Chemistry Lecture ’94 B



Chemistry Lecture ’94 B. Rife CHS

Text: Modern Chemistry; Holt, Rinehart & Winston 1993

Solutions Chapter 14

Homework: DUE DATE

1 Section Reviews (pg 410,419,425,433)

2 Reviewing Concepts: (all) (pg 434-5)

3 Problems (all black / red EC) (pg 435-7)

4 Chapter/Section Review (Handout)

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

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 SATURATED SOLUTION WITH A SPECIFIC AMOUNT OF SOLVENT AT A SPECIFIED TEMPERATURE. (g solute / 100 g of solvent @ oC)

14.2C Explain the meaning of “like dissolves like” in terms of polar and nonpolar substances. ( )

POLAR SOLVENTS (WATER) DISSOLVES POLAR SOLUTES (SALTS, ALCOHOL)

NONPOLAR SOLVENTS (CCl4) 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 LARGE (+) LARGE (+)

SOLUTE-SOLVENT LARGE (-) SMALL (-)

HEAT OF SOLUTION (-) EXOTHERMIC (+) 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.

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 (H2O2) IN WATER. ASSUMING THAT THE DENSITY OF THE SOLUTION IS 1.0 g/cm3 CALCULATE THE MOLARITY, MOLALITY, AND THE MOLE FRACTION OF H2O2.

ASSUME ONE LITER OF SOLUTION THUS 30 g OF H2O2

30 g H2O2 ( 1 MOL ) = 0.88 mol / L or 0.88 M H2O2

34 g

1 L = 1000 g - 30 g H2O2 = 970 g H2O = 0.970 kg H2O

0.88 mol H2O2 = 0.91 mol / kg or 0.91 m H2O2

0.970 kg H2O

MOLE FRACTION = MOLS OF SOLUTE / TOTAL MOLS OF SOLUTION

970 g H2O = 53.9 0.88 mol = 1.6 x 10-2

18 g 53.9 + 0.88 mol

HOW MANY MILLILITERS OF 18.0 M H2SO4 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:

3H2SO4 + 2Al(OH)3 ---> Al2(SO4)3 + 6H2O

# 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 H2SO4 ) ( 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.

PSOLN < PSOLV

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 (_ tf) - IS THE DIFFERENCE BETWEEN THE FREEZING POINTS OF THE PURE SOLVENT AND THE SOLUTION. _ tf = Kf m

THE VAPOR PRESSURES OF ICE AND LIQUID WATER ARE THE SAME AT 0oC (FREEZING POINT). A SOLUTION WILL NOT FREEZE AT 0oC BECAUSE THE WATER IN THE SOLUTION HAS A LOWER VAPOR PRESSURE THAN THAT OF PURE ICE.

BOILING POINT DEPRESSION (_ tb) - IS THE DIFFERENCE BETWEEN THE BOILING POINTS OF THE PURE SOLVENT AND THE SOLUTION. _ tb = Kb m

THE VAPOR PRESSURE OF LIQUID WATER AND THE ATMOSPHERE ARE THE SAME AT 100oC (BOILING POINT). A SOLUTION WILL NOT BOIL AT 100oC 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 C12H22O11 in 400 g H2O

MOLS OF C12H22O11 = 77 g (1 mol) = 0.225 mol

342 g

MOLALITY = 0.225 mol = 0.563 m

0.40 kg

_ tf = Kf m = -1.86 (0.563 m) = -1.05 Co

_ tb = Kb m = 0.51 (0.563 m) = 0.286 Co

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 Kf and Kb.

2. OBSERVE THE FREEZING POINT DEPRESSION OF BOILING POINT ELEVATION.

3. CALCULATE THE MOLALITY OF THE SOLUTION. m = _ tf / Kf

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 ( Kf ) FOR ETHER IS

-1.79 Co/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 Co. 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 = _ tf / Kf = -0.568 Co / -1.79 Co/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

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download