301°C

PRACTICE
What is the molality of barium sulfate
(Ba2SO4) with a freezing a point of
1.12°C?

ANSWER
Molality = 0.20 m

PRACTICE
A solution of 3.39 g of an unknown compound in
10.00 g of water has a freezing point of 7.31°C.
The solution does not conduct electricity. What is the molar mass of the compound?

ANSWER
Molar mass = 86.26 g/mol

Boiling
Point
Elevation

Boiling Point Elevation
Temperature difference between a solution’s