11.5 Colligative Properties of Solutions
Comparing the Properties of a Pure Solvent with Those of a Solution The vapor of a solution is lower. The b...
Comparing the Properties of a Pure Solvent with Those of a Solution The vapor of a solution is lower. The boiling point of a solution is higher. The freezing point of a solution is lower.
Phase Diagrams of Solvent and Solution Phase diagrams of solvent and solution.
Boiling Point Elevation The boiling point of a solution is higher than the boiling point of the pure solvent (for a nonvolatile solute). The difference between the boiling point of the solution and the boiling point of the pure solvent is directly proportional to the molal concentration of solute particles.
∆Tb = Kb • m The proportionality contant is called the Boiling Point Elevation Constant Kb.
The units of Kb are (ºC•kg)/mol or ºC/m
Freezing Point Depression The freezing point of a solution is lower than the freezing point of the pure solvent (for a nonvolatile solute). The difference between the freezing point of the solution and the freezeng point of the pure solvent is directly proportional to the molal concentration of solute particles.
∆Tb = Kf • m The proportionality contant is called the Freezing Point Depression Constant Kf.
The units of Kb are (ºC•kg)/mol or ºC/m
Molal Boiling Point-Elevation Constants (Kb) and Molal Freezing-PointDepression Constants (Kf) for Some Common Substances
Solute Concentration: Molality Changes in boiling point/freezing point of solutions depends on molality:
Preferred concentration unit for properties involving temperature changes because it is independent of temperature For typical solutions: molality > molarity
Calculating Molality from mass of solute and volume of solvent 1/M= molar mass
gsolute X Lsolvent X
molsolute gsolute kgsolvent Lsolvent
molsolute kgsolvent
= molality
density of solvent
Calculating Molality without these
molality =
molsolute kgsolvent
from mass of solute and volume of solution from molarity of a solution (Msolution = molsolute /Lsolution)
You need the density of the solution !!
Calculating Molality from mass of solute and volume of solution 1/M= molar mass-1
gsolute X
molsolute gsolute
Lsolution X
kgsolution Lsolution
molsolute kgsolvent
= molality
= kgsolution
gsolute (kgsolution)-(kgsolute) = (kgsolvent)
density of solution
Calculating Molality from molarity of a solution (Msolution = molsolute /Lsolution) assume 1.00 L
molsolute Lsolution Lsolution X
X
Lsolution
kgsolution Lsolution
molarity (M)
molsolute kgsolvent = kgsolution
= molality
assume 1.00 L
density of solution molarity (M)
assume 1.00 L
(kgsolution)-(kgsolute) = (kgsolvent)
Practice: Molality #1 2) Calculate the molality of a solution containing 28.4 g of glucose (C6H12O6, M=180.2) in 0.355 kg of water. gsolute X
Lsolvent X
28.4gglucose X
molsolute gsolute kgsolvent Lsolvent
molglucose 180.2 gglucose
molsolute kgsolvent
= molality
= 0.158 moleglucose
0.158 molsolute 0.355 kgsolvent
= 0.455 m
Practice: Molality #2 3) A sample of KNO3 solution contains 72.5 g of KNO3 in 2.00 L of solution. If the density of the solution is 1.005 g/mL, what is the molality of KNO3 in the solution? molsolute = molality kgsolvent gsolute
Practice: Molality #3 4) Seawater contains 0.558 M NaCl at the surface at 25oC. If the density of sea water is 1.022 g/mL, what is the molality of NaCl in sea water?
Osmosis Osmosis is the flow of solvent through a semi-permeable membrane from solution of low concentration to solution of high concentration . The amount of pressure needed to keep osmotic flor from taking place is called the osmotic pressure . Osmotic pressure, π , is directly proportional to the molarity of the solute particles.
π = MRT
Osmosis
Osmosis
Osmosis
Osmosis — Van’t Hoff Factor
Van’t Hoff Factors at 0.05 m in Aqueous Solution
Jacobus Henricus van 't Hoff
Van’t Hoff Factor and Concentration
(a) The experimentally measured freezing point of a 0.010 m solution of NaCl is the same as the theoretical value. This agreement means that the solution behaves ideally and little or no ion pairing takes place.
Van’t Hoff Factor and Concentration
(b) The experimentally measured freezing point of a 0.10 m solution is about 0.04°C higher than the theoretical value because some of the Na and Cl– ions form ion pairs. The formation of ion pairs causes the concentration of solute particles to be less than the theoretical number.
Van’t Hoff Factors
FIGURE 11.22 Theoretical and experimentally measured values for the van ’t Hoff factors for 0.1 m solutions of several electrolytes and the nonelectrolyte ethanol. The higher the charge on the ions, the greater the difference between theoretical and experimentally measured values.
11.6 Molar Mass from Colligative Properties
Let’s rearrange the Colligativee Property Relationships !!!
ΔTf = (m)(Kf)
M = g/mol
ΔTf =
(M)=
m = mol/kg solvent
mol = g/M
g/(M) kg solvent
g (kg solvent)(ΔTf )
(Kf)
(Kf)
mol = g/(g/mol)
ΔTf =
(M)=
mass/(M) kg solvent
(Kf)
mass (kg solvent)(ΔTf )
(Kf)
M = g/mol
mol = g/M
mol = g/(g/mol)
π = MRT = (mol/L)(R)(T)
( π= (
g/M L
)(R)(T) M=(
π= (
mass (V)(π)
L atm mol K
)
mass/M
V
)(R)(T)
)(R)(T)
Molal Boiling Point-Elevation Constants (Kb) and Molal Freezing-PointDepression Constants (Kf) for Some Common Substances
Practice: Molar Mass from Colligative Properties 5) The freezing point of a solution prepared by dissolving 1.50 × 102 mg of caffeine in 10.0 g of camphor is 3.07 ºC lower than that of pure camphor (Kf = 39.7oC/m). What is the molar mass of caffeine? mass (g) of compound
(M)=
(M)=
g (kg solvent)(ΔTf )
(0.150 g)
(Kf) (39.7 ºC/m)
(0.010 kg)(3.07 ºC)
M = 194 g/mol
Example-Molar Mass from Osmotic Pressure
6) An aqueous solution contains 0.97g/L of an organic compound. The osmotic pressure of this solution is 62.9 torr, at 25°C. What is the molar mass of this compound?
