- Vapor pressure is the pressure of a vapor above its liquid in a sealed container where vapor and liquid are in dynamic equilibrium. - According to Raoult's law, the vapor pressure of a solution is lower than the vapor pressure of the pure solvent. This lowering of vapor pressure (ΔP) can be calculated using Raoult's law. - Freezing point depression (ΔTf) and boiling point elevation (ΔTb) are colligative properties that depend on the number of solute particles in solution. ΔTf and ΔTb increase with increasing molality of the solute.

1. Colligative Properties – Vapor Pressure XII Grade
The Lowering Vapor Pressure

2. Vapor Pressure
Vapor Pressure is the pressure of a vapor
above its liquid in a sealed container, where
the vapor and liquid are in a dynamic
equilibrium

3. Diagram

4. Raoult’s Law
The value of ∆𝑃 can be calculated from the
vapor pressure of solution explained by
Francois Raoult and known as Raoult’s
Law.
𝑷 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 = 𝑷° 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝑿 𝒔𝒐𝒍𝒗𝒆𝒏𝒕
∆𝑷 = 𝑷° 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝑿 𝒔𝒐𝒍𝒖𝒕𝒆
∆𝑷 = 𝑷° 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 − 𝑷° 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏

5. Example
5 g of wax C22H46 is dissolved in 50 g of carbon
tetrachloride (CCl4) at a temperature of 23 ℃
work out the vapor pressure of the solution
and the lowering of the vapor pressure of the
solution. (P° CCl4 at 23 ℃ = 0.131 atm; Mr
C22H46 = 310; and Mr CCl4 = 154)

6. Moles of CCl4 =
𝑔𝑟
𝑀𝑟
=
50𝑔
154 𝑔/𝑚𝑜𝑙
= 0.325 mol
Moles of C22H46 =
𝑔𝑟
𝑀𝑟
=
5𝑔
311 𝑔/𝑚𝑜𝑙
= 0.016
mol
∆𝑃 = 𝑃° 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑋𝑠𝑜𝑙𝑢𝑡𝑒
=0.131 atm ×
0.016 𝑚𝑜𝑙
0.016 𝑚𝑜𝑙+0.325
=0.006 atm

7. Colligative Properties Grade XII Semester 1
P – T DIAGRAM OF (∆TF) AND (∆𝑇𝑏)

8. Freezing Point Depression

9. • is the decrease of the freezing point of a
solution relative to the freezing point of
its pure solvent
• Freezing point depression (∆Tf)

10. ∆𝑇𝑓 = 𝑇𝑓 − 𝑇𝑓°
∆𝑇𝑓 = 𝐾𝑓 . 𝑚
• ∆Tf : freezing Point depression (℃)
• Tf° : freezing temperature of the pure solvent
(℃)
• Tf : freezing temperature of the solution (℃)
• Kf : molal freezing point depression constant
(℃ kg/mol)
• m : molality of the solute (mol/L)

11. Boiling Point Elevation
• What is boiling point elevation?
• How does the influence of solute to
boiling point elevation explain?

12. In the activity above, we can observe that adding
NaCl into water as a pure solvent causes the
solution to possess a higher boiling point
Boiling Point Elevation

13. Definition
Boiling point elevation (∆𝑇𝑏) is the increase
of the boiling point of a solution relative to
the boiling point of its pure solvent

14. ∆𝑇𝑏 = 𝑇𝑏 − 𝑇𝑏°
∆𝑇𝑏 = 𝐾𝑏 . 𝑚
• ∆Tb : boiling Point elevation (℃)
• Tb° : boiling temperature of the pure solvent (℃)
• Tb : boiling temperature of the solution (℃)
• Kb : molal boiling point elevation constant (℃ kg/mol)
• m : molality of the solute (mol/L)

15. Example
8 g of glucose (C6H12O6) is dissolved in 60 g
of water. Calculate the boiling point
elevation of the solution (Kb water = 0.512
°C kg/mol)

16. ∆𝑇𝑏 = 𝐾𝑏 𝑚
∆𝑇𝑏 = 𝐾𝑏
𝑔
𝑚𝑟
𝑥
1000
𝑝
∆𝑇𝑏 = 0.512 ℃
𝑘𝑔
𝑚𝑜𝑙
8 𝑔
180
𝑔
𝑚𝑜𝑙
𝑥
1000𝑔
𝑘𝑔
60𝑔
∆𝑇𝑏 = 0.38 ℃

18. Have you ever
heard about
“osmosis”?
Osmosis is a solvent transfer from a low concentration to a
high concentration area through a semipermeable membrane.

20. Where
𝜋 : Osmotic Pressure (atm)
M : Molarity of solution (mol/L)
R : Gas constant (0.0821 L atm/mol K)
T : Temperature (K)
𝜋 = 𝑀 𝑅 𝑇

21. Work out the osmotic pressure of a solution
containing 36 grams of glucose (Mr = 180) in
1 L of solution at 27 ° C (R = 0.082 L.atm.mol-
1K-1).
Example

22. Given: mass of glucose = 36 gram
Mr = 180
V = 1 L
R = 0,0821 L atm/mol K
T = 27 °C = 27 + 273K = 300 K
Asked : 𝜋?
Mol of glucose =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒
𝑀𝑟
=
36 𝑔
180 𝑔/𝑚𝑜𝑙
=0.2 mol
𝑀 =
𝑛
𝑉
=
0.2 𝑚𝑜𝑙
1𝐿
= 0.2
𝑚𝑜𝑙
𝐿
𝜋 = 𝑀 𝑅 𝑇
𝜋 = 0.2
mol
L
0.0821
L. atm
mol. K
300 𝐾
𝜋 = 4.926 𝑎𝑡𝑚

23. Aplication

24. a) Hypotonic Solution is one of two
solutions that has a lower
concentration of a solute
b) Hypertonic Solution is one of two
solutions that has a haigher
concentration of a solute
c) Isotonic Solution are two solutions
that have the same concentration of
a solute

25. Colligative Properties – Vapor Pressure XII Grade

26. • If electrolytes solution disolve in solvent. So
particles of electrolytes solution have a
dissociation/ionization process.
• So for the same concentration, electrolyte
solutions have greater colligative properties
than nonelectrolyte solution.
• For example, NaCl breaks into 2 particles
NaCl Na+ + Cl-
Colligative Properties of Electrolyte
Solution

27. So, the value of the colligative properties of
the electrolyte solutions is influenced by the
Van’t Hoff, that can be determined as
follows:
𝑖 = {1 + 𝛼 𝑛 − 1 }
Where: 𝑖 : Van’t Hoff factor
n : the number of ions
𝛼 : degree of ionization/dissociation

28. Colligative
Properties
Non -
Electrolyte
Electrolyte
Lowering of
Vapor Pressure
∆𝑃 = 𝑃°𝑋 𝐵 ∆𝑃 = 𝑃°𝑋 𝐵 𝑖
Boiling Point
Elevation
∆𝑇𝑏 = 𝐾𝑏 𝑚 ∆𝑇𝑏 = 𝐾𝑏 𝑚 𝑖
Freezing Point
Decrease
∆𝑇𝑓 = 𝐾𝑓 𝑚 ∆𝑇𝑓 = 𝐾𝑓 𝑚 𝑖
Osmotic
Pressure
𝜋 = 𝑀 𝑅 𝑇 𝜋 = 𝑀 𝑅 𝑇 𝑖

29. Calculate the van’t Hoff factor of a 0.1 m
solution of HF in the freezing point of the
solution is -0.197°C. (Kf water = 1.86°C
kg/mol)
Example

30. Given :
Molal of HF: 0.1 m
Freezing point of HF: -0.197 ℃
Asked:
𝑖 of HF solution?
∆𝑇𝑓 = 𝑇𝑓° − 𝑇𝑓
∆𝑇𝑓 = 0℃ − −0.197 ℃ = 0.197℃
∆𝑇𝑓 = 𝐾𝑓 𝑚 𝑖
𝑖 =
∆𝑇𝑓
𝐾𝑓 𝑚
=
0.197℃
1.86℃
𝑘𝑔
𝑚𝑜𝑙
(0.1 𝑚𝑜𝑙/𝑘𝑔)
= 1.059