2. Colligative Properties

3. Colligative Properties
The properties of solvent that depend ONLY on the amount of solute added to it
NOT depend on the identity of solute.

4. Colligative Properties
VAPOR PRESSURE
BOILING POINT
FREEZING POINT
OSMOTIC PRESSURE

5. Understanding Colligative Properties
To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

6. 1. Vapor Pressure

7. Understanding Colligative Properties
To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution.

8. Vapor Pressure Lowering
Figure 14.12

9. Understanding Colligative Properties
VP of H2O over a solution depends on the number of H2O molecules per solute molecule.
Psolvent proportional to Xsolvent
Psolvent = Xsolvent • Posolvent
VP of solvent over solution = (Mol frac solvent)•(VP pure solvent)
RAOULT’S LAW

10. Raoult’s Law PA = XA • PoA
An ideal solution obeys Raoult’s law.
PA = XA • PoA
mole fraction A (XA): always less than 1, PA is always less than PoA.
The vapor pressure of solvent over a solution is always LOWERED!

11. Raoult’s Law Pwater = 29.7 mm Hg Mole fraction Xg = 0.0672
A solution contains 62.1 g of glycol (1 mol) in 250. g of water. What is the vapor pressure of water over the solution at 30 oC?
(The VP of pure H2O is 31.8 torr)
Solution
Mole fraction Xg =
Know Xg + XH2O = 1.000
XH2O = = 0.933
PH2O = XH2O • PoH2O = (0.933)(31.8 torr)
Pwater = 29.7 mm Hg

12. Changes in Freezing and Boiling Points of Solvent
See Figure 14.12

13. Elevation of Boiling Point
∆TBP = KBP•m
(where KBP is characteristic of solvent)
∆TBP

15. Change in Boiling Point
What is the BP of the solution with 62.1 g of glycol (1.00 mol) dissolved in 250. g of water?
KBP = oC/molal for water
Solution
1. Calculate solution molality = 4.00 m
2. ∆TBP = KBP • m
∆TBP = oC/molal (4.00 molal)
∆TBP = oC
new BP = oC = oC

16. Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor
= # of particles produced / formula unit.
Compound Value of i
glycol 1
NaCl 2
CaCl2 3

17. Boiling Point D6C32 The addition of solute raises the boiling point.
The larger the “m” the greater the increase in temperature: NOT matter what you add

18. WHY?

19. 3. Freezing Point:
The addition of
solute lowers the
freezing point.
Salt is added to streets
ice cream freezes at different temp (depends
on flavor, more you add the lower the temp)

20. Change in Freezing Point
Ethylene glycol/water
solution
Pure water
The freezing point of a solution is LOWER than that of the pure solvent.
FP depression = ∆TFP = KFP•m

22. Boiling Point Elevation and Freezing Point Depression
∆T = K•m•i
i = van’t Hoff factor
= # of particles produced / formula unit.
Compound Value of i
glycol 1
NaCl 2
CaCl2 3

23. Freezing Point Depression
Calculate the FP of a 4.00 molal glycol/water solution.
KFP = oC/molal (Table 14.4)
Solution
∆TFP = KFP • m
∆TFP = (-1.86 oC/molal) (4.00 m)
∆TFP = oC
FP = oC = oC
∆TBP = ˚C for this solution.

24. Freezing Point Depression
How much NaCl (g) must be dissolved in 1.00 kg of water to lower FP to oC?.
Calc. required molality
∆TFP = KFP • m
oC = (-1.86 oC/molal) • m
Conc = molal ions

25. Freezing Point Depression
Conc req’d = 5.38 molal ions
We need 5.38 mol dissolved particles / kg of solvent.
NaCl(aq) --> Na+(aq) + Cl-(aq)
need
5.38 mol / 2 = mol NaCl / kg
2.69 mol NaCl / kg ---> 157 g NaCl / kg

26. Boiling Point
Vapor Pressure
Freezing Point

27. 4. Osmosis
The movement of solvent through a membrane from an area of higher to lower solvent concentration.
Semipermeable Membrane- barrier allowing solvent through, but block larger solute particles.

28. Osmosis at the Particulate Level
Figure 14.15

29. 4. Osmosis
Osmotic Pressure- Pressure developed across a membrane that allows the diffusion of solvent particles.
Chemistry behind your kidneys and making pickles

30. Osmosis
Driving force is entropy
The semi-permeable membrane allows only the movement of solvent molecules.
Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute.

31. Process of Osmosis

32. Osmosic Pressure, ∏ (c is conc. in mol/L)
Equilibrium is reached when pressure — the OSMOTIC PRESSURE, ∏ — produced by extra solution counterbalances pressure of solvent molecules moving thru the membrane.
∏ = cRT
(c is conc. in mol/L)
Osmotic pressure

33. Osmosis Calculating a Molar Mass
Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. ∏ measured to be 10.0 torr at 25 ˚C. Calc. molar mass of hemoglobin.
Solution
(a) Calc. ∏ in atmospheres
∏ = torr • (1 atm / 760 mmHg)
= atm
(b) Calc. concentration

34. Osmosis Calculating a Molar Mass
Solution
(b) Use ∏ = cRT
c = 5.39 x 10-4 mol/L
(c) Calc. molar mass
Molar mass = 35.0 g / 5.39 x 10-4 mol/L
Molar mass = 65,100 g/mol

35. Osmosis
Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC — they have the same concentration.
Osmotic pressure in living systems: FIGURE 14.16

36. Osmosis and Living Cells

37. Reverse Osmosis Water Desalination

38. Colligative Properties
On adding a solute to a solvent, the props. of the solvent are modified.
Vapor pressure decreases
Melting point decreases
Boiling point increases
Osmosis is possible (osmotic pressure)
They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.