1. A.P. Chemistry Solutions Chapter 11 ↑ ↑ ↓ ↑ ↑ ↑ Solute Solvent M m + +
Separating them…
Chromatography
Chapter 11
making
Homogeneous Mixtures
Colligative
Properties
(Where Conc. Affects Them)
Changes Phase
No Set Ratio
B.P. Elevation
Solute
parts
Solvent
V.P. Lowering
F.P. Depression
Solvation Process
Concentration Descriptions
Raoult’s Law
V.P.(sol’n) = C(solv) V.P.(solv)
+ C(solute) V.P.(solute) DT
= Kmi M
1. Expand Solute
volatile solutes
Ideal
2. Expand Solvent m
DH= +
3. Intermix
DH=
Non-volatile solutes + C
van’t Hoff Factor, i
Non-Ideal
DH= -
DH=
hydration DHhyd=
mass %
Small Deviations
Large Deviations
+/-
Neg. Deviations DH=
Pos. Deviations DH=
Low [ ]
High [ ] - +
Like Dissolves Like (generally)
Low charges
High charges
Heat of solution
DHsol
Phases Involved
Effect (Generally) ↑ ↑
Sol. in Liq.
T sol.
(Dictates the degree to which you can dissolve a solute in a solvent)
Temp. ↓ + ↑ -
DH= +
DH=
DH=
Gas in Liq.
T sol.
affected by:
large
Entropy driven ↑ ↑
Press.
Gas in Liq.
P sol.
( )
solubility
From the Gas
P = kC →
Henry’s Law
insoluble
soluble
(≥0.10 M)

2. End of Chapter Problem:
You’re in the lab…in front of you are four substances. Two are labeled solutes…they are 25.0 grams of TTE (C2Cl3F3) and 2.69 grams of CsOH. Two are labeled solvents…they are grams of benzene (C6H6) and grams of water. Which substances would you combine to make solutions in which the solute would be considered soluble in the solvent?
Looking at the one that produces a water solution, calculate the heat of hydration if the heat of solution is -72 kJ/mol and the lattice energy for the solute is -724 kJ/mol.
Calculate the molarity, molality, and mole fraction (with respect to the solvent) of this solution given that the density of the solution is 1.03 g/mL
Find the freezing pt. of this solution assuming ideal behavior.
Find the vapor pressure of this solution.
basic, non-volatile, ionic solute
mol
6.83 mol
TTE in benzene
(both nonpolar)
CsOH in water
DH=
+ or – and small
DH1=
724 kJ
DH2 +
DH3 =
DHhyd=
DHsol – DH1=
-796 kJ
C = mol/(6.83 mol mol) = 0.995
Vol of sol’n = g/(1.03 g/mL)
M = mol/0.122 L = molar
m = nol/ kg = molal
=122 mL = L
-0.543°C
DT = Kfmi = (1.86 K kg/mol)(0.146 mol/kg)(2)=
0.543 K
Would the measured VP be higher or lower than 23.7?
Lower, DH is neg; negative deviation
VPsolution = Csolvent VPsolvent= (0.995)(23.8 torr)=
23.7 torr