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Chemistry Lecture ’94 B. Rife CHS
Text: Modern Chemistry; Holt, Rinehart & Winston 1993
2
3
4
Solutions Chapter 14
Homework:
1 Section Reviews (pg 410,419,425,433)
Reviewing Concepts: (all)
Chapter/Section Review (Handout)
(pg 434-5)
Problems (all black / red EC) (pg 435-7)
DUE DATE
Exam Date _
14.1 Types of Mixtures
14.1A Distinguish between heterogeneous and homogeneous mixtures. ( )
HETEROGENEOUS - IS A MIXTURE WHOSE COMPOSITION AND PROPERTIES ARE NOT UNIFORM BUT
THAT DIFFER FROM POINT TO POINT IN THE MIXTURE.
HOMOGENEOUS - IS A MIXTURE WHOSE COMPOSITION AND PROPERTIES ARE UNIFORM
THROUGHOUT THE MIXTURE ALSO CALLED A SOLUTION.
14.1B Distinguish between electrolytes and nonelectrolytes. ( )
ELECTROLYTE - IS A SUBSTANCE THAT DISSOLVES IN WATER TO GIVE A SOLUTION THAT
CONDUCTS ELECTRIC CURRENT. THE SOLUTION USUALLY CONTAINS IONS SUCH AS H + , Na + ,
Cl -
NONELECTROLYTE - IS A SUBSTANCE THAT DISSOLVES IN WATER TO GIVE A SOLUTION THAT
DOES NOT CONDUCT AN ELECTRIC CURRENT. (SUGAR)
14.1C Compare the properties of suspensions, colloids, and solutions. ( )
SOLUTIONS - IS A HOMOGENEOUS MIXTURE OF TWO OR MORE SUBSTANCES IN A SINGLE PHASE.
THE PARTICLES OF SOLUTE ARE MOLECULAR IN SIZE. (VERY SMALL)
SOLUTE - IN A SOLUTION, IS THE SUBSTANCE THAT IS DISSOLVED IN THE SOLVENT.
SOLUBLE - MEANS CAPABLE OF BEING DISSOLVED.
SOLVENT - IS THE DISSOLVING MEDIUM IN A SOLUTION.
TYPES OF SOLUTIONS :
1. GASEOUS SOLUTIONS - TWO OR MORE GASES ARE MIXED EXAMPLES: AIR, WATER VAPOR
2. LIQUID SOLUTIONS - CONTAIN A LIQUID SOLVENT AND A SOLID, LIQUID, OR GAS SOLUTE.
EXAMPLES: SALT WATER, ANTIFREEZE, SOFT DRINKS
3. SOLID SOLUTIONS - ALLOYS ARE SOLID SOLUTIONS IN WHICH THE ATOMS OF TWO OR MORE
METALS ARE UNIFORMLY MIXED.
EXAMPLES: BRASS IS A SOLUTION OF 66% COPPER AND 33% ZINC.
SUSPENSIONS - IS A HETEROGENEOUS MIXTURE OF A SOLVENT-LIKE SUBSTANCE WITH PARTICLES
THAT SLOWLY SETTLE OUT.
THE PARTICLES IN SUSPENSION CAN BE SEPARATED FROM THE HETEROGENEOUS MIXTURE
BY FILTRATION. THE PARTICLES OF SOLUTE ARE OVER 1000 nm IN SIZE. (LARGE)
COLLOIDS - IS A MIXTURE CONTAINING PARTICLES THAT ARE INTERMEDIATE IN SIZE (1 nm - 1000 nm) BETWEEN THOSE IN SOLUTIONS AND SUSPENSIONS.
THE COLLOIDAL PARTICLES MAKE UP THE DISPERSED PHASE, AND SOLVENTLIKE PHASE IS
CALLED THE DISPERSING MEDIUM.
THE PARTICLES CANNOT BE SEEN WITH THE NAKED EYE BUT ARE LARGE ENOUGH TO
SCATTER LIGHT. (TYNDALL EFFECT) UNDER THE MICROSCOPE, THE RANDOM MOTION IS
CALLED BROWNIAN MOTION. EXAMPLES: FOAM, FOG

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14.2 The Solution Process
14.2A List and explain three factors that influence the rate of dissolving of a solid in a liquid. ( )
1. IN GENERAL, THE MORE FINELY DIVIDED A SUBSTANCE IS, THE GREATER THE SURFACE AREA PER
UNIT MASS AND THE MORE QUICKLY IT WILL DISSOLVE.
2. STIRRING OR SHAKING HELPS TO DISPERSE THE SOLUTE PARTICLES AND BRINGS FRESH SOLVENT
INTO CONTACT WITH THE SURFACE.
3. AS THE TEMPERATURE OF THE SOLVENT INCREASES, SOLVENT MOLECULES MOVE AROUND
FASTER WHICH CAUSE SOLUTE MOLECULES TO LEAVE THE SURFACE AT A FASTER RATE.
14.2B Explain solution equilibrium and distinguish among saturated, unsaturated, and supersaturated solutions. ( )
SOLUTION (DYNAMIC) EQUILIBRIUM - IS THE PHYSICAL STATE IN WHICH THE OPPOSING
PROCESSES OF DISSOLVING AND CRYSTALLIZING OF A SOLUTE OCCUR AT EQUAL RATES.
SATURATED SOLUTION - IS A SOLUTION THAT CONTAINS THE MAXIMUM AMOUNT OF DISSOLVED
SOLUTE.
UNSATURATED SOLUTION - IS A SOLUTION THAT CONTAINS LESS SOLUTE THAN A SATURATED
SOLUTION.
SUPERSATURATED - IS A SOLUTION THAT CONTAINS MORE DISSOLVED SOLUTE THAN A
SATURATED SOLUTION.
SOLUBILITY - OF A SUBSTANCE IS THE AMOUNT OF THAT SUBSTANCE REQUIRED TO FORM A
14.2C Explain the meaning of “like dissolves like” in terms of polar and nonpolar substances. ( )
POLAR SOLVENTS (WATER) DISSOLVES POLAR SOLUTES (SALTS, ALCOHOL)
SATURATED SOLUTION WITH A SPECIFIC AMOUNT OF SOLVENT AT A SPECIFIED
TEMPERATURE. (g solute / 100 g of solvent @ o C)
NONPOLAR SOLVENTS (CCl
4
) DISSOLVES NONPOLAR SOLUTES (FATS, OILS)
SUBSTANCES THAT ARE NOT SOLUBLE IN EACH OTHER ARE IMMISCIBLE. (OIL AND WATER)
TWO SUBSTANCES THAT ARE MUTUALLY SOLUBLE IN ALL PROPORTIONS ARE SAID TO BE MISCIBLE.
HENRY’S LAW - THE SOLUBILITY OF A GAS IN A LIQUID IS DIRECTLY PROPORTIONAL TO THE
PARTIAL PRESSURE OF THAT GAS ON THE SURFACE OF THE LIQUID.
14.2D List three interactions that contribute to the heat of solution, and explain what causes dissolving to be exothermic or endothermic.. ( )
EFFERVESCENCE - IS THE RAPID ESCAPE OF A GAS FROM A LIQUID IN WHICH IT IS DISSOLVED.
INCREASING TEMPERATURE USUALLY DECREASES GAS SOLUBILITY.
MOST OFTEN, INCREASING THE TEMPERATURE INCREASES THE SOLUBILITY OF SOLIDS.
14.2E Compare the effects of temperature and pressure on solubility. ( )
ENTHALPY OR HEAT OF SOLUTION - IS THE AMOUNT OF HEAT ENERGY ABSORBED OR RELEASED
WHEN A SOLUTE DISSOLVES IN A SPECIFIC AMOUNT OF SOLVENT.
SOLUTE-SOLUTE LARGE (+) LARGE (+)
SOLVENT-SOLVENT
SOLUTE-SOLVENT
LARGE (+)
LARGE (-)
HEAT OF SOLUTION (-) EXOTHERMIC
LARGE (+)
SMALL (-)
(+) ENDOTHERMIC
IF SOLUTE-SOLUTE AND SOLVENT-SOLVENT IS VERY LARGE AND SOLUTE-SOLVENT IS VERY SMALL
THE SUBSTANCES ARE IMMISCIBLE
14.3 Concentrations of Solutions
14.3A Define concentration using molarity, molality, and percent by mass. ( )
MASS PERCENT (WEIGHT PERCENT) - IS THE PERCENT BY MASS OF THE SOLUTE IN THE SOLUTION.

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MASS PERCENT = MASS OF SOLUTE x 100 %
MASS OF SOLUTION
MOLARITY (M) - IS THE NUMBER OF MOLES OF SOLUTE PER LITER OF SOLUTION.
MOLARITY (M) = MOLES OF SOLUTE
LITERS OF SOLUTION
MOLALITY (m) = IS THE NUMBER OF MOLES OF SOLUTE PER KILOGRAM OF SOLVENT.
MOLALITY = MOLES OF SOLUTE x 100 %
KG OF SOLVENT
14.3B Given the concentration of a solution, find the amount of solute in a given amount of solution. ( )
14.3C Given the concentration of a solution, find the amount of solution that contains a given amount of solute.
A COMMONLY PURCHASED DISINFECTANT IS A 3.0% (BY MASS) SOLUTION OF HYDROGEN PEROXIDE
(H
2
O
2
) IN WATER. ASSUMING THAT THE DENSITY OF THE SOLUTION IS 1.0 g/cm 3 CALCULATE
THE MOLARITY, MOLALITY, AND THE MOLE FRACTION OF H
2
O
2
.
ASSUME ONE LITER OF SOLUTION THUS 30 g OF H
2
O
2
30 g H
2
O
2
( 1 MOL ) = 0.88 mol / L or 0.88 M H
2
O
2
34 g
1 L = 1000 g - 30 g H
2
O
2
= 970 g H
2
O = 0.970 kg H
2
O
0.88 mol H
2
0.970 kg H
2
O
2
= 0.91 mol / kg or 0.91 m H
2
O
O
2
MOLE FRACTION = MOLS OF SOLUTE / TOTAL MOLS OF SOLUTION
970 g H
2
O = 53.9 0.88 mol = 1.6 x 10 -2
18 g 53.9 + 0.88 mol
HOW MANY MILLILITERS OF 18.0 M H
2
SO
4
WOULD BE REQUIRED TO REACT WITH 250 mL OF 2.50
M Al(OH)
3
IF THE PRODUCTS ARE ALUMINUM SULFATE AND WATER?
BALANCED EQUATION:
3H
2
SO
4
+ 2Al(OH)
3
---> Al
2
(SO
4
)
3
+ 6H
2
O
# OF MOLES GIVEN
0.250 L x 2.50 mol/L Al(OH)
3
= 0.625 mol Al(OH)
3
#OF MOLS AND VOLUME REQUIRED
0.625 mol Al(OH)
3
(3 mol H
2
SO
4
) ( L ) = 0.0521 L
2 mol Al(OH)
3
18 mol = 52.1 mL
14.4 Colligative Properties of Solutions
COLLIGATIVE PROPERTY - IS A PROPERTY THAT DEPENDS ON THE NUMBER OF SOLUTE PARTICLES
BUT IS INDEPENDENT OF THEIR NATURE.
NONVOLATILE - IS A SUBSTANCE THAT HAS LITTLE TENDENCY TO BECOME A GAS UNDER THE
EXISTING CONDITIONS.
THE PRESENCE OF A NONVOLATILE SOLUTE LOWERS THE VAPOR PRESSURE OF A SOLVENT.
P
SOLN
< P
SOLV
14.4A List three colligative properties and describe how each is caused. ( )
14.4B Write the expression for freezing-point depression and boiling-point elevation. Define and give the units for each term in the expressions. ( )
FREEZING POINT DEPRESSION (

t f
) - IS THE DIFFERENCE BETWEEN THE FREEZING POINTS OF THE
PURE SOLVENT AND THE SOLUTION. _ t f
= K f
m

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THE VAPOR PRESSURES OF ICE AND LIQUID WATER ARE THE SAME AT 0 o C (FREEZING POINT). A
SOLUTION WILL NOT FREEZE AT 0 o C BECAUSE THE WATER IN THE SOLUTION HAS A LOWER
VAPOR PRESSURE THAN THAT OF PURE ICE.
BOILING POINT DEPRESSION (_ t b
) - IS THE DIFFERENCE BETWEEN THE BOILING POINTS OF THE
PURE SOLVENT AND THE SOLUTION. _ t b
= K b
m
THE VAPOR PRESSURE OF LIQUID WATER AND THE ATMOSPHERE ARE THE SAME AT 100 o C (BOILING
POINT). A SOLUTION WILL NOT BOIL AT 100 o C BECAUSE THE WATER IN THE SOLUTION HAS A
LOWER VAPOR PRESSURE THAN THAT OF THE ATMOSPHERE.
MOLAR MASS = SOLUTE MASS IN GRAMS / MOLS OF SOLUTE
MOLALITY = MOLS OF SOLUTE / kg OF SOLVENT
MOLS OF SOLUTE = MOLALITY x kg OF SOLVENT
14.4C Given appropriate information, calculate freezing-point depression, boiling-point elevation, or solution molality.
( )
DETERMINE THE FREEZING POINT DEPRESSION, BOILING POINT ELEVATION, AND SOLUTION
MOLALITY OF A SOLUTION WITH 77 g of C
12
H
22
O
11
in 400 g H
2
O
MOLS OF C
12
H
22
O
11
= 77 g (1 mol) = 0.225 mol
342 g
MOLALITY = 0.225 mol = 0.563 m
0.40 kg
_ t f
= K f
m = -1.86 (0.563 m) = -1.05 C o
_ t b
= K b
m = 0.51 (0.563 m) = 0.286 C o
14.4D Describe an experimental method for determining molar mass using a colligative property. ( )
1. ADD A KNOWN QUANTITY (IN GRAMS) OF SOLUTE TO A KNOWN MASS OF SOLVENT WITH KNOWN
K f
and K b
.
2. OBSERVE THE FREEZING POINT DEPRESSION OF BOILING POINT ELEVATION.
3. CALCULATE THE MOLALITY OF THE SOLUTION. m = _ t f
/ K f
4. CALCULATE THE MOLES OF SOLUTE PRESENT IN THE SOLUTION.
MOLS OF SOLUTE = MOLALITY x kg OF SOLVENT
5. MOLAR MASS EQUALS MASS OF SOLUTE DIVIDED BY MOLS OF SOLUTE.
14.4E Calculate molar mass from freezing-point depression or boiling-point elevation data. ( )
THE MOLAL FREEZING POINT CONSTANT ( K f
) FOR ETHER IS
-1.79 C o /MOLAL. WHEN 21.2 g OF A SOLUTE IS DISSOLVED IN 740 g OF ETHER, THE FREEZING
POINT OF THE ETHER IS LOWERED BY 0.568 C o . WHAT IS THE MOLAR MASS OF THE SOLUTE?
MOLAR MASS = SOLUTE MASS IN GRAMS / MOLS OF SOLUTE
SOLUTE MASS IN GRAMS = 21.2 g
MOLALITY m = _ t f
/ K f
= -0.568 C o / -1.79 C o /MOLAL
= 0.3173 m
CALCULATE THE MOLES OF SOLUTE PRESENT IN THE SOLUTION.
MOLS OF SOLUTE = MOLALITY x kg OF SOLVENT
= 0.3173 m x 0.740 kg = 0.235 mol
MOLAR MASS = 21.2 g / 0.235 mol = 90.3 g/mol