CHEM II Exam 1

MOLARITY increasing temperature

With increasing temperature the volume changes CHANGING MOLARITY.

Moles of water

1000 X density of solution

Molality

mole of solute / mass of solvent in Kg

g to kg

move decimal 3 space to the left

L to mL

decimal 3 spaces back

mass with given density and volume

volume X density

molarity

mol of solute / liter of solvent

volume of solution

1000g / density = ml to L move decimal 3 places to the front

solution volume with given mol

given mol / molarity

Solution process (liquid)

Molecules of the solid are separated from one another, and each solid molecule is surrounded by liquid molecules (solvation)

“like dissolves like”

Substances with similar intermolecular forces tend to be soluble in one another because the solute-solute interactions are similar in magnitude to the solute-solvent interactions.

Endothermic

requires energy be added to the system because the solvent-soluteinteractions release less energy than the energy needed to separate the solute and solvent particles. A solution is still formed due to the increase in entropy of the system.

dissolving solid lead to increase entropy (disorder)

packed particles are free to move throughout large volume of solvent

IMF forces in order strong to weak

1. Ion-dipole

2. Hydrogen bonding (H bond to NOF)

3. Dipole-dipole

4. Ion-dipole

5. London dispersion (ALL)

C-C

Length increases = less polar = solubility decreases

O-H

can form strong hydrogen bonds w/ water but property diminishes in importance as chain length increases

Solubility of ionic compounds Pressure and Temp

Pressure: No impact

Temperature: solubility increases with temperature

Solubility of GAS in a LIQUID

Temperature: increasing temp decreases solubility of MOST gases

Pressure: increasing pressure increases solubility of gas

molar mass

mass/ mol

Raoul’s Law

the partial pressure of a substance over a solution is equal to the mole fraction of the substance times its pure vapor pressure. P1=x1P2

Vapor pressure of solution

mole fraction X vapor pressure given

vapor pressures with 2 volatile solutions

mole fraction X vapor pressure given of both solutions

mol fractions of 2 volatile solutions

solution 1 = partial pressure 1 / partial pressure 1 + 2

Solutiion 2= 1- solution 1

Volality

vaporize easily. HIGH vocality = low boiling point high vapor pressure

Fractional distillation

more volatile components go to the top; less volatile component sink

Volatile solutions

weaker IMFS

HIGH vapor pressure

LOW boiling point

Solute dissolved in solvent

INCREASE in vapor pressure

DECREASE in freezing point

BC needs more energy to raise the vapor pressure to be equivalent to the atmospheric pressure