Chemistry 101
Chapter 12
BOILING POINT ELEVATION

Normal Boiling Point (BP) of a liquid is the temperature at which the vapor pressure of the liquid
equals 1.00 atm
I. Effect of a Dissolved Solute on the BP of a Solution
What happens if a nonvolatile solute such as ethylene glycol (C2H6O2) is added to pure water ?
Vapor Pressure of Solution  Vapor Pressure of Pure Solvent (water)
P(A + B)  PA

It follows:
1. The normal BP of solution (A+B) is the temperature at which P(A+B) is equal to 1.00 atm
2. This requires a temperature higher than for the pure liquid (water)
3. BP(solution)  BP (pure solvent)  B.P. Elevation
Tb = Boiling Point Elevation
Tb

= BP (solution)  BP (solvent)
Boiling point elevation is a Colligative Property
 depends on the concentration of the Solution
 does not depend on the nature of the Solute
Tb = KbCm
Cm = Concentration in molality
Kb = Boiling Point Elevation Constant (units of C/m)
18
Chemistry 101
Chapter 12
PHASE DIAGRAM FOR WATER AND AN
AQUEOUS SOLUTION OF ETHYLENE GLYCOL
1.00 atm
Freezing Curve
for Pure Solvent
Freezing Curve
for Solution
Solid Vapor
Pressure
t (0C)
FP of
BP of
Solvent
FP of
Solution
Solvent
BP of
Solution
Tf
Tb
19
Chemistry 101
Chapter 12
Examples:
1. At what temperature would a a 5.00 molal solution of ethylene glycol boil? (Kb(water) = 0.512 oC/m)
Tb = (0.512 oC/m) (5.00 m) = 2.56 oC
BP(solution) = 100.00 oC + 2.56 oC = 102.56 oC
2. A solution was made of eugenol in diethyl ether (ether). If the solution was 0.575 m eugenol in
ether, what was the Boiling Point of the solution ?
BP(diethyl ether) = 34.6 oC
Kb(ether) = 2.02 oC/m)
Tb = (2.02 oC/m) (0.575 m) = 1.16 oC
BP(solution) = 34.6 oC + 1.16 oC = 35.8 oC
20
Chemistry 101
Chapter 12
FREEZOMG POINT DEPRESSION



From the phase diagram, the vapor pressure curve for the solid solution is unchanged.
As the temperature of a solution is lowered, the pure solvent freezes out of solution. (Sea ice is
almost pure water)
FP of the solution shifts toward a lower temperature (F. P. Depression)
Tf = Freezing Point Depression
Tf =

FP(solvent) 
FP (solution)
Freezing point depression is a Colligative Property
 depends on the concentration of the Solution
 does not depend on the nature of the Solute
Tf
= KfCm
Cm = Concentration in molality
Kf = Freezing Point Depression Constant (units of C/m)
Practical Applications
1.
Use of ethylene glycol as antifreeze in car radiators:
 lower the FP of coolant
 elevates the BP of coolant (prevents it from boiling away)
2.
NaCl poured over icy roads lowers the FP of ice below thetemperature of the surrounding air. As
a result, the salted ice melts.
3.
To obtain Molecular Weights of compounds
21
Chemistry 101
Chapter 12
Examples:
1. How many grams of ethylene glycol (C2H6O2) must be added to 37.8 g of water to give a Freezing
Point of  0.150 oC ? (Kf = 1.858 C/m)
Tf = FP (solvent) – FP(solution) = 0.000 oC – (  0.150 oC ) = 0.150 oC
Tf
= Kf Cm
Cm =
? g EG = 37.8 g water x
Tf
0.150 C
=
= 0.0807 m
Kf
1.858 C/m
0.0807 mol EG 62.08 g
x
= 189 g
1000 g water
1 mol
2. An aqueous solution of a molecular compound freezes at – 0.0860C. What is the molality of the
solution? (Kf = 1.858 0C/m)
Tf = 0.0000C  (0.0860C) = 0.0860C
Tf = Kf x m
Tf
m = 
Kf
0.0860C
=  = 4.6 x 102 m
1.858 0C/m
22
Chemistry 101
Chapter 12
OSMOSIS & OSMOTIC PRESSURE

Osmosis is the phenomenon of solvent flow through a semipermeable membrane to equalize the
solute”s concentrations on both sides of the membrane.

A semipermeable membrane allows solvent molecules to pass through but not solute molecules.

For example, in the diagram (a) below, water flows through the cellophane membrane to equalize
glucose’s concentration on both sides of the membrane.

Similarly, in diagram (b), water flows through the membrane on the funnel to equalize the
concentration of glucose in the funnel.

Osmotic pressure, π, is equal to the pressure that, when applied, just stops osmosis.

Osmotic pressure can be determined from the following relationship:
=MRT
where,
M = concentration in molarity (mol/L)
R= Universal gas constant (0.0821 Latm/molK)
T = absolute temperature (K)
(a)
(b)
23
Chemistry 101
Chapter 12
Examples:
1. Dextran, a polymer of glucose units, is produced by bacteria growing in sucrose solutions.
Solutions of dextran in water have been used as a blood plasma substitute. At 21°C, what is the
osmotic pressure (in mmHg) of a solution containing 1.50 g of dextran dissolved in 100.0 mL of
aqueous solution, if the average molecular mass of the dextran is 4.0  108 amu?
24
Chemistry 101
Chapter 12
COLLIGATIVE PROPERTIES OF IONIC SOLUTONS

When evaluating colligative properties of ionic solutions, the total concentration of the ions must be
considered.

The number of ions produced from each formula unit is designated i.

NaCl → Na+ + Cl–
i=2
MgCl2 → Mg2+ + 2 Cl–
i=3
The colligative properties can be calculated with the following modifications:
∆P = i XB PA
∆Tb = i m Kb
∆Tf = i m Kf
π = i MRT
Examples:
1.
Calculate the freezing point of 0.010 m solution of aluminum sulfate, Al (SO ) . ( K =1.86 ºC/m)
2
2. Which of the following solutions will have the lowest freezing point?
0.15 M NaCl
0.25 M C6H12O6
0.10 M Fe (NO3)3
25
4 3
f