Chapter 8 Solutions - Riverside City Collegewebsites.rcc.edu/grey/files/2012/02/Chapter-8.pdf · Chapter 8 Solutions Solutions to In-Chapter Problems ... 8.6 Magnesium salts are not

Chapter 8–1

Chapter 8 Solutions Solutions to In-Chapter Problems 8.1 A heterogeneous mixture does not have a uniform composition throughout a sample.

A solution is a homogeneous mixture that contains small particles. Liquid solutions are transparent. A colloid is a homogeneous mixture with larger particles, often having an opaque appearance.

a. Cherry Garcia ice cream: heterogeneous mixture b. mayonnaise: colloid c, d, e. seltzer water, nail polish remover, and brass: solutions

8.2 A substance that conducts an electric current in water is called an electrolyte. If a solution

contains ions, it will conduct electricity. A substance that does not conduct an electric current in water is called a nonelectrolyte. If a solution contains neutral molecules, it will not conduct electricity. a. KCl in H2O: electrolyte c. KI in H2O: electrolyte b. sucrose (C12H22O11) in H2O: nonelectrolyte

8.3 Use the general solubility rule: “Like dissolves like.” Generally, ionic and polar compounds are soluble in water. Nonpolar compounds are soluble in nonpolar solvents.

NH2OH

CH4

C CH

HOH

HOH

H

KBr

a.

b.

c.

d.

NaNO3e.

ionic compoundwater soluble

nonpolar compoundNOT water soluble

polar compoundwater soluble

ionic compoundwater soluble

polar compoundwater soluble

8.4 a. Benzene (C6H6) and hexane (C6H14): both nonpolar compounds—would form a solution. b. Na2SO4 and H2O: one ionic compound and one polar compound—would form a solution. c. NaCl and hexane (C6H14): one ionic and one nonpolar compound—cannot form a solution. d. H2O and CCl4: one polar and one nonpolar compound—cannot form a solution.

8.5 Identify the cation and anion and use the solubility rules to predict if the ionic compound is water

soluble.

a. Li2CO3: cation Li+ with anion CO32–—water soluble.

b. MgCO3: cation Mg2+ with anion CO32–—water insoluble.

c. KBr: cation K+ with anion Br–—water soluble. d. PbSO4: cation Pb2+ with anion SO4

2–—water insoluble. e. CaCl2: cation Ca2+ with anion Cl–—water soluble. f. MgCl2: cation Mg2+ with anion Cl–—water soluble.

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Solutions 8–2

8.6 Magnesium salts are not water soluble unless the anion is a halide, nitrate, acetate, or sulfate. The interionic forces between Mg2+ and OH– must be stronger than the forces of attraction between the ions and water. Therefore, milk of magnesia is a heterogeneous mixture, rather than a solution.

8.7 A soft drink becomes “flat” when CO2 escapes from the solution. CO2 comes out of solution

faster at room temperature than at the cooler temperature of the refrigerator because the solubility of gases in liquids decreases as the temperature increases.

8.8 For most ionic and molecular solids, solubility generally increases as temperature increases.

The solubility of gases decreases with increasing temperature. Pressure changes do not affect the solubility of liquids and solids. The higher the pressure, the higher the solubility of a gas in a solvent.

Na2CO3(s) N2(g) a. increasing temperature increased solubility decreased solubility b. decreasing temperature decreased solubility increased solubility c. increasing pressure no change increased solubility d. decreasing pressure no change decreased solubility

8.9 Use the formula in Example 8.2 to solve the problem; 525 mg = 0.525 g bismuth subsalicylate.

Answer(w/v)% = 0.525 g bismuth

15 mL solution100%x = 3.5% (w/v) bismuth subsalicylate

8.10 Use the formula in Example 8.2 to solve the problems.

Answer(w/v)% = 4.3 g ethanol

30. mL solution100%x = 14% (w/v) ethanol

Answer(w/v)% = 0.021 g antiseptic

30. mL solution100%x = 0.070% (w/v) antiseptic

8.11 Use the formula in Example 8.3 to solve the problem.

(v/v)% = 21 mL ethanol250 mL mouthwash

100%x = 8.4% (v/v) ethanolAnswer


Chapter 8–3

8.12 Use the stepwise analysis in Example 8.4 to solve the problem.

known quantities

? mL ethanol

desired quantity

8.0% (v/v) mouthwash30. mL

8.0 mL ethanol100 mL mouthwash8.0 mL ethanol

100 mL mouthwashor

x =Answer

30. mL mouthwash 2.4 mL ethanol8.0 mL ethanol

100 mL mouthwash


known quantities

? mL solution

desired quantity

0.50% (w/v) vitamin C1,000. mg vitamin C

100 mL solution1 g1000 mg

or1000 mg

1 g

mg–g conversion factors

Choose the conversion factors with the unwanted units—mg and g—in the denominator.

0.50 g vitamin C100 mL solution

or

g–mL solution conversion factors

0.50 g vitamin C

x =Answer0.50 g vitamin C

100 mL solution1 g1000 mg

1000. mg vitamin C x 2.0 x 102 mL solution


known quantities

? mg of each drug

desired quantity

2.0% (w/v) guaifenisin3.0 tsp = 15 mL cough syrup

x =Answer1 g

1000 mg0.20 g dextromethorphan100 mL cough syrup

15 mL cough syrup x 30. mg dextromethorphan

0.20% (w/v) dextromethorphan

x =Answer1 g

1000 mg2.0 g guaifenisin100 mL cough syrup

15 mL cough syrup x 3.0 x 102 mg guaifenisin


Solutions 8–4

8.15 Use the formula, ppm = (g of solute)/(g of solution) × 106 to calculate parts per million as in Example 8.6.

0.042 mg DDT1 g

1000 mgx = 0.000 042 g DDT

1,400 g plankton106x = 0.030 ppm DDT

Answer

a.0.000 042 g DDT

5 x 10–4 g DDT1000 g1.0 kg

x = 1,000 g tissue1,000 g tissue

106x = 0.5 ppm DDTAnswer

b. 1.0 kg tissue

2.0 mg DDT1 g

1000 mgx = 0.0020 g DDT 1,000 g tissue106x = 2.0 ppm DDT

Answerc.

0.0020 g DDT

225 µg DDT1 g

1,000,000 µgx = 0.000 225 g DDT 1.0 x 103 g breast milk106x = 0.23 ppm DDT

Answerd.

0.000 225 g DDT

8.16 Calculate the molarity of each aqueous solution as in Example 8.7.

1.0 mol NaCl=

Answer0.50 L solutionM

moles of solute (mol)=

molarity

= 2.0 MV (L)

a.

2.0 mol NaCl

Answer0.25 L solutionM =

molarity

= 8.0 Mb. 250 mL1000 mL

x1 L

= 0.25 L

0.050 mol NaCl


molarity

= 10. Mc. 5.0 mL1000 mL

x1 L

= 0.0050 L

0.205 mol NaCl


molarity

=d. 12.0 g NaCl58.44 g NaCl

x1 mol NaCl = 0.205 mol NaCl 0.10 M

0.418 mol NaCl


molarity

=

e. 24.4 g NaCl58.44 g NaCl

x1 mol NaCl = 0.418 mol NaCl

1.2 M

350 mL1000 mL

x1 L

= 0.35 L

1.03 mol NaCl


molarity

=

f. 60.0 g NaCl58.44 g NaCl

x1 mol NaCl = 1.03 mol NaCl

1.4 M

750 mL1000 mL

x1 L

= 0.75 L


Chapter 8–5

8.17 Calculate the molarity of each solution to determine which has the higher concentration.

0.250 mol NaOH

Answermore concentrated

0.15 L solutionM =

molarity

=

10.0 g NaOH40.00 g NaOH

x 1 mol NaOH = 0.250 mol NaOH

1.7 M

150 mL1000 mL

x1 L

= 0.15 L

0.375 mol NaOH


molarity

=

15.0 g NaOH40.00 g NaOH

x 1 mol NaOH = 0.375 mol NaOH

1.5 M

250 mL1000 mL

x1 L

= 0.25 L

8.18 Use the molarity as a conversion factor to determine the number of milliliters as in Sample

Problem 8.9.

desired quantity

1.5 M solution0.15 mol glucose

? V (L) solution

known quantities

V (L)M

=moles of solute (mol)

= 0.15 mol glucose

1.5 mol/L0.10 L solution=

0.10 L solution 1000 mL1 L

x = 1.0 x 102 mL glucose solutionAnswer

a.

V (L)

M= moles of solute (mol)

= 0.020 mol glucose

1.5 mol/L0.013 L solution= 1000 mL

1 Lx = 13 mL glucose solution

Answer

b.

V (L)M

= moles of solute (mol)

= 0.0030 mol glucose


1 Lx = 2.0 mL glucose solution

Answer

c.

V (L)

M= moles of solute (mol)

= 3.0 mol glucose


1 Lx = 2.0 x 103 mL glucose solution

Answer

d.

8.19 Use the molarity as a conversion factor to determine the number of moles.

mol = V (L) • M = (2.0 L)(2.0 mol/L) 4.0 mol NaCl=a.

mol = V (L) • M = (2.5 L)(0.25 mol/L) 0.63 mol NaCl=b.


Solutions 8–6

mol = V (L) • M = (0.025 L)(2.0 mol/L) 0.050 mol NaCl=c.

mol = V (L) • M = (0.25 L)(0.25 mol/L) 0.063 mol NaCl=d.

8.20 Use the molarity to convert the volume of the solution to moles of solute. Then use the molar mass to convert moles to grams as in Example 8.8.

=1.25 mol NaCl

1 L0.10 L solution x

volume molarity

0.125 mol NaCl x 58.44 g NaCl1 mol NaCl

= 7.3 g NaClAnswer

molar massconversion factor

a.

=1.25 mol NaCl

1 L2.0 L solution x 2.5 mol NaCl x 58.44 g NaCl

1 mol NaCl= 150 g NaCl

Answer

b.

=1.25 mol NaCl

1 L0.55 L solution x 0.69 mol NaCl x 58.44 g NaCl

1 mol NaCl= 40. g NaCl

Answer

c.

1.25 mol NaCl1 L

50. mL solution x x 58.44 g NaCl1 mol NaCl

= 3.7 g NaClAnswer

d. 1 L1000 mL

x

8.21 Use the molar mass to convert grams to moles. Then use the molarity to convert moles of solute

to volume of solution. Sample Problem 8.10 is similar, but the order of steps is different.

molarityconversion factor

molar massconversion factor

0.25 mol sucrose

1 L0.500 g sucrose xx

342.3 g sucrose1 mol sucrose = 5.8 mL solution

Answer

a.1 L

1000 mLx

0.25 mol sucrose

1 L2.0 g sucrose xx

342.3 g sucrose1 mol sucrose = 23 mL solution

Answer

b.1 L

1000 mLx

0.25 mol sucrose

1 L1.25 g sucrose xx

342.3 g sucrose1 mol sucrose = 15 mL solution

Answer

c.1 L

1000 mLx

0.25 mol sucrose

1 L50.0 mg sucrose xx

342.3 g sucrose1 mol sucrose = 0.58 mL solution

Answer

d.1 L

1000 mLx1000 mg

1 g x


Chapter 8–7

8.22 Calculate a new molarity (M2) using the equation, M1V1 = M2V2.

V2

M1V1=M2 =(3.8 M)(25.0 mL)

(275 mL)0.35 M glucose solutionAnswer

=

8.23 Calculate the volume needed (V1) using the equation, M1V1 = M2V2.

M1

M2V2=V1 =(2.5 M)(525 mL)

6.0 M220 mLAnswer

=a.

=(4.0 M)(750 mL)

6.0 M5.0 x 102 mLAnswer

=b.M1

M2V2=V1

M1

M2V2=V1 =(0.10 M)(450 mL)

6.0 M7.5 mLAnswer

=c.

M1

M2V2=V1 =(3.5 M)(25 mL)

6.0 M15 mLAnswer

=d.

8.24 Calculate the new concentration of ketamine, and determine the volume needed to supply 75 mg.

V2

C1V1=C2 =(100. mg/mL)(2.0 mL)

10.0 mL20. mg/mL=

75 mg x20. mg

1 mL3.8 mLAnswer

=

8.25 Determine the number of “particles” contained in the solute. Use 0.51 oC/mol as a conversion factor to relate the temperature change to the number of moles of solute particles, as in Example 8.10.

0.51 oCmol particles x 2.0 mol 1.0 oC=

a. 2.0 mol of sucrose molecules

Boiling point = 100.0 °C + 1.0 °C = 101.0 °C

Boiling point =

b. 2.0 mol of KNO3

0.51 oCmol particles x 2.0 mol KNO3 2.0 oC=

100.0 °C + 2.0 °C = 102.0 °C

x 2 mol particlesmol KNO3


Solutions 8–8

c. 2.0 mol of CaCl2

0.51 oCmol particles x 2.0 mol CaCl2 3.1 oC=

100.0 °C + 3.1 °C = 103.1 °C

x 3 mol particlesmol CaCl2

Boiling point = d. 20.0 g of NaCl

0.51 oCmol particles

x 0.342 mol NaCl 0.35 oC=

100.0 °C + 0.35 °C = 100.35 °C rounded to 100.4 °C

1 mol NaCl58.44 g NaClx20.0 g NaCl

x 2 mol particlesmol NaCl

= 0.342 mol NaCl

Boiling point = 8.26 Determine the number of “particles” contained in the solute. Use 1.86 oC/mol as a conversion

factor to relate temperature change to the number of moles of solute particles, as in Sample Problem 8.14.

Melting point =mol particles

x 2.0 mol 3.7 oC=

a. 2.0 mol of sucrose molecules

–3.7 °C1.86 oC

Melting point =

b. 2.0 mol of KNO3

mol particlesx 2.0 mol KNO3 7.4 oC= –7.4 oCx 2 mol particles

mol KNO3

1.86 oC

Melting point =

c. 2.0 mol of CaCl2

mol particlesx 2.0 mol CaCl2 11 oC= –11 oCx 3 mol particles

mol CaCl2

1.86 oC

Melting point =

d. 20.0 g of NaCl

mol particlesx 0.342 mol NaCl 1.27 oC= –1.27 oC

1 mol NaCl58.44 g NaClx20.0 g NaCl

x 2 mol particlesmol NaCl

= 0.342 mol NaCl

1.86 oC

8.27 Determine the number of moles of ethylene glycol. Use 1.86 oC/mol as a conversion factor to

relate temperature change to the number of moles of solute particles, as in Sample Problem 8.14.

1.86 oCtemperature decrease

=mol particles

x 4.0 mol C2H6O2 – 7.4 °C=

1 mol250 g ethylene glycol x62.07 g

4.0 mol ethylene glycol=


Chapter 8–9

8.28 The solvent (water) flows from the less concentrated solution to the more concentrated solution.

a. A 5.0% sugar solution has greater osmotic pressure than a 1.0% sugar solution. b. 4.0 M NaCl has greater osmotic pressure than 3.0 M NaCl. c. 0.75 M NaCl has greater osmotic pressure than 1.0 M glucose solution because NaCl has two

particles for each NaCl, whereas glucose has only one. 8.29 a. Water flows across the membrane.

b. More water initially flows from the 1.0 M side to the more concentrated 1.5 M side. When equilibrium is reached there is equal flow of water in both directions.

c. The height of the 1.5 M side will be higher and the height of the 1.0 M NaCl will be lower. 8.30 A hypotonic solution has a lower osmotic pressure than body fluids.

A hypertonic solution has a higher osmotic pressure than body fluids.

a. A 3% (w/v) glucose solution is hypotonic, so it will cause hemolysis. b. A 0.15 M KCl solution is isotonic, so no change results. c. A 0.15 M Na2CO3 solution is hypertonic, so it will cause crenation.

Solutions to End-of-Chapter Problems 8.31 A shows KI dissolved in water, since the K+ and I– ions are separated when KI is dissolved. The

solution contains ions, so it conducts an electric current. B shows KI as a molecule, without being separated into ions, a situation that does not occur.

8.32 B shows NaF dissolved in water, since the Na+ and F– ions are separated when NaF is dissolved.

The solution contains ions, so it conducts an electric current. A shows NaF as a molecule, without being separated into ions, a situation that does not occur.

8.33 Use the definitions from Answer 8.1 to classify each substance as a heterogeneous mixture, a

solution, or a colloid.

a. bronze (alloy of Sn and Cu): solution d. household ammonia: solution b. diet soda: solution e. gasoline : solution c. orange juice with pulp: heterogeneous mixture f. fog: colloid

8.34 Use the definitions from Answer 8.1 to classify each substance as a heterogeneous mixture, a

solution, or a colloid. a. soft drink: solution d. lava rock: heterogeneous mixture b. cream: colloid e. bleach: solution c. wine: solution f. apple juice: solution 8.35 A solution that has less than the maximum number of grams of solute is said to be unsaturated.

A solution that has the maximum number of grams of solute that can dissolve is said to be saturated. A solution that has more than the maximum number of grams of solute is said to be supersaturated.


Solutions 8–10

a. Adding 30 g to 100 mL of H2O at 20 oC: unsaturated b. Adding 65 g to 100 mL of H2O at 50 oC: saturated c. Adding 20 g to 50 mL of H2O at 20 oC: saturated d. Adding 42 g to 100 mL of H2O at 50 oC and slowly cooling to 20 oC to give a clear solution

with no precipitate: supersaturated 8.36 A solution that has less than the maximum number of grams of solute is said to be unsaturated.

A solution that has the maximum number of grams of solute that can dissolve is said to be saturated. A solution that has more than the maximum number of grams of solute is said to be supersaturated.

a. Adding 200 g to 100 mL of H2O at 20 °C: unsaturated b. Adding 245 g to 100 mL of H2O at 50 °C: unsaturated c. Adding 110 to 50 mL of H2O at 20 °C: saturated

d. Adding 220 g to 100 mL of H2O at 50 °C and slowly cooling to 20 °C to give a clear solution with no precipitate: supersaturated

8.37 Use the general solubility rule: “Like dissolves like.” Generally, ionic and polar compounds are

soluble in water. Nonpolar compounds are soluble in nonpolar solvents.

LiClCC

CC C

C

H

H

H H

CH3

H

C CO

CH

HH

HH

HNa3PO4a. b. c. d.

ionicwater soluble

nonpolarNOT water soluble

polarwater soluble

ionicwater soluble

8.38 Use the general solubility rule: “Like dissolves like.” Generally, ionic and polar compounds are

soluble in water, whereas nonpolar compounds are soluble in nonpolar solvents.

C5H12 CH3Bra. b. c. d.

polarwater soluble

polarwater soluble

nonpolarNOT water soluble

C N HH

HH H

CaCl2ionic

water soluble

8.39 Methanol will hydrogen bond to water.

hydrogen bond


Chapter 8–11

8.40 Dimethyl ether can hydrogen bond with water.

H

OH

hydrogen bond

8.41 Water-soluble compounds are ionic or are small polar molecules that can hydrogen bond with the water solvent, but nonpolar compounds, such as oil, are soluble in nonpolar solvents.

8.42 A bottle of salad dressing that contains oil and vinegar has two layers because the oil is nonpolar

and the vinegar is polar. A heterogenous mixture is formed because the oil and vinegar are insoluble in each other.

8.43 Iodine would not be soluble in water but is soluble in CCl4 since I2 is nonpolar and CCl4 is a

nonpolar solvent. 8.44 Glycine is soluble in water because the nitrogen and oxygen atoms can form hydrogen bonds to

water. 8.45 Cholesterol is not water soluble because it is a large nonpolar molecule with a single OH group. 8.46 a. KCl and CCl4 do not form a solution. b. 1-Propanol (C3H8O) and H2O do form a solution. c. Cyclodecanone (C10H18O) and H2O do not form a solution. d. Pentane (C5H12) and hexane (C6H14) do form a solution. 8.47 The solubility of gases decreases with increasing temperature. The H2O molecules are omitted in

each representation.

a. b.

Less O2 is dissolved at 50 °C. More O2 is dissolved at 10 °C.


Solutions 8–12

8.48 The solubility of most ionic solids increases with increasing temperature and decreases with decreasing temperature. The H2O molecules are omitted in each representation.

= Cl= Na+

a.

More NaCl is dissolved at 50 oC

b.

Less NaCl is dissolved at 10 oC 8.49 For most ionic and molecular solids, solubility generally increases as temperature increases.

The solubility of gases decreases as temperature increases. Pressure changes do not affect the solubility of liquids and solids. The higher the pressure, the higher the solubility of a gas in a solvent.

NaCl(s) a. increasing temperature increased solubility b. decreasing temperature decreased solubility c. increasing pressure no change d. decreasing pressure no change

8.50 For most ionic and molecular solids, solubility generally increases as temperature increases.

The solubility of gases decreases as temperature increases. Pressure changes do not affect the solubility of liquids and solids. The higher the pressure, the higher the solubility of a gas in a solvent.

He(g) a. increasing temperature decreased solubility b. decreasing temperature increased solubility c. increasing pressure increased solubility d. decreasing pressure decreased solubility

8.51 A decrease in temperature (a) increases the solubility of a gas and (b) decreases the solubility of a

solid. 8.52 A decrease in pressure (a) decreases the solubility of a gas and (b) has no affect on the solubility

of a solid. 8.53 Many ionic compounds are soluble in water because the ion–dipole interactions between ions and

water provide the energy needed to break apart the ions from the crystal lattice. The water molecules form a loose shell of solvent around each ion.

8.54 Some ionic compounds are insoluble in water because the attractions between the ions in the

crystalline solid are stronger than the forces of attraction between the ions and water.


Chapter 8–13

8.55 Identify the cation and anion and use the solubility rules to predict if the ionic compound is water soluble. a. K2SO4: cation K+ with anion SO4

2–—water soluble. b. MgSO4: cation Mg2+ with anion SO4

2–—water soluble. c. ZnCO3: cation Zn2+ with anion CO3

2–—not water soluble. d. KI: cation K+ with anion I–—water soluble. e. Fe(NO3)3: cation Fe3+ with anion NO3

–—water soluble. f. PbCl2: cation Pb2+ with anion Cl–—not water soluble. g. CsCl : cation Cs+ with anion Cl–—water soluble. h. Ni(HCO3)2: cation Ni2+ with anion HCO3

–—not water soluble. 8.56 Identify the cation and anion and use the solubility rules to predict if the ionic compound is water

soluble. a. Al(NO3)3: cation Al3+ with anion NO3

–—water soluble. b. NaHCO3: cation Na+ with anion HCO3

–—water soluble. c. Cr(OH)2: cation Cr2+ with anion –OH—not water soluble. d. LiOH: cation Li+ with anion –OH—water soluble. e. CuCO3: cation Cu2+ with anion CO3

2–—not water soluble. f. (NH4)2SO4: cation NH4

+ with anion SO42–—water soluble.

g. Fe(OH)3: cation Fe3+ with anion –OH—not water soluble. h. (NH4)3PO4: cation NH4

+ with anion PO43–—water soluble.

8.57 Write two conversion factors for each concentration.

5% (w/v)100 mL 5 g

6.0 M 1.0 L 6.0 mol

10 ppm106 g 10 g

a.

b.

c.5 g

6.0 mol

10 g 106 g100 mL

1.0 L

8.58 Write two conversion factors for each concentration.

15% (v/v)100 mL 15 mL

12.0 M 1.0 L 12.0 mol

15 ppm106 g 15 g

a.

b.

c.15 mL

12.0 mol

15 g 106 g100 mL

1.0 L


Answer(w/v)% = 10.0 g LiCl

750 mL solution100%x = 1.3% (w/v) LiCla.

Answer(w/v)% = 25 g NaNO3

150 mL solution100%x = 17% (w/v) NaNO3b.

Answer(w/v)% = 40.0 g NaOH

500. mL solution100%x = 8.00% (w/v) NaOHc.


Solutions 8–14


Answer(w/v)% = 5.5 g LiCl

550 mL solution100%x = 1.0% (w/v) LiCla.

Answer(w/v)% = 12.5 g NaNO3

250 mL solution100%x = 5.0% (w/v) NaNO3b.

Answer(w/v)% = 20.0 g NaOH

400. mL solution100%x = 5.00% (w/v) NaOHc.


(v/v)% = 25 mL ethyl acetate150 mL solution

100%x = 17% (v/v) ethyl acetateAnswer


(v/v)% = 75 mL acetone250 mL solution

100%x = 30. % (v/v) acetoneAnswer


3.5 mol KCl=

Answer1.50 L solutionM


molarity

= 2.3 MV (L)

a.

0.44 mol NaNO3=Answer0.855 L solution

Mmoles of solute (mol)=

molarity

=V (L)

b. 0.855 L solution1 L1000 mL

855 mL solution =x

0.51 M

0.428 mol NaCl=Answer0.65 L solution


molarity

=V (L)

c.

0.65 L solution1 L1000 mL

650 mL solution =x

0.66 M

0.428 mol NaCl1 mol NaCl

58.44 g25.0 g NaCl =x

=Answer3.3 L solution


molarity

=V (L)

d.

0.036 M

0.119 mol NaHCO31 mol NaHCO3

84.01 g10.0 g NaHCO3 =x

0.119 mol NaHCO3


Chapter 8–15


2.4 mol NaOH=Answer1.50 L solution


molarity

= 1.6 MV (L)

a.

0.48 mol KNO3=Answer0.75 L solutionM


molarity

=V (L)

b. 0.75 L solution1 L1000 mL

750 mL solution =x

0.64 M

0.335 mol KCl=Answer0.65 L solution


molarity

=V (L)

c.

0.65 L solution1 L1000 mL

650 mL solution =x

0.52 M

0.335 mol KCl1 mol KCl

74.55g25.0 g KCl =x

=Answer3.8 L solution


molarity

=V (L)

d.

0.025 M

0.0943 mol Na2CO31 mol Na2CO3

105.99 g10.0 g Na2CO3 =x

0.0943 mol Na2CO3

8.65

a. Add 12 g of acetic acid to the flask and then water to bring the volume to 250 mL.

(w/v)% = x g acetic acid250 mL solution

100%x = 4.8% (w/v) acetic acidAnswer

x = 12 g acetic acid

b. Add 55 mL of ethyl acetate to the flask and then water to bring the volume to 250 mL.

(v/v)% = x mL ethyl acetate250 mL solution

100%x = 22% (v/v) ethyl acetateAnswer

x = 55 mL ethyl acetate

c. Add 37 g of NaCl to the flask and then water to bring the volume to 250 mL.

Answer

(0.25 L solution)


molarity

=

V (L)

(2.5 M)

1 mol NaCl58.44 g =x0.63 mol NaCl

0.63 mol NaCl

37 g NaCl

moles of solute (mol) =(M)(V) =


Solutions 8–16

8.66 a. Add 5.0 g of KCl to the flask and then water to bring the volume to 250 mL.

(w/v)% = x g KCl

250 mL solution100%x = 2.0% (w/v) KCl

Answerx = 5.0 g KCl

b. Add 85 mL of ethanol to the flask and then water to bring the volume to 250 mL.

(v/v)% = x mL ethanol250 mL solution

100%x = 34% (v/v) ethanolAnswer

x = 85 mL ethanol

c. Add 58 g of NaCl to the flask and then water to bring the volume to 250 mL.

Answer

(0.25 L solution)


molarity

=

V (L)

(4.0 M)

1 mol NaCl58.44 g =x1.0 mol NaCl

1.0 mol NaCl

58 g NaCl

moles of solute (mol) =(M)(V) =

8.67 Calculate the moles of solute in each solution.

Answer(0.15 L solution)= (0.25 M) 0.038 mol NaNO3moles of solute (mol) (M)= (V) =a.

Answer(0.045 L solution)= (2.0 M) 0.090 mol HNO3moles of solute (mol) (M)= (V) =b.

Answer(2.5 L solution)= (1.5 M) 3.8 mol HClmoles of solute (mol) (M)= (V) =c.

8.68 Calculate the moles of solute in each solution.

Answer

(0.25 L solution)= (0.55 M) 0.14 mol NaNO3moles of solute (mol) (M)= (V) =a.

Answer(0.145 L solution)= (4.0 M) 0.58 mol HNO3moles of solute (mol) (M)= (V) =b.

Answer(6.5 L solution)= (2.5 M) 16 mol HClmoles of solute (mol) (M)= (V) =c.

8.69 Calculate the number of grams of each solute.

Answer0.038 mol NaNO3a.

1 mol NaNO3

85.00 g NaNO3x = 3.2 g NaNO3


Chapter 8–17

Answer0.090 mol HNO3b.

1 mol HNO3

63.02 g HNO3x = 5.7 g HNO3

Answer3.8 mol HClc.

1 mol HCl36.46 g HCl

x = 140 g HCl

8.70 Calculate the number of grams of each solute.

Answer0.14 mol NaNO3a.

1 mol NaNO3

85.00 g NaNO3x = 12 g NaNO3

Answer0.58 mol HNO3b.

1 mol HNO3

63.02 g HNO3x = 37 g HNO3

Answer16 mol HClc.

1 mol HCl36.46 g HCl

x = 580 g HCl

8.71 Calculate the number of milliliters of ethanol in the bottle of wine.

(v/v)% = x mL ethanol750 mL solution

100%x = 11.0% (v/v) ethanolAnswer

x = 83 mL ethanol

8.72 Use the density and molar mass to calculate the molarity.

Answer1 mL ethanol0.790 g ethanol

x = 3.43 M46.07 g ethanolmol ethanol

x1 L

1000 mLx100 mL

20.0 mL ethanol

8.73 Calculate the amount of acetic acid (number of grams and number of moles) in the solution and

convert to molarity.

(w/v)% = x g acetic acid1890 mL solution

100%x = 5.0% (w/v) acetic acid x = 95 g acetic acida.Answer

95 g acetic acid x

1 mol acetic acid60.05 g acetic acid

=Answer

1.6 mol acetic acidb.

c.1.6 mol acetic acid


molarityV (L)

=1.89 L solution Answer

0.85 M=

8.74 Calculate the amount of glucose and convert to molarity.

Answerx = 0.83 M

180.18 g glucosemol glucose

x1 L

1000 mL100 mL

15 g glucose


Solutions 8–18

8.75 Use the formula, ppm = (g of solute)/(g of solution) × 106, to calculate parts per million as in Example 8.6.

80 µg CHCl31 g

1,000,000 µgx = 0.000 08 g CHCl3 1,000 g106x = 0.08 ppm CHCl3

Answera.

0.000 08 g CHCl3

700 µg glyphosate1 g

1,000,000 µgx = 0.0007 g glyphosate

1,000 g106x = 0.7 ppm glyphosate

Answer

b.

0.0007 g glyphosate

8.76 Use the formula ppm = (g of solute)/(g of solution) × 106, to calculate parts per million as in

Example 8.6.

1,300 µg Cu1 g

1,000,000 µgx = 0.001 3 g Cu 1,000 g106x = 1.3 ppm Cu

Answera.

0.001 3 g Cu

10 µg As1 g

1,000,000 µgx = 0.000 01 g As 1,000 g106x = 0.01 ppm As

Answerb.

0.000 01 g As

100 µg Cr1 g

1,000,000 µgx = 0.000 1 g Cr 1,000 g106x = 0.1 ppm Cr

Answerc.

0.000 1 g Cr

8.77 When solution X is diluted, the volume will increase, but the amount of solute will stay the same

(B).

X A B CIncreased volumedecreased solute

Increased volumesame solute

Same volumedecreased solute

8.78 C is the most concentrated and A is the least concentrated.


Chapter 8–19

8.79

A B

Add H2O.B has half as much NaCl, and is therefore half the molarity:0.05 M.

M1V1 = (0.1 M)(50.0 mL)M2V2 = (0.05 M)(x mL)=

100 mL=x

a.

b.

Since the original volume was 50 mL, 50 mL of water is added. 8.80

C D

Add H2O.2.0 M

M1V1 = (2.0 M)(100. mL)M2V2 = (1.2 M)(x mL)=

1.7 x 102 mL=x

a.

b.

Since the original volume was 100 mL, 70 mL of water is added.

x 35

= 1.2 M

8.81 Concentration is the amount of solute per unit volume in a solution. Dilution is the addition of

solvent to decrease the concentration of solute. 8.82 Dilution is the addition of solvent to decrease the concentration of solute. The amount of solute is

a constant. The addition of solvent would decrease the molar concentration of the 2.5 M NaOH solution, not increase it, so it is impossible to prepare 200 mL of a 5.0 M NaOH solution by diluting a 2.5 M NaOH solution.

8.83 Use the equation, (C1)(V1) = (C2)(V2), to calculate the new concentration after dilution.

V2

C1V1=C2 =(30.0%)(100. mL)

200. mL15.0% (w/v)=a.

V2

C1V1=C2 =(30.0%)(100. mL)

500. mL6.00% (w/v)=b.

V2

C1V1=C2 =(30.0%)(250 mL)

1500 mL5.0% (w/v)=c.

V2

C1V1=C2 =(30.0%)(350 mL)

750 mL14% (w/v)=d.


Solutions 8–20

8.84 Use the equation (C1)(V1) = (C2)(V2) to calculate the new concentration after dilution.

(w/v)% = 1.00 g10.0 mL solution

100%x = 10.0% (w/v)a.

V2

C1V1=C2 =(10.0%)(1.0 mL)

10.0 mL1.0% (w/v)=b. [1]

V2

C1V1=C2 =(10.0%)(1.0 mL)

2.5 mL4.0% (w/v)=b. [2]

V2

C1V1=C2 =(10.0%)(1.0 mL)

50.0 mL0.20% (w/v)=b. [3]

V2

C1V1=C2 =(10.0%)(1.0 mL)

120 mL0.083% (w/v)=b. [4]

8.85 Use the equation, (M1)(V1) = (M2)(V2), to calculate the new molarity after dilution.

V2

M1V1=M2 =(12.0 M)(125 mL)

850 mL1.8 M=

8.86 Use the equation (M1)(V1) = (M2)(V2) to calculate the new molarity after dilution.

V2

M1V1=M2 =(6.0 M)(250 mL)

450 mL3.3 M=

8.87 Use the equation, (M1)(V1) = (M2)(V2), to calculate the volume needed to prepare each solution.

M1

M2V2=V1 =(1.0 M)(25 mL)

2.5 M10. mL=a.

M1

M2V2=V1 =(0.75 M)(1500 mL)

2.5 M450 mL=b.

M1

M2V2=V1 =(0.25 M)(15 mL)

2.5 M1.5 mL=c.

M1

M2V2=V1 =(0.025 M)(250 mL)

2.5 M2.5 mL=d.

8.88 Use the equation (M1)(V1) = (M2)(V2) to calculate the volume needed to prepare each solution.

M1

M2V2=V1 =(4.0 M)(45 mL)

5.0 M36. mL=a.


Chapter 8–21

M1

M2V2=V1 =(0.5 M)(150 mL)

5.0 M20 mL=b.

M1

M2V2=V1 =(0.025 M)(1,200 mL)

5.0 M6.0 mL=c.

M1

M2V2=V1 =(1.0 M)(750 mL)

5.0 M150 mL=d.

8.89 Ocean water contains nonvolatile dissolved salts, which increase the boiling point. 8.90 Osmotic pressure is the pressure that prevents the flow of additional solvent into a solution on one

side of a semipermeable membrane. Pure water will not have osmotic pressure since water is on both sides of the semipermeable membrane and there will be no pressure difference.

8.91 Determine the number of “particles” contained in the solute. Use 0.51 oC/mol as a conversion

factor to relate temperature change to the number of moles of solute particles, as in Example 8.10.


x 3.0 mol 1.5 oC=

a. 3.0 mol of fructose molecules

100.0 °C + 1.5 °C = 101.5 °C

b. 1.2 mol of KI


x 1.2 mol KI 1.2 oC= 100.0 °C + 1.2 °C = 101.2 °Cx 2 mol particlesmol KI

c. 1.5 mol Na3PO4


x 1.5 mol Na3PO4 3.1 oC= 100.0 °C + 3.1 °C = 103.1 °Cx 4 mol particlesmol Na3PO4

8.92 Determine the number of “particles” contained in the solute. Use 1.86 °C/mol as a conversion

factor to relate temperature change to the number of moles of solute particles, as in Example 8.10.


x 3.0 mol 5.6 oC=

a. 3.0 mol of fructose molecules

0.0 °C – 5.6 °C = –5.6 °C

b. 1.2 mol of KI


x 1.2 mol KI 4.5 oC= 0.0 °C – 4.5 °C = – 4.5 °Cx 2 mol particlesmol KI


Solutions 8–22

c. 1.5 mol Na3PO4


x 1.5 mol Na3PO4 11 oC= 0.0 °C – 11 °C = –11 °Cx 4 mol particlesmol Na3PO4

8.93 Determine the number of “particles” contained in the solute. Use 1.86 oC/mol as a conversion


150 g ethylene glycol

(C2H6O2)

mol particlesx 2.4 mol 4.5 oC= Melting point = –4.5 oC

1 mol ethylene glycol62.07 g ethylene glycol

x = 2.4 mol ethylene glycol

1.86 oC

8.94 Determine the number of “particles” contained in the solute. Use 1.86 °C/mol as a conversion


5.4 mol ethylene glycol

mol particlesx x mol 10. oC=

1 mol ethylene glycol62.07 g ethylene glycol

x = 330 g ethylene glycol

1.86 oC

!T = 0.0 oC – 10. oC = –10. oC

x mol = 5.4 mol

8.95 a. The NaCl solution has a higher boiling point because it contains two particles per mole,

whereas the glucose solution contains one per mole. b. The glucose solution has the higher melting point because it contains one particle per mole,

whereas the NaCl solution contains two particles per mole. c. The NaCl solution has the higher osmotic pressure because it contains two particles per mole,

whereas the glucose solution contains one per mole. d. The glucose solution has a higher vapor pressure at a given temperature because it contains one

particle per mole, whereas the NaCl solution contains two particles per mole. 8.96 a. The CaCl2 solution has a higher boiling point because it contains three particles per mole,

whereas the NaCl solution contains two particles per mole. b. The NaCl solution has the higher melting point because it contains two particles per mole,

whereas the CaCl2 solution contains three particles per mole. c. The CaCl2 solution has the higher osmotic pressure because it contains three particles per mole,

whereas the NaCl solution contains two particles per mole. d. The NaCl solution has a higher vapor pressure at a given temperature because it contains two

particles per mole, whereas the CaCl2 solution contains three particles per mole.


Chapter 8–23

8.97 Determine the number of “particles” contained in the solute. If necessary, use 1.86 oC/mol as a conversion factor to relate the temperature change to the number of moles of solute particles.

a. A 0.10 M glucose solution has a higher melting point than 0.10 M NaOH even though the

solutions have the same molar concentration, because the NaOH solution contains twice as many particles. Therefore the NaOH solution will have its melting point reduced by twice as much as the glucose solution.

b. A 0.20 M NaCl solution has a higher melting point than a 0.15 M CaCl2 solution.

mol particles x 0.20 mol NaCl 0.74 oC= Melting point = –0.74 oCx 2 mol particlesmol NaCl

1.86 oC

mol particles x 0.15 mol CaCl2 0.84 oC= Melting point = –0.84 oCx 3 mol particlesmol CaCl2

1.86 oChigher melting point

c. A 0.10 M Na2SO4 solution has a higher melting point than 0.10 M Na3PO4 even though the solutions have the same molar concentration, because the Na2SO4 solution contains fewer particles (three vs. four).

d. A 0.10 M glucose solution has a higher melting point than a 0.20 M glucose solution since it

has a lower molar concentration, which causes less melting point depression. 8.98 Determine the number of “particles” contained in the solute. If necessary, use 0.51 °C/mol as a

conversion factor to relate the temperature change to the number of moles of solute particles.

a. A 0.10 M NaOH solution has a higher boiling point than 0.10 M glucose, even though the solutions have the same molar concentration, because the NaOH solution contains twice as many particles. Therefore, the NaOH solution will have its boling point increased by twice as much as the glucose solution.

b. A 0.15 M CaCl2 solution has a higher boiling point than a 0.20 M NaCl solution.

mol particles x 0.20 mol NaCl 0.20 oC= Boiling point = 100.20 oCx 2 mol particlesmol NaCl

0.51 oC

mol particles x 0.15 mol CaCl2 0.23 oC= Boiling point = 100.23 oCx 3 mol particlesmol CaCl2

0.51 oC

higher boiling point c. A 0.10 M Na3PO4 solution has a higher boiling point than 0.10 M Na2SO4, even though the

solutions have the same molar concentration, because the Na2SO4 solution contains fewer particles (three vs. four).

d. A 0.20 M glucose solution has a higher boiling point than a 0.10 M glucose solution since it

has a higher molar concentration, which causes an increased boiling point elevation.


Solutions 8–24

8.99 a. A > B. Since solution A contains a dissolved solute, water will flow from compartment B to compartment A.

b. B > A. Since solution B is more concentrated, water will flow from compartment A to compartment B.

c. No change will occur since the solutions have an equal number of dissolved particles. d. A > B. Solution A contains more dissolved particles than solution B. Therefore, water will

flow from compartment B to compartment A. e. No change will occur since the solutions have an equal number of dissolved particles (NaCl

has two particles per mole, making it equivalent to the glucose solution). 8.100 a. Diagram 2 represents the final level of the liquid since solution B is more concentrated. Thus,

water will flow from compartment A to compartment B. b. Diagram 3 represents the final level of the liquid since solution A contains more dissolved

particles than solution B. Therefore, water will flow from compartment B to compartment A. c. Diagram 2 represents the final level of the liquid since solution B contains a dissolved solute.

Thus, water will flow from compartment A to compartment B. d. Diagram 3 represents the final level of the liquid since solution A contains a dissolved solute.

Thus, water will flow from compartment B to compartment A. e. Diagram 3 represents the final level of the liquid since solution A is more concentrated. Thus,

water will flow from compartment B to compartment A. 8.101 At warmer temperatures, CO2 is less soluble in water and more is in the gas phase and escapes as

the can is opened and pressure is reduced. 8.102 Sugar is more soluble at higher temperatures, so more sugar dissolves in hot coffee than iced

coffee. 8.103 Calculate the weight/volume percent concentration of glucose, and the molarity. (w/v)% = 0.09 g glucose

100 mL blood100%x = 0.09% (w/v) glucose

Answer

0.09 g glucose x 1 mol glucose180.2 g glucose

= 0.0005 mol glucose 0.0005 mol glucose =0.1 L blood

0.005 MAnswer

8.104 Convert L to mL and mg to g.

90. mg glucose100 mL blood

xx 1000 mL1 L

= 4.5 g glucoseAnswer

5.0 L 1 g1000 mg

x

8.105 a. 280 mL of mannitol solution would have to be given.

(w/v)% = 70. g mannitolx mL solution

100%x = 25% (w/v) mannitolAnswer

x = 280 mL

b. The hypertonic mannitol solution draws water out of swollen brain cells and thus reduces the

pressure on the brain.


Chapter 8–25

8.106 Calculate the mass of glucose and then convert to moles.

10. g glucose100 mL solution

750 mL solutionx = 75 g glucoseAnswer

75 g glucose x 1 mol glucose180.2 g glucose

= 0.42 mol glucoseAnswer

a.

b.

8.107 When a cucumber is placed in a concentrated salt solution, water moves out of the cells of the

cucumber to the hypertonic salt solution, so the cucumber shrinks and loses its crispness. 8.108 When a raisin is placed in water, water moves into the raisin (hypotonic solution), so the raisin

swells. 8.109 NaCl, KCl, and glucose are found in the bloodstream. If they weren’t in the dialyzer fluid, they

would move out of the bloodstream into the dialyzer, and their concentrations in the bloodstream would fall.

8.110 Pure water is not used in the solution contained in a dialyzer during hemodialysis because water

would move out of the dialyzer and into the bloodstream and dilute the blood. 8.111 Convert ounces to milliliters, and then calculate the weight/volume percent concentration.

(w/v)% = 15 g complex carbohydrates240 mL solution

100%x = 6.3% (w/v) complex carbohydratesAnswer

8.0 oz1 oz

29.6 mL240 mL=x

8.112 Convert ounces to millimeters, and then calculate the ppm concentration.

ppm =

25 mg Mg

x

= 0.025 g Mg

Answer

8.0 oz1 oz

29.6 mL 240 mL=x1.0 mL

1.0 gx = 240 g solution

1000 mg

1 gx

0.025 g Mg

240 g solution106 = 1.0 x 102 ppm

8.113 Use the molarity to determine the number of moles of HCl, and then convert this to grams.

0.10 M HCl 2.0 L 0.20 mol HCl=x

Answer0.20 mol HCl

1 mol HCl36.46 g HClx = 7.3 g HCl

mol = M x V =


Solutions 8–26

8.114 When a red blood cell is placed in pure water, water will move into the blood cell, causing it to swell and eventually rupture.

8.115

(w/v)% = x g ethanol5.0 x 103 mL blood

100%x = 0.08% (w/v) ethanol x = 4 g ethanol

Answer4 g ethanol x = 4,000 mg ethanol

1000 mg1 g

5.0 L = 5.0 x 103 mL blood

8.116

40. mL ethanol100 mL solution

250 mL solutionx =Answer

1.0 x 102 mL ethanol

8.117

1 mL106x = 15 ppm acetaminophen

x g acetaminophen

0.000 015 g acetaminophen

0.000 015 g acetaminophen x1 mol

151.2 g1 mL blood5000 mL blood x = 0.000 50 mol acetaminophen

Answer

106 µg1 g

= 15 µg acetaminophen, which is in the therapeutic rangex

x = 0.000 015 g acetaminophena.

b.

8.118 The two values are equivalent because 5.2 × 10–3 M is the same as 200. mg/dL to 2 significant

figures.

Lx5.2 x 10–3 mol

mole386.74 g x

1 g1000 mg x

10 dL1 L = 2.0 x 102 mg/dL


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Chapter 8 Solutions - Riverside City Collegewebsites.rcc.edu/grey/files/2012/02/Chapter-8.pdf · Chapter 8 Solutions Solutions to In-Chapter Problems ... 8.6 Magnesium salts are not