CH. 12 – PHYSICAL PROPERTIES OF SOLUTIONS
I. Some Types of Solutions
A. solvent – solution component that is present in the greatest amount and determines the state of matter of the solution
B. solute – other component
C. solutions are by no means limited to the liquid state
II. Solution Concentration
Molarity – relates an amount of solute in moles and a solution volume in liters
M = amount of solute (mol)
Volume of solution (L)
A. percent by mass, percent by volume, and mass/volume percent
1. Express solution concentration through percentage composition
2. If a precise quantity of solute is required, measure out a mass or volume of solution
3. Solution concentration is often expressed as percent by volume
4. mass/volume percent is expressed in milligrams per deciliter (mg/dL)
B. parts per million, parts per billion, and parts per trillion
1. For extremely dilute solutions, concentrations are expressed in ppm, ppb, or ppt
2. For liquid solutions, ppm, ppb, and ppt are generally based on mass
a. 1 ppm of solute in a solution is equal to 1 g solute per 1x106 g solution
3. for gaseous solutions, they are generally based on numbers of molecules or on volumes
a. 1 ppm of a substance in air is 1 molecule of the substance per 1x106 molecules of air or 1 L of that substance per 1x106 L of air
1 ppm = 1 mg/L1 ppb = 1 µg/L1 ppt = 1 ng/L
C. molality
1. Molarity varies with temperature
a. when volume increases as temperature increases, molarity decreases
2. concentration units independent of temperature and based only on mass, not volume:
a. percent by mass
b. percent by volume and mass/volume percent are not
3. molality
a. independent of temperature
b. number of moles of solute per kilogram of solvent
m = amount of solute (mol)
mass of solvent (kg)
D. mole fraction and mole percent
1. mole fraction (xi) – fraction of all the molecules in the solution that are molecules of i
xi = amount of component i (mol) = ni
total amount of solution components (mol) ntotal
a. has no unit; it is dimensionless
b. the sum of the mole fractions is equal to 1
x1 + x2 + x3 + ··· = 1
2. mole percent – mole fraction multiplied by 100%
III. Energetics of Solution Formation
A. enthalpy of solution
1. move the molecules of solvent apart to make room for the solute molecules
a. requires work to overcome the intermolecular forces
b. ΔH1 > 0
2. separate the molecules of solute to the distances found between them in the solution
a. work is required
b. ΔH2 > 0
3. allow the separated solute and solvent molecules to mix randomly
a. energy is release because now there are forces of attraction between the solute and solvent molecules
1) pure solvent separated solvent moleculesΔH1
2) pure solute separated solute moleculesΔH2
3) separated solvent and solute molecules solutionΔH3
Net: pure solvent + pure solute solutionΔHsoln = ΔH1 + ΔH2 + ΔH3
4. whether the formation is an endothermic (ΔHsoln > 0) or exothermic (ΔHsoln > 0) process depends on the relative values of the enthalpy changes of the hypothetical steps
B. intermolecular forces in solution formation