EXTRA HOMEWORK 3A - Saddleback College

EXTRA HOMEWORK 3A

1. Draw pictures on the atomic level for a solid, a liquid, and a gas.

2. What must be true about the kinetic energy of the particles making up a liquid if the liquid is to turn
into a gas?

3. What causes the high surface tension and high viscosity of molasses?

4. Draw pictures on the atomic level of a crystalline solid and an amorphous solid.

5. Define ionic bond.

6. Define metallic bond.

7. Define covalent bond.

8. Identify the particles that make up each of the following types of matter, and identify the interparticle
attractive forces found in each:

(a) Cr (b) CaBr2 (c) SiO2

9. Draw atomic level pictures of the three substances in question 8.

10. Given unlabeled samples of the three substances in question 8, how could they be identified
experimentally?

*11. Identify each substance as metallic, ionic, or macromolecular.

Substance Melting Electrical Conduction Water
Point (C) Solubility
1 As A Solid As A Liquid
2 845 high
3 1244 no yes low
1585 yes yes low
no no

EXTRA HOMEWORK 3B

1. Define each of the following:
(a) London dispersion force
(b) Dipole-dipole interaction

2. Identify the particles that make up each of the following types of matter, and identify the interparticle
attractive forces found in each:

(a) CH3OH (b) S8 (c) H2 (d) Cl2O
(e) U (f) NaNO3 (g) B (h) CH3CH2F

3. Draw atomic level pictures of the eight substances in question 2.

4. Given unlabeled samples of solid C12H22O11 and C14H10, how could they be identified experimentally?

*5. Identify each substance as metallic, ionic, macromolecular, polar molecular, or nonpolar molecular.

Substance Melting Electrical Conduction Water Hexane
Point (C) As A Solid As A Liquid Solubility Solubility
1
2 45 no no low high
3 112 no no high low
4 1475 no no low low
5 1545 no yes high low
2480 yes yes low low

EXTRA HOMEWORK FOR UNIT CELLS (OPTIONAL)
1. Nickel has a face-centered-cubic unit cell. The density of nickel is 6.84 g/cm3. Calculate the atomic

radius of nickel.
2. Tungsten metal exists in a body-centered-cubic structure. The atomic radius of tungsten is 0.139 nm.

Calculate the density of solid tungsten.
3. The radius of gold atoms is 0.144 nm and the density of gold is 19.32 g/cm3. Determine if gold has a

face-centered-cubic structure or a body-centered-cubic structure.
4. Unit cells for xenon fluoride and nickel arsenide are shown below. Determine their empirical

formulas.
(a) (b)

5. Unit cells for calcium chloride and lithium oxide are shown below. Determine their empirical
formulas.

6. Draw pictures on the atomic level of three layers of atoms that are cubic closest packing and
hexagonal closest packing.
(continued on next page)

*7. The unit cell for sodium chloride is shown below.

where the chloride ions create a face centered cubic unit cell pattern, and the sodium ions occupy the
octahedral holes in the unit cell. An ionic substance will crystallize in this pattern – with the smaller
ions occupying the octahedral holes – as long as the smaller ions are large enough to prevent the larger

ions from touching each other, and therefore repelling. How large must the radii of the smaller ions

be, in relationship to the larger ions, in order for the ionic substance to crystallize in this pattern?

EXTRA HOMEWORK 3C

1. Define electrical conduction.

2. Draw a band diagram for each of the following. (b) a nonmetallic network
(a) a metallic network (d) a p-type conductor
(c) an n-type conductor

*3. Aluminum antimonide is a semiconductor and is used as a laser. When energy is applied to an
aluminum antimonide crystal, electrons are promoted from the valence band to the conduction band.
When the electrons return to the valence band, the crystal emits light with a wavelength of 730. nm.
Calculate the energy of the band gap in aluminum antimonide.

EXTRA HOMEWORK 3D

1. Predict the ionic substance in each pair that has the greater melting point, and explain why:

(a) MgCl2 or AlCl3 (b) K2S or K2O (c) AlP or CsI

2. Predict the metallic substance in each pair that has the greater melting point, and explain why:

(a) Mg or Sr (b) Sc or V (c) Mn or Co

3. Assuming that silicon and germanium form macromolecular substances, tell which would have the
greater melting point, and explain why.

4. Draw Lewis structures and bond-orbital- models for each of the following molecular substances, and
predict whether they are polar or nonpolar.

(a) C2H2 (b) C2H4 (c) CH3CN (d) CH3NH2

5. Predict the molecular substance in each pair that has the greater melting point, and explain why:
(a) P4 or As4
(b) CO2 or SO2
(c) CH3NH2 or CH3CN
(d) Li2O or Br2O

6. Explain the boiling points of the following three substances:

C3H8 -42ºC C4H10 -1ºC C3H7OH 97ºC

7. For each set of substances, rank them from lowest to highest melting points, and explain why based
upon the type of particles and interparticle attractive forces found in each.

(a) NaF, LiF, HF (b) C6H6, C6H5CO2H, C10H8
(c) H2O, H2S, H2Se (d) Li, Be, C

*8. Predict the substance in each pair that has the greater melting point, and explain why:
(a) BCl3 or AlCl3

EXTRA HOMEWORK 3E

1. Explain how an increase in each of the following would affect an equilibrium between liquid water
and water vapor in a closed container:

(a) temperature (b) surface area

2. Calculate the amount of heat it would take to convert 50.0 g of water at 20.0C to steam at 150.0C.
The specific heat capacity for ice is 2.22 J/gC, the specific heat capacity for water is 4.18 J/gC, and
the specific heat capacity for steam is 2.01 J/gC. The heat of fusion for ice is 335 J/g and the heat of
vaporization for water is 2260 J/g.

*3. Calculate the amount of heat it would take to convert 50.0 g of methanol (CH3OH) at 20.0C to
130.0C. The specific heat capacity for solid methanol is 48.5 J/molC, the specific heat capacity for
liquid methanol is 79.5 J/molC, and the specific heat capacity for gaseous methanol is 52.3 J/molC.
The heat of fusion for methanol is 4.28 kJ/mol and the heat of vaporization for methanol is 37.4
kJ/mol. Methanol has a normal melting point of -97.0ºC and a normal boiling point of 64.7ºC

EXTRA HOMEWORK 3F

1. Consider Phase Diagram 1 for iodine below. Answer each of the following questions:
(a) What is the normal boiling point for iodine?
(b) What is the melting point for iodine at 1 atm?
(c) What phase is present at room temperature and normal atmospheric pressure?
(d) What phase is present at 186ºC and 1.0 atm?

Phase Diagram 1 Phase Diagram 2

2. Consider Phase Diagram 2 for sulfur above. The rhombic and monoclinic phases are two solid
allotropic phases. Answer each of the following questions:

(a) Below what pressure will solid sulfur sublime?

(b) Which of the two solid phases of sulfur is most dense?

EXTRA HOMEWORK 3G

1. Indicate which solvent, toluene (C7H8) or methanol (CH3OH), would be more suitable for dissolving
each of the following:

(a) CH3CH2CH2OH (b) LiNO3 (c) CCl4 (d) I2

2. For each of the following pairs, indicate which substance would be more soluble in water:

O O (b) CH4 or CH3F

|| ||

(a) CH3 C–H or CH3 C–OH

(c) vitamin A or vitamin C

3. Why does C5H11OH dissolve both polar and nonpolar substances?

4. Indicate each as a strong electrolyte, a weak electrolyte, or a nonelectrolyte:

(a) HCl (b) KI (c) C2H5OH (d) HF

5. State whether each of the following is a strong electrolyte, weak electrolyte, or nonelectrolyte, and
complete each equation. Showing how the solute would be expressed accurately in solution:

(a) Fe(NO3)3 (s) → (b) CH2OHCH2OH (l) →

(c) H2SO4 (g) → (d) H3PO4 (g) →

6. Complete the following table:

Soluble in H2O Ionizes or Dissociates in H2O
(Yes or No) (No, a little, a lot)

C3H6O3
C3H8
LiOH
NaI
HCN
HClO3

EXTRA HOMEWORK 3H

1. Calculate the mass percent of solute in each of the following solutions.
(a) 15.3 grams of sodium chloride dissolved in 155.0 grams of water.
(b) 19 grams of sulfur dissolved in 158 grams of benzene.

2. Calculate the mole fraction of solute in each of the following solutions.
(a) 0.25 moles of sucrose dissolved in 32.50 moles of water.
(b) 27.0 grams of ethanol (C2H5OH) dissolved in 175.0 grams of water.

3. Calculate the molarity of solute in each of the following solutions.
(a) 2.36 moles of dextrose dissolved in enough water to make 2.50 liters of solution.
(b) 18.5 g of sodium acetate trihydrate dissolved in enough water to make 750. milliliters of solution.

4. Calculate the molality of solute in each of the following solutions.
(a) 2.35 moles of naphthalene dissolved in 0.750 kilograms of carbon tetrachloride.
(b) 17.2 g of ethylene glycol (C2H6O2) dissolved in 500. grams of water.

5. Calculate the molarity of each ion present in each of the following solutions.

(a) 1.30 M potassium carbonate (b) 0.120 M aluminum sulfate

6. Calculate the number of grams of nickel (II) chloride hexahydrate needed to prepare 500.0 mL of a
0.100 M nickel (II) chloride solution.

7. A solution is prepared by dissolving 5.0 g of toluene (C7H8) in 225 g of benzene (C6H6). The density
of the solution is 0.876 g/mL. Fill out the unshaded regions in the table below, then calculate the
following concentration units

Mass (g) Quantity of Matter (mol) Volume (mL)

Solute
Solvent
Solution (total)

(a) mass percent of toluene (b) molarity of toluene
(c) molality of toluene (d) mole fraction of toluene

(continued on next page)

*8. A solution is prepared by dissolving 15.0 g of sodium nitrate in enough water to make the solution’s
volume 100.0 mL. The solution’s density was determined to be 1.12 g/mL. Calculate the

(a) mass percent of sodium nitrate (b) molarity of sodium nitrate

(c) molality of sodium nitrate (d) mole fraction of sodium nitrate

*9. The term “proof” is defined as twice the percent by volume of pure ethanol (C2H5OH) in solution.
Thus, a solution that is 95% ethanol by volume is 190 proof. What is the molarity of ethanol in a 151
proof ethanol/water solution? The density of pure ethanol is 0.79 g/mL and the density of pure water
is 1.00 g/mL.

*10. A solution is prepared by dissolving 516 mg of oxalic acid (H2C2O4) to make 100.0 mL of solution. A
10.00 mL portion of this solution is diluted to 250.0 mL. The density of the diluted solution is 1.07
g/mL. Fill out the table and then solve for the concentration units below.

Mass (g) Quantity of Matter (mol) Volume (ml)

Solute
Solvent
Solution (total)

(a) What is the molarity of the oxalic acid in the diluted solution?

(b) What is the mass percent of the oxalic acid in the diluted solution?

(c) What is the mole fraction of the oxalic acid in the diluted solution?

(d) What is the molality of the oxalic acid in the diluted solution?

(e) Suppose you wanted to make another batch of the oxalic acid solution with the same molality as
the diluted solution above, but you only had 45.0 mg of oxalic acid. What mass of water would
you have to use?

EXTRA HOMEWORK FOR CONCENTRATION UNITS (OPTIONAL)

1. Calculate the mass percent of solute for each of the following solutions.
(a) 31.0 grams of potassium chloride dissolved in 152.0 grams of water.
(b) 4.5 grams of toluene dissolved in 29.0 grams of benzene.
(c) 5.5 grams of sodium bromide dissolved to make 78.2 grams of solution.
(d) 22.0 grams of glucose (C6H12O6) dissolved in 115 mL of ethanol. The density of ethanol is
0.791 g/mL.

2. Calculate the mole fraction of solute for each of the following solutions.

(a) 38.4 grams of lithium nitrate dissolved in 175 grams of water.

(b) 33.0 g of sodium phosphate dissolved in 253 grams of water.

(c) 9.95 grams of anthracene (C12H10) dissolved in 100. mL of hexane (C6H14). The density of hexane
is 0.655 g/mL.

3. Calculate the molarity for each of the following solutions.
(a) 10.4 grams of calcium chloride dissolved in enough water to make 220. mL of solution.
(b) 6.57 grams of methanol (CH3OH) dissolved in enough water to make 150. mL of solution.
(c) 7.82 grams of naphthalene (C10H8) dissolved in enough benzene to make 85.2 mL of solution.
(d) 143 grams of barium chloride dihydrate dissolved in enough water to make 750. mL of solution.

4. Calculate the molality for each of the following solutions.

(a) 26.2 grams of magnesium chloride dissolved in 425 grams of water.

(b) 14.3 grams of sucrose (C12H22O11) dissolved in 676 grams of water.
(c) 7.82 grams of naphthalene (C10H8) dissolved in 85.0 mL of pentane (C5H12). The density of

pentane is 0.626 g/mL.

5. A solution is prepared by dissolving 16.45 g of magnesium chloride in enough water to make
100.0 mL of solution. Calculate the

(a) molarity of magnesium chloride (b) molarity of each ion present in the solution

6. A solution is prepared by dissolving 22.26 g of aluminum iodide in enough water to make 200.0 mL
of solution. Calculate the

(a) molarity of the aluminum iodide (b) molarity of each ion present in the solution

(continued on next page)

7. A solution is prepared by dissolving 12.5 g of iron (II) sulfate in 95.0 g of water. The volume of the
solution is 105.0 mL. Fill out the unshaded regions in the table below, then calculate the following
concentration units

Mass (g) Quantity of Matter (mol) Volume (mL)

Solute
Solvent
Solution (total)

(a) mass percent of iron (II) sulfate (b) molarity of iron (II) sulfate
(c) molality of iron (II) sulfate (d) mole fraction of iron (II) sulfate

8. A solution is prepared by dissolving 15.0 g of sodium nitrate in enough water to make the solution’s
volume 100.0 mL. The solution’s density was determined to be 1.12 g/mL. Fill out the table below,

then calculate the following concentration units

Mass (g) Quantity of Matter (mol) Volume (mL)

Solute
Solvent
Solution (total)

(a) mass percent of sodium nitrate (b) molarity of sodium nitrate
(c) molality of sodium nitrate (d) mole fraction of sodium nitrate

9. Concentrated hydrochloric acid is 37.0% HCl by mass and has a solution density of 1.19 g/mL. Fill
out the unshaded regions of the table below, then calculate the following concentration units

Mass (g) Quantity of Matter (mol) Volume (mL)

Solute
Solvent
Solution (total)

(a) molarity of HCl (b) molality of HCl
(c) mole fraction of HCl

EXTRA HOMEWORK 3I

1. Calculate the vapor pressure of each of the following solutions at 24C if the vapor pressure of pure
water is 22.4 torr at 24C.
(a) 0.45 moles of fructose (C6H12O6), a nonelectrolyte, dissolved in 20.00 moles of water.
(b) 0.45 moles of aluminum chloride dissolved in 20.00 moles of water.

2. Calculate the vapor pressure of each of the following solutions at 22C if the vapor pressure of pure
water is 19.8 torr at 22C.
(a) 21.5 grams of fructose (C6H12O6), a nonelectrolyte, dissolved in 175 g water.
(b) 21.5 grams of aluminum chloride dissolved in 175 g water.

3. Heptane (C7H16) and octane (C8H18) form an ideal solution. At 25C, the vapor pressures of heptane
and octane are 45.8 torr and 10.9 torr, respectively. A solution is made by mixing 50.0 mL of heptane
(density = 0.684 g/mL) with 50.0 mL of octane (density = 0.703 g/mL).
(a) Calculate the mole fractions of heptane and octane in the solution
(b) Calculate the equilibrium vapor pressure of the solution

4. Consider the following four liquids and/or solutions.
pure water
a 0.1 m solution of sucrose in water
a 0.1 m solution of potassium nitrate in water
a 0.1 m solution of calcium nitrate in water

Place the liquids and/or solutions in order of lowest vapor pressure to highest vapor pressure

*5. A solution is made by dissolving 28.5 grams of glycerin (C3H8O3), a nonelectrolyte, in 125 mL
methanol (CH3OH). Calculate the vapor pressure of the solution at 21.2C if the vapor pressure of
pure methanol is 100. torr at 21.2C, and methanol’s density is 0.792 g/mL.

*6. Calculate the vapor pressure at 25ºC of an aqueous solution that is 5.50% sodium chloride by mass.
The vapor pressure of pure water is 23.8 torr at 25C.

*7. 12.5 grams of a nonelectrolyte is dissolved in 75.0 grams of water at 22C, and the vapor pressure of
the solution is 18.6 torr. If the vapor pressure of pure water at 22C is 19.8 torr, calculate the molar
mass of the nonelectrolyte.

*8. Anthracene is 94.3% carbon and 5.7% hydrogen by mass. When 8.25 grams of anthracene is dissolved
in 38.50 grams of hexane (C6H12) at 25C, and the vapor pressure of the solution is 137 torr. If the
vapor pressure of pure hexane at 25C is 151 torr, determine the molar mass and the molecular
formula of anthracene.

EXTRA HOMEWORK 3J

1. Consider the following four liquids and/or solutions.

pure water
a 0.3 m solution of glucose in water
a 0.2 m solution of lithium bromide in water
a 0.1 m solution of magnesium bromide in water

Place the liquids and/or solutions in order of

(a) lowest freezing point to highest freezing point

(b) lowest boiling point to highest boiling point

(c) lowest osmotic pressure to highest osmotic pressure

2. A solution is made by dissolving 0.75 moles of lithium bromide in 0.30 kilograms of water. Calculate

(a) the freezing point of the solution (b) the boiling point of the solution.

3. A solution is made by dissolving 21.5 grams of copper (II) chloride in 425 grams water. Calculate

(a) the freezing point of the solution (b) the boiling point of the solution.

4. An organic solid is found to be 54.5% carbon, 9.2% hydrogen, and 36.3% oxygen by mass. A solution
containing 17.32 grams of the organic solid dissolved in 115 grams of carbon disulfide has a boiling
point of 50.2C. Determine the molecular formula of organic solid.

5. A 1.00 g sample of an organic compound is dissolved in 8.50 g of benzene, and the freezing point of
the solution is 0.3°C. If the compound is known to be 62.0% carbon, 10.4% hydrogen, and 27.6%
oxygen by mass, calculate the molecular formula of the compound.

*6. A 0.243 g sample of an organic compound is dissolved in 25.0 g water, and the freezing point of the
solution is -0.201°C.

(a) If the compound is known to be 53.31% carbon, 11.18% hydrogen, and 35.51% oxygen by mass,
calculate the molecular formula of the compound.

(b) Draw a Lewis structure for the compound with this formula that is capable of forming hydrogen
bonds

(c) Draw a Lewis structure compound with this formula that is not capable of forming hydrogen
bonds

*7. Cyclohexane freezes at 6.5ºC, and a solution of 3.57 grams of styrene (C8H8) dissolved in 27.67 grams
of cyclohexane freezes at -18.4ºC. A second solution is prepared by dissolving 4.23 grams of
phenylacetylene in 26.34 grams of cyclohexane, and this solution freezes at -25.1ºC. Phenylacetylene
is 94.1% carbon and 5.9% hydrogen by mass. Determine the

(a) molal freezing point constant for cyclohexane

(b) molecular formula of phenylacetylene

*8. A aqueous 2.00 m hydrofluoric acid solution has a normal boiling point of 101.09°C. Find the percent
ionization of hydrofluoric acid in the solution.

EXTRA HOMEWORK 3K

1. A solution is prepared by dissolving 8.92 grams of potassium bromide in enough water to make the
total volume 500.0 mL. Calculate the osmotic pressure of this solution at 27ºC.

2. A 5.87 milligram sample of a protein is dissolved in water at 25C to make 10.0 mL of solution. The
osmotic pressure of the solution is 2.45 torr. Calculate the molar mass of the protein.

3. Before refrigeration was common, many fruits were preserved by mixing them with large amounts of
sugar. Explain how sugar acts as preservatives.

*4. Explain how the Tyndall effect can be used to distinguish between a colloidal suspension and a true
solution.

EXTRA HOMEWORK FOR COLLIATIVE PROPERTIES (OPTIONAL)

1. Calculate the equilibrium vapor pressures for each of the following solutions.
(a) 396 grams of sucrose (C12H22O11) dissolved in 624 grams of water at 30ºC, if the equilibrium
vapor pressure of pure water at 30ºC is 31.8 torr.
(b) 24.6 grams of camphor (C10H16O) dissolved in 98.5 grams of benzene at 26.1ºC, if the
equilibrium vapor pressure of pure benzene at 26.1ºC is 100.0 torr.
(c) 396 grams of sodium chloride dissolved in 315 grams of water at 25ºC, if the equilibrium vapor
pressure of pure water at 25ºC is 23.8 torr.

2. Calculate the equilibrium vapor pressures for each of the following solutions.
(a) 40.0 grams of methanol (CH3OH) dissolved in 60.0 grams of acetone (C3H6O) at 20ºC, if the
equilibrium vapor pressure of pure methanol at 20ºC is 96.9 torr and the equilibrium vapor
pressure of pure acetone at 20ºC is 184.6 torr.
(b) 75.5 milliliters of ethanol (C2H5OH, density = 0.789 g/mL) dissolved in 24.5 milliliters of water
(density = 1.00 g/mL) at 20ºC, if the equilibrium vapor pressure of pure ethanol at 20ºC is 43.6
torr and the equilibrium vapor pressure of pure water at 20ºC is 17.5 torr.

3. Calculate the boiling points for each of the following solutions.
(a) 125 grams of calcium bromide dissolved in 675 grams of water.
(b) 16.8 grams of elemental sulfur (S8) dissolved in 90.0 grams of carbon tetrachloride.

4. Calculate the freezing points for each of the following solutions.
(a) 40.7 grams of lithium nitrate dissolved in 245 grams of water.
(b) 9.85 grams of elemental iodine dissolved in 30.0 grams of benzene.

5. Calculate the osmotic pressures for each of the following solutions.
(a) water from the Dead Sea, 5.8 M sodium chloride, at 19ºC
(b) 31.2 grams of potassium sulfate dissolved to make 500. mL of solution at 22ºC

(continued on next page)

6. Calculate the molar mass of each solute based upon the given experimental data.

(a) 4.278 grams of an organic nonelectrolyte dissolved in 25.6 grams of carbon disulfide, and the
boiling point of the solution is found to be 48.3ºC

(b) 13.61 grams of a nonelectrolyte dissolved in 75.6 grams of water, and the freezing point of the
solution is found to be -5.58ºC

(c) 1.22 grams of an organic nonelectrolyte dissolved in water to make 262 mL of solution, and the
osmotic pressure of the solution is found to be 30.3 torr at 35ºC.

EXTRA 3A ANSWERS

1. solid liquid gas

2. The kinetic energy of the particles must be greater than the attractive forces between the particles

3. Strong attractive forces between the particles that make up molasses (the sugar molecules and the
water molecules)

4. crystalline amorphous

5. The electrostatic attraction between positive and negative ions

6. The attraction between the nuclei of metal atoms and the valence electrons in delocalized molecular
orbitals

7. The attraction between the nuclei of nonmetal atoms and the valence electrons in molecular orbitals

8. (a) atoms; metallic bonds
(b) positive and negative ions; ionic bonds
(b) atoms; covalent bonds

(continued on next page)

9. (a) (b) (c)

10. Cr is the only conductor as a solid because it is metallic, CaBr2 is the only water soluble substance
because it is ionic, and the other will be SiO2 which is macromolecular

*11. 1 – ionic, 2 – metallic, 3 – macromolecular

EXTRA 3B ANSWERS

1. (a) the attractive force that exists between molecules, caused by the polarizing of electron clouds in
neighboring molecules, producing temporary dipole moments

(b) the attractive force that exists between polar molecules, caused by the attraction of the permanent
positive and negative ends of neighboring molecules

2. (a) molecules; hydrogen bonds (b) molecules; London dispersion forces
(c) molecules; London dispersion forces (d) molecules dipole-dipole interactions
(e) atoms; metallic bonds (f) positive and negative ions; ionic bonds
(g) atoms; covalent bonds (h) molecules; dipole-dipole interactions

3. (a) (b) (c) (d)

(e) (f) (g) (h)

4. Solid C12H22O11 will dissolve in water because it is a polar molecular substance with hydrogen
bonding, C14H10 will not because it is a nonpolar molecular substance

*5. 1 – nonpolar molecular, 2 – polar molecular, 3 – macromolecular, 4 – ionic, 5 – metallic

EXTRA UNIT CELL ANSWERS (c) NiAs
1. 1.36  10-8 cm (c) Li2O
2. 18.5 g/mL
3. face-centered cubic
4. (a) XeF2
5. (a) CaCl2
6. cubic closest packing

hexagonal closest packing

*7. > 41.4% of the larger ions

EXTRA 3C ANSWERS

1. The movement of charged particles in one direction.

2. (a) (b) (c) (d)

*3. 2.72  10-19 J

EXTRA 3D ANSWERS

1. (a) AlCl3 - Both are ionic substances with ionic bonds between the ions. The aluminum ion has a
higher charge than the magnesium ion, so it attracts more strongly to the chloride ions.

(b) K2O - Both are ionic substances with ionic bonds between the ions. The oxide ion is smaller than
the sulfide ion, so it can get closer to the potassium ion, attracting more strongly.

(c) AlP - Both are ionic substances with ionic bonds between the ions. The aluminum and phosphide
ions have higher charges than the cesium and iodide ions, so they attract more strongly. And, the
aluminum and phosphide ions are smaller than the cesium and iodide ions, so they can get closer
together, attracting more strongly.

2. (a) Mg – Both are metallic substances with metallic bonds between the atoms. They both have the
same number of bonding electrons per atom (2), but Mg atoms are smaller, so the bonding
electrons are in smaller molecular orbitals, which attract to the bonding nuclei more strongly

(b) V - Both are metallic substances with metallic bonds between the atoms. V has more net bonding
electrons per atom than Sc (5 to 3), which attract to the bonding nuclei more strongly making the
metallic bonds stronger.

(c) Mn - Both are metallic substances with metallic bonds between the atoms. Mn has more net
bonding electrons per atom than Co (Mn has 6 bonding electrons – 1 antibonding electron = 5 net
bonding electrons per atom, Co has 6 bonding electrons – 3 antibonding electrons = 3 net bonding
electrons per atom), which attract to the bonding nuclei more strongly making the metallic bonds
stronger.

3. Si – If both are macromolecular substances, their attractive forces would be covalent bonds. Si atoms
are smaller, so the bonding electrons are in smaller molecular orbitals, which attract to the
bonding nuclei more strongly

4. (a) (b) (c) (d)

nonpolar nonpolar polar polar

(continued on next page)

5. (a) As4 – Both are nonpolar molecular substances, with only London dispersion forces between the
molecules. As4 has more electrons, so it is more polarizing, making its London dispersion forces
stronger.

(b) SO2 – SO2 is a polar molecular substance, with London dispersion forces and dipole-dipole
interactions between the molecules. CO2 is a nonpolar molecular substance, having only London
dispersion forces between the molecules.

(c) CH3NH2 – CH3NH2 is a polar molecular substance, with London dispersion forces, dipole-dipole
interactions, and hydrogen bonding between the molecules. CH3CN is a polar molecular
substance, but only has London dispersion forces and dipole-dipole interactions between the
molecules.

(d) Li2O – Li2O is an ionic substance with strong ionic bonds between the ions. Br2O is a polar
molecule substance, with only London dispersion forces and dipole-dipole interactions between
the molecules.

6. Both C3H8 and C4H10 are nonpolar molecular, with only London dispersion forces. C4H10 has more
electrons, so it is more polarizing, making its London dispersion forces stronger. C3H7OH is polar
molecular, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding.

7. (a) HF < NaF < LiF – HF is a polar molecular substances, with weak London dispersion forces,
dipole-dipole interactions, and hudrogen bonds between the molecules. Both NaF and LiF are
ionic substances with ionic bonds between the ions. The lithium ion is smaller than the sodium
ion, so it can get closer to the fluoride ion, attracting more strongly.

(b) C6H6 < C10H8 < C6H5CO2H – C6H6 and C10H8 Both are nonpolar molecular substances, with only
London dispersion forces between the molecules, but C10H8 has more electrons, so it is more
polarizing, making its London dispersion forces stronger. C6H5CO2H is a polar molecular
substance, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding
between the molecules.

(c) H2S < H2Se < H2O – H3S amd H2Se are polar molecular substances, with London dispersion
forces and dipole-dipole interactions between the molecules, but H2Se has more electrons, so it is
more polarizing, making its London dispersion forces stronger. H2O is a polar molecular
substance, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding
between the molecules.

(d) Li < Be < C – Both Li and Be are metallic substances with metallic bonds between the atoms, but
Be has more net bonding electrons per atom than Li (2 to 1), which attract to the bonding nuclei
more strongly making the metallic bonds stronger. C is a macromolecular substance, and because
of the low number of net bonding electrons per atom for the two metals, the covalent bonds in C
are stronger.

*8. AlCl3 > BCl3

AlCl3 is an ionic network with strong ionic bonds, BCl3 is nonpolar molecular, with weak London
dispersion forces

EXTRA 3E ANSWERS

1. (a) An increase in temperature would increase the rate of evaporation. As more water vapor
accumulates in the closed container, the rate of condensation would increase. Eventually, the rate
of condensation will equal the rate of evaporation, and equilibrium is reestablished. This new
equilibrium will now have a greater amount of water vapor, so its equilibrium vapor pressure
would now be higher.

(b) An increase in surface area would increase the rate of evaporation and condensation equally, so
there would be no change in the equilibrium.

2. 135,000 J

*3. 69,200 J

EXTRA 3F ANSWERS

1. (a) 184.4ºC (b) 113.6ºC (c) solid (d) gas
2. (a) 0.027 mm Hg (b) rhombic

EXTRA 3G ANSWERS

1. (a) methanol (b) methanol (c) toluene (d) toluene

2. (a) CH3COOH – both oxygens and the hydrogen bonded to the oxygen are capable of hydrogen
bonding to water molecules. CH3CHO only has one oxygen capable of hydrogen bonding to water
molecules.

(b) CH3F – the molecule is polar, and the fluorine is capable of hydrogen bonding to water molecules.
CF4 is a nonpolar molecule.

(c) Vitamin C – more OH groups to hydrogen bond to water molecules

3. The molecule has a nonpolar end (C5H11-) that attracts nonpolar solvents and a polar end (-OH) that
attracts polar solvents

4. (a) strong electrolyte (b) strong electrolyte (c) nonelectrolyte (d) weak electrolyte

(continued on next page)

5. (a) strong electrolyte Fe(NO3)3 (s) → Fe3+ (aq) + 3NO3- (aq)
(b) nonelectrolyte CH2OHCH2OH (l) → CH2OHCH2OH (aq)
(c) strong electrolyte H2SO4 (g) → 2H+ (aq) + SO42- (aq)
(d) weak electrolyte H3PO4 (g) → H3PO4 (aq)

6. C3H6O3 Yes No

C3H8 No -

LiOH Yes A Lot

NaI Yes A Lot

HCN Yes A Little

HClO3 Yes A Lot

EXTRA 3H ANSWERS

1. (a) 8.98% (b) 11%

2. (a) 0.0076 (b) 0.0569

3. (a) 0.944 M (b) 0.181 M

4. (a) 3.13 m (b) 0.554 m
5. (a) 2.60 M K+, 1.30 M CO32- (b) 0.240 M Al3+, 0.360 M SO42-
6. 11.9 g

7. (a) 2.2% toluene (b) 0.21 M toluene (c) 0.24 m toluene (d) 0.018 toluene

*8. (a) 13.4% NaNO3 (b) 1.76 M NaNO3 (c) 1.82 m NaNO3 (d) 0.0317 NaNO3

*9. 1.3 M C2H5OH

*10. (a) 0.00229 M H2C2O4 (b) 0.000193 H2C2O4 (c) 0.0000386 H2C2O4 (d) 0.00214 m H2C2O4
(e) 233 g H2O

EXTRA CONCENTRATION UNITS ANSWERS

1. (a) 16.9% KCl (b) 13% toluene (c) 7.0% NaBr (d) 19.5% glucose
2. (a) 0.0542 (d) 0.781 M
3. (a) 0.426 M (b) 0.0141 (c) 0.0782
4. (a) 0.647 m (d) 0.0154
5. (a) 1.728 M (b) 1.37 M (c) 0.716 M (d) 0.032
6. (a) 0.2730 M AlI3
7. (a) 11.6% (b) 0.0618 m (c) 1.15 m C10H8
8. (a) 13.4%
9. (a) 12.1 M (b) 1.728 M Mg2+, 3.456 M Cl-

(b) 0.2730 M Al3+, 0.8190 M I-

(b) 0.784 M (c) 0.866 m

(b) 1.76 M (c) 1.8 m

(b) 10.1 m (c) 0.225

EXTRA 3I ANSWERS

1. (a) 21.9 torr (b) 20.6 torr

2. (a) 19.6 torr (b) 18.6 torr

3. (a) 0.526 heptane, 0.474 octane (b) 29.3 torr

4. (a) 0.1 m calcium nitrate, 0.1 m potassium nitrate, 0.1 m sucrose, pure water

*5. 90.9 torr

*6. 23.0 torr

*7. 47 g/mol

*8. C14H10

EXTRA 3J ANSWERS

1. (a) 0.2 m lithium bromide, 0.3 m glucose and 0.1 m magnesium bromide (tie), pure water
(b) pure water, 0.3 m glucose and 0.1 m magnesium bromide (tie), 0.2 m lithium bromide
(c) pure water, 0.3 m glucose and 0.1 m magnesium bromide (tie), 0.2 m lithium bromide

2. (a) -9.3ºC (b) 102.6ºC

3. (a) -2.10ºC (b) 100.58ºC

4. C4H8O2 (c)

5. C6H12O2

*6. (a) C4H10O2
(b)

*7. (a) 20.1 C°/m
(b) C8H6

*8. 10% (7% without rounding to the correct number of significasnt figures)

EXTRA 3K ANSWERS

1. 7.39 atm

2. 4,450 g/mol

3. The large amount of sugar dissolves in water that is in contact with fruit, making a concentrated sugar
solution. Bacteria cells come in contact with this concentrated sugar solution, and through the cell
membranes, water flows out of the bacteria cells into the sugar solution faster than water flows into
the bacteria cells from the sugar solution. This dehydrates the bacteria and kills them.

*4. A beam of light passing through a colloidal suspension will be scattered, but the same beam of light
passing through a true solution will not be scattered.

EXTRA COLLIATIVE PROPERTIES ANSWERS

1. (a) 30.8 torr (b) 88.6 torr (c) 13.4 torr
2. (a) 136.6 torr (b) 30.2 torr (c) 2,950 g/mol
3. (a) 101.4ºC (b) 80.2ºC
4. (a) -8.96ºC (b) -1.1ºC
5. (a) 280 atm (b) 26.0 atm
6. (a) 190 g/mol (b) 60.0 g/mol