SOLUTIONS
A homogeneous
mixture in which the
components are
uniformly intermingled
Terms
Solvent –
The substance present in the largest amount
in a solution.
The substance that does the dissolving.
Solute –
The other substance or substances in a
solution.
The substance that is dissolved.
ELECTROLYTES



Substances that break up in water to
produce ions.
These ions can conduct electric current
Examples: Acids, Bases and Salts (ionic
compounds)
SOLUBILITY

“Like dissolves Like”
–
–
Polar molecules dissolve polar molecules
Nonpolar molecules dissolve nonpolar molecules
SOLUBILITY RULES





All common salts of Group I elements and
ammonium are soluble
All common acetates and nitrates are soluble
All binary compounds of Group 7 with metals are
soluble except those of silver, mercury I and lead
All sulfates are soluble except those of barium,
strontium, calcium, silver, mercury I and lead
Except for those in Rule 1, carbonates, hydroxides,
oxides, sulfides and phosphates are insoluble
Terms

Saturated
–

Unsaturated
–

When the solution contains more solute than a saturated
solution will hold at that temperature
Concentrated
–

When a solvent can dissolve more solute
Supersaturated
–

When a solution contains the maximum amount of solute
When a relatively large amount of solute is dissolved
Dilute
–
When a relatively small amount of solute is dissolved
Factors Affecting the Rate of
Dissolution

Surface Area

Stirring

Temperature
Temperature vs Solubility
MOLARITY

Molarity-the number of moles of solute per
liters of solution

M = molarity = moles of solute
liter of solution

Calculate the molarity of a solution prepared
by dissolving 11.5 g of NaOH in enough
water to make a 1.50 L solution.

Calculate the molarity of a solution prepared
by dissolving 1.56 g of HCl into enough water
to make 26.8 ml of solution.

Calculate the number of grams of sodium
phosphate required to make 150. ml of a
2.5M solution.

How many liters of solution are needed to
dissolve 5.0 g of hydrochloric acid to make a
3.0 M hydrochloric acid solution?

What is the concentration of each ion in a
0.50 M solution of Co(NO3)2?

What is the concentration of each ion in a
0.25 M solution of aluminum sulfate?

How many moles of Ag+ ions are present in
25.0 ml of a 0.75 M AgNO3 solution?

Calculate the number of moles of Cl- ions in
1.75 L of 1.0 x 10-3M AlCl3

To analyze the alcohol content of a certain
wine, a chemist needs 1.00 L of an aqueous
0.200 M K2Cr2O7 (molar mass is 294.2g/mol)
How much K2Cr2O7 must be weighed out to
make this solution?
DILUTIONS

M1 x V1
= M2 x V2

What volume of 16 M sulfuric acid must be
used to prepare 1.5 L of a 0.10 M H2SO4

What volume of 12 M HCl must be used to
prepare 0.75 L of a 0.25 M HCl?

When barium nitrate and potassium
chromate react in aqueous solution, the
yellow solid barium chromate is formed.
Calculate the mass of barium chromate that
forms when 3.50 x 10-3 mole of solid barium
nitrate is dissolved in 265 ml of 0.0100 M
potassium chromate solution.
MOLALITY

A unit of concentration equal to the number
of moles of solute per kilogram of solvent
m = moles of solute
kg solvent

98.0 g RbBr in 824 g water

85.2 g SnBr2 in 1.40 x 102 g water
Phase Change Diagram
Definition
Point – when the vapor
pressure of the liquid is equal
to the atmospheric pressure
 Boiling
Freezing Point Depression/
Boiling Point Elevation

Colligative property – a solution property that
depends on the number of solute particles
present (ie – f.p. and b.p.)
–
–
Freezing Point Depression
Boiling Point Elevation
Calculating Boiling Points

Kbp = boiling point constant
–
–
Water 0.515OCkg/mol
1 mole of a solute particle will raise the bp of 1 kg of water
by 0.515OC

1m solution of sugar water
1(0.515OC)
100.515OC

1m solution of NaCl water
2(0.515OC)
101.03OC

1m solution of CaCl2 water
3(0.515OC)
101.545OC
Calculating Freezing Points

Kfp = freezing point constant
–
–
Water 1.853OCkg/mol
1 mole of a solute particle will lower the fp of 1 kg of water
by 1.853OC

1m solution of sugar water
1(1.853OC)
-1.853OC

1m solution of NaCl water
2(1.853OC)
-3.706OC

1m solution of CaCl2 water
3(1.853OC)
-5.559OC



ΔTfp = im Kfp
Kfp = 1.853oCkg/mol
ΔTbp = im Kbp
Kbp = 0.515oCkg/mol
If 26.4 grams of nickel II bromide are
dissolved in 224 grams of water, what will
be the new boiling point and freezing point
of the resulting solution?

If 25.0 grams of calcium chloride are
dissolved in 500 grams of water, what will
be the new boiling point and freezing point
of the resulting solution?
MASS PERCENT

A unit of concentration equal to the mass of
solute per mass of solution
part
x
100
whole

A solution is prepared by mixing 1.00 g of
ethanol with 100.0 g of water. Calculate the
mass percent of ethanol in this solution.

A 135 g sample of seawater is evaporated to
dryness, leaving 4.73 g of salt. Calculate the
mass percent of salt in the saltwater.
moles of solute
1.Molarity (M) =
liters of solution
mass of solute
 100%
2.Mass (weight) percent =
mass of solution
molesA
3.Mole fraction (A) =
total moles in solution
moles of solute
4.Molality (m) =
kilograms of solvent
Sol’n is prepared by adding 5.84 g of
formaldehyde (H2CO) to 100.0 g water. Final
vol of solution is 104.0 mL. Calculate the
molarity, molality, mass % and .
M = 1.87 M H2CO
m = 1.94 m H2CO
Mass %= 5.52 % H2CO
 = .0338
Molecular Mass Determination


If 99.0 g of a nonionizing solute dissolved in
669 grams of water and the freezing point of
the resulting solution is -0.960oC, what is the
molecular mass of the solute?
ΔTfp = im Kfp
m = ΔTfp
Kfp


If 64.3 g of a nonionizing solute dissolved in
390. grams of water and the boiling point of
the resulting solution is 100.680oC, what is
the molecular mass of the solute?
ΔTbp = im Kbp
m = ΔTbp
Kbp
Anthraquinone contains only carbon,
hydrogen and oxygen and has an empirical
formula of C7H4O. When 15.93g of
anthraquinone are added to 1 kg of water
the freezing point depression was
determined to be 0.240oC. Calculate the
molar mass of the biomolecule (Kf for
chloroform is 4.70oCkg/mol)