Kiel Benedick M. Gianan
12 – Canary
Chapter 1 – What Hold Solids and Liquids?
Practice Exercises
Practice Exercise 1.1
What type of intermolecular force will act in the following substances/solution? Justify your answer.
1. carbon dioxide (CO2 )
London dispersion force; because CO2 is composed of one carbon and two oxygen atoms, and both
carbon and oxygen are nonmetals It also contains covalent bonds.
2. neon (Ne) gas
London dispersion force; They are the weakest sort of intermolecular force since they are only
temporary, yet their total influence is sufficient to generate significant attraction between particles.
Dispersion forces are caused by the random mobility of electrons within the atom.
3. magnesium chloride (MgCl2 ) dissolved in water
Ion-dipole force; Magnesium chloride and water have an ion-dipole interaction. This interaction is
caused by polar water molecules interacting with a magnesium ion. A water molecule's oxygen atom
has a slight negative charge, which attracts the positively charged magnesium ion. These are
significantly weaker forces than covalent or ionic bonding.
Self-checks

What phase of water exists at 100°C and 5 atm? at 200°C and 1 atm?
The phase of water that exists at 100°C and 5 atm is liquid, and at 200°C and 1 atm is gas.

What phase of carbon dioxide exists at - 120°C and 10 atm? at 80°C and 100 atm?
The phase of carbon dioxide that exists at - 120°C and 10 atm is solid, and at 80°C and 100 atm is
supercritical fluid.
Kiel Benedick M. Gianan
12 – Canary
Chapter 2 – Solutions and Their Properties
Practice Exercises
Practice Exercise 2.1
1. Calculate the mass (in grams) of magnesium chloride (MgCl2 ) that would be needed to prepare 150
mL of a 20% by mass aqueous solution of the salt. The density of the solution is 1.1 g/mL.
Let x be the mass of solute in g.
percent by mass =
=
x
× 100
165 g
(20%) (165 g)
100
= 𝟑𝟑 𝐠
2. Calculate the percent by mass of the solution containing 8.60 g of salt in 95.0 g of solution.
percent by mass =
mass of solute in grams
× 100
mass of solution in grams
percent by mass =
8.60 g
× 100
95.0 g
= 𝟗. 𝟎𝟓 %
3. A solution is prepared by dissolving 20 mL of pure hydrogen peroxide (H2 O2 ) in enough water to make
200 mL of solution. What is the concentration of the H2 O2 solution?
volume of solute
× 100
volume of solutio
20 mL
percent by volume =
× 100
200 mL
percent by volume =
= 𝟏𝟎 %
Practice Exercise 2.2
1. Calculate the molality of a solution containing 10.8 g of ethylene glycol (C2 H6 O2 ) in 360 g of water.
Solution:
mole C2 H6 O2 =
molality =
10.8 g
= 0.17 mol
62.06 g/mol
mol C2 H6 O2 0.17 mol
=
= 0.47 m
kg H2 O
0.36 kg
2. What is the molality of a 60.5% by mass of nitric acid (HNO3 ) solution?
Solution:
mass of H2 O = 100 g
mass of HNO3 = 60.5
= 100 g − 60.5 = 0.04 kg
mole HNO3 =
molality =
60.5 g
= 0.96 mol
63.01 g/mol
mol HNO3
0.96 mol
=
= 24 m
kg HNO3
0.04 kg
3. Calculate the molarity of a solution containing 2.80 moles of ethyl alcohol (C2 H6 O) in 500 mL of
solution.
Solution:
molarity =
mol C2 H6 O 2.80 mol
=
= 5.6 M
L solution
0.5 L
4. Determine the molarity of a solution containing 2.40 g of sodium chloride (NaCl) in 40.0 mL of
solution. (molar mass of NaCl = 58.45 g/mol)
Solution:
mole NaCl =
molarity =
2.40 g
= 0.04 mol
58.44 g/mol
mol NaCl 0.04 mol
=
=1M
L solution
0.04 L
SELF-CHECK
Express the concentration of a solution (in % w/w, % v/v, and % w/v) containing 5 g of NaCl and 50 mL
of water. (𝜌 of NaCl = 2.16 g/mL)
Solution:
Mass of Solute (NaCl)= 5g
Volume of solvent (H2 O) = 50 mL
Density of NaCl= 2.16 g/ml
Density of water= 1 g/mL
5g
Volume of solute (NaCl) = 2.16 g/mL = 2.31 mL Mass of solvent (H2 O) = 50 × 1 = 50 g
mass of solute
a.) Percent by mass (% w/w) = mass of solution × 100
mass of solute
= mass of solute+mass of solvent × 100
=
5g
× 100
50 g+5g
5g
= 55 g × 100
= 𝟗. 𝟎𝟗%
volume of solute
b.) percent by volume (%v/v) = volume of solution × 100
volume of solute
= volume of solute+volume of solvent × 100
2.31 mL
= 2.31 mL+50 mL × 100
=
2.31 mL
×
52.31 mL
100
= 𝟒. 𝟒𝟐%
mass of solute in grams
c.) percent by mass − volume (%w/v) = volume of solution in mL × 100
5g
= 52.31 mL × 100
= 𝟗. 𝟔𝟎%
Kiel Benedick M. Gianan
12 – Canary
Chapter 3 – Energy in Transit
Practice Exercises
Practice Exercise 3.1
1. Exothermic, combustion is used to liberate energy.
2. Exothermic, the formation of salt and water is involved.
3. Exothermic, as stated in number one, combustion has occurred.
4. Endothermic, because it is poorly soluble in water and necessitates the use of heat energy.
5. Exothermic, because it emits heat in some way.
Practice Exercise 3.1
1.
a. dw = work done
b. heat released
du = Δ in interval
dq = Δ in heat
ΔQ = Δu + ΔW
= 23 J + (-45 J)
dw = du - dq = 12 – 47 = -35 J
ΔQ = 22 J
work done by the system dw = -35 J
2.
a. W = -(P)(ΔV)
W = -(1 atm) (0.5L)
W = -0.5 L ⋅ atm ⋅ 101.325 J
1L ⋅ atm
W = -50.6625 J work done
b. q = 50.6625 J heat absorbed
Love of Lab
1. N2(g) + O2 (g) → 2NO
ΔH = 180 KJ
2NO2(g) → 2NO (g) + O2 (g)
ΔH = 112 KJ
2NO2 → N2 + 2O2
ΔH = 68 KJ
2. m (CH2N2) = 4.00
n (CH2N2) = m (CH2N2) = 4.0 = 0.09 mol
M (CH2N2) = 46.1
Q (reactant) =113.013 KJ or 113.013 x 103 J
Q (reactant/mol) = 113.013 x 103 J / 0.09 mol = 1.30 x 106 J/mol
M (CH2N2) = 46.1
Kiel Benedick M. Gianan
12 – Canary
Test Yourself
Chapter 1
1. C
6. B
11. C
2. D
7. A
12. B
3. C
8. B
13. A
4. D
9. B
14. C
5. D
10. D
15. A
1. A
6. C
11. B
2. A
7. A
12. A
3. C
8. D
13. D
4. C
9. C
14. B
5. C
10. D
15. B
Chapter 2
Chapter 3
1. D
6. 3051 J
11. B
2. C
7. C
12. D
3. B
8. C
13. B
4. D
9. B
14. A
5. C
10. C
15. C