Chapter 11 - Solutions and Colligative Properties
11.1
Molarity = moles solute/liter of solution
Molality = moles solute/kilograms of solvent
Mole Fraction = (chi) => mole fraction of a component of solution
Component “A”= χA= nA/ nA+ nB
(mixture of A and B in solution)
Mass Percent = (grams solute/ grams of solution) × 100
Sample Calculations
eg: You are given a solution of HCl which is a 38% concentration and has a
density of 1.19g/cm3. Calculate molality, molarity, and mole fraction.
Molarity= 38g HCl × 1.19g soln × 103 cm3 × 1 mol HCl = 12 mol/ L
(mol/L) 100g soln
cm3
1L
36.5g HCl
Molality= 30% (By Mass!)= 38g HCl × 103g H20 × 1 mol HCl = 17 mol/ kg
62g H20 1 kg H20 36.5 HCl
Mole Fraction= 38g HCl × 1 mol= 1.04 mol
36.5g
So χHCl =
1.04 mol
=0.23
1.04 mol + 3.44 mol
62g H20 × 1 mol = 3.44 mol
18.0g
(χ H20=0.77)
Student eg: Given a 40.0% solution of C2H602 with a known density of 1.05g/cm3,
determine the molarity, molality, and mole fraction.
Molarity= 40.0g C2H602 × 1.05g soln × 103 cm3 × 1 mol
= 6.7M
100.g soln
1 cm3
1L
62.1g C2H602
Molality= 40.0g C2H602 × 1 mol C2H602 × 103g H20= 10.7 mol/ kg
6.00g H20
62.1g C2H602 1 kg H20
Mole Fraction= 40.0g C2H602× 1 mol = .644 mol
62.1C2H602
60.0g H20 × 1 mol = 33.3 mol
18.0g
χ C2H602=
.644 mol
= .162 C2H602
.644 + 3.33 mol
11.2
Energies of Solution Formation- Recall: “Like dissolves like” –Why?
A. The three steps of solution formation- Examine p. 517
1st = ∆H1: Energy needed to overcome to interparticle forces within solute
2nd= ∆H2: “
“
“
“
“
“solvent
2nd= ∆H3: Solute to solvent interaction
∆Hsoln=∆H1+ ∆H2+ ∆H3
*For solutions that form…∆H1 and ∆H2 are endothermic while ∆H3 is exothermic
*Examine p. 517…Table 11.2
B. In general- DISORDER is favorable to order.
11.3
Solubility and Factors Affecting Solubility
A. Structure- Structure/ geometry determines polarity- “Like dissolves like”
B. Pressure-(Solubility of gas in a liquid) Pressure is directly proportional to
solubility.
Henry’s Law: C=kP
P=partial pressure of a gas above the soln
C=conc. of dissolved gas
K=constant
eg.Vacuum chamber and scuba diving (or Lake Nyos)
C. Temperature- ↑ Temp ↓ solid solution - (some exceptions!)
↑ Temp ↓ gas solubility – ex: thermal pollution of lakes
*Examine examples from books problems # 21, 26
11.4
Effect of Solute on Vapor Pressure
A. The pressure of a nonvolatile solute lowers the vapor pressure of a solvent by
decreasing the number of solvent molecules found at the surface where
vaporization (evaporation) occurs. (diagram)
B. Raoult’s Law
P soln = χsolvent P○solvent
Psoln = vapor pressure of solution; P○= vapor pressure of pure solvent
Raoult’s Law is in the form of a linear equation (y=mx+b) so a graph of this
should yield.
NOTE: χ is for solvent not solute…so the less solute dissolves…the larger the
χ solvent…the higher the vapor pressure!
Raoult’s Law example: You are given a solution made by dissolving 50.0g
glucose into 600.g of water. The pressure of pure water is 23.8 torr at this
temperature. Determine the vapor pressure of this solution.
*Determine moles of each: 50.0g C2H602 × 1 mol = 0.278 mol
180.1g
6.00g H20 × 1 mol = 33.3 mol
18.0g
χ H20 = 33.3 mol =0.991
so
Psoln = 0.991 ×23.8torr= 23.6torr
33.3 + .278
*NOTE: The χ should be relatively close to 1.00 as the solvent is the part of
the soln. in
*Raoult’s Law may be given as a calculation OR essay… example #49 in
book
C. Ideal Solutions vs Nonideal Solutions
An ideal solution follows Raoult’s Law and has a graph like:
(1)IF THE SOLUTE /SOLVENT AFFINITY IS VERY HIGH… you will
get a negative deviation from Raoult’s Law and a lower than expected
upper pressure interation which yields a (-) deviation:
∆H1 & ∆H2 < ∆H3 = -∆Hsoln
(2)WEAK SOLUTE/WOLVENT INTERACTIONS indicate higher than
expected vapor pressures and thus a positive deviation from Raoult’s Law.
∆H1 & ∆H2 > ∆H3 = (+) ∆Hsoln
(Look for more volatile solutes in solvents)
(3) GRAPH:
11.5
Boiling Point Equation/ Freezing Point Equation
*Addition of a molecule solute extends that liquid phase of the solvent.
The solution will have a lower freezing point and a higher boiling.
* Explain this at the atomic/ molecular level!
eg: You dissolve 4.9g sucrose in 175g H20. Calculate the new boiling and
freezing point of this solution.
*this is a nonvolatile solute so the boiling point should be greater while
the freezing point is lower.
EQUATIONS:
∆T=Kbm
and ∆T=Kf m
∆T=change in temp
Kb=molar boiling pt. constant
m= molality
so: molality = 4.4g sucrose × 1 mol sucrose × 103g = 0.082 molal
175g solvent
342g
1kg
∆T=Kbm = 0.51○C × 0.082 m = 0.042○C
m
Tb= 100.042○C
∆T=Kfm= 1.86○C × 0.082m = 1.15○C
m
Tf= -0.15○C
11.6 Osmotic Pressure
↑ = MRT
T= Kelvin Temp; ↑= osmotic pressure in atm; M= molarity; R= gas law constant
*This is often used to calculate the molar mass of an unknown
EXAMIN EXERCISES 11.11 and 11.12 from book
11.6
Colligative Preoperties of Electolyte Solutions
Ionic compounds, when dissolved, into water, produce more than one particle pre
molecule dissolved…
van’t Hoff factor = “i” where i= moles of particles in solution
moles of solute dissolved
eg. NaCl = 2 mol ions = 2 = i
* This is the expected van’t Hoff factor, however
1 mol NaCl
however, the actual value will be slightly less
due to ion pairing in solution.
*Ion pairing is most important in concentrated solutions, and in solutions with
highly charged ions, (both of these favor strong or frequent ion–ion attractions).
“i” is a correction factor for freezing/boiling point changes
Eg. ∆T=imK
=iMRT