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CHEM 120 Week 3–6 Practice Exam: 100 Questions with Answers
Section 1: Solutions, Molarity & Concentration (Questions 1–20)
Q1. What is the molarity of a solution containing 2.5 moles of NaCl dissolved in enough
water to make 0.500 L of solution?
A. 0.20 M
B. 1.25 M
C. 5.0 M
D. 12.5 M
Answer: C
Rationale: Molarity (M) = moles of solute / liters of solution = 2.5 mol / 0.500 L = 5.0 M .
Q2. How many grams of glucose (C₆H₁₂O₆, molar mass 180.16 g/mol) are needed to prepare
250.0 mL of a 0.300 M solution?
A. 5.40 g
B. 13.5 g
C. 54.0 g
D. 135 g
Answer: B
Rationale: Moles = M × L = 0.300 mol/L × 0.250 L = 0.0750 mol. Grams = 0.0750 mol × 180.16
g/mol = 13.51 g ≈ 13.5 g .
Q3. What formula would you use to prepare a dilute solution from a concentrated solution?
A. M = mol/L
B. C₁V₁ = C₂V₂
C. P₁V₁ = P₂V₂
D. PV = nRT
Answer: B
Rationale: The dilution equation C₁V₁ = C₂V₂ relates the concentration and volume of the
concentrated and diluted solutions .
Q4. You are preparing a solution from 750 mL of a 3.0 M NaCl solution. How many mL of a
1.5 M NaCl solution could you produce by diluting the original solution?
A. 750 mL
B. 1500 mL
C. 2250 mL
D. 3000 mL
Answer: B
Rationale: C₁V₁ = C₂V₂ → (3.0 M)(750 mL) = (1.5 M)(V₂) → V₂ = 1500 mL .
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Q5. How many ions would you expect MgBr₂ to break down into in water?
A. 1
B. 2
C. 3
D. 4
Answer: C
Rationale: MgBr₂ dissociates into one Mg²⁺ ion and two Br⁻ ions, for a total of 3 ions .
Q6. A chemist dissolves FeBr₃ in water to make a 2.0 M solution. What is the osmolarity of
this solution?
A. 2.0 Osm/L
B. 4.0 Osm/L
C. 6.0 Osm/L
D. 8.0 Osm/L
Answer: D
Rationale: FeBr₃ dissociates into 1 Fe³⁺ and 3 Br⁻ ions (4 particles total). Osmolarity = 2.0 M ×
4 = 8.0 Osm/L .
Q7. Which of the following is a strong base?
A. NH₃ (ammonia)
B. KOH (potassium hydroxide)
C. CH₃OH (methanol)
D. H₂CO₃ (carbonic acid)
Answer: B
Rationale: KOH is a strong base that completely dissociates into K⁺ and OH⁻ in water. NH₃ is a
weak base .
Q8. A solution has a pH of 3.5. What is the hydronium ion concentration [H₃O⁺]?
A. 3.5 × 10⁻¹ M
B. 3.2 × 10⁻⁴ M
C. 3.2 × 10⁻¹¹ M
D. 1.0 × 10⁻⁷ M
Answer: B
Rationale: [H₃O⁺] = 10⁻ᵖᴴ = 10⁻³·⁵ = 3.16 × 10⁻⁴ M .
Q9. A 0.500-g sample of magnesium is burned completely in oxygen to form magnesium
oxide. The balanced equation is 2Mg + O₂ → 2MgO. If 0.825 g of MgO is collected, calculate
the percent yield (Mg = 24.3 g/mol, MgO = 40.3 g/mol).
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A. Approximately 50%
B. Approximately 75%
C. Approximately 83%
D. Approximately 99%
Answer: D
Rationale: Theoretical yield = 0.500 g Mg × (1 mol/24.3 g) × (1 mol MgO/1 mol Mg) × (40.3
g/1 mol) ≈ 0.830 g. Percent yield = (0.825/0.830) × 100 ≈ 99% .
Q10. What is the molar mass of CaSO₄?
A. 100.09 g/mol
B. 136.14 g/mol
C. 172.17 g/mol
D. 88.06 g/mol
Answer: B
Rationale: Ca (40.08) + S (32.07) + 4 O (4 × 16.00 = 64.00) = 136.15 g/mol ≈ 136.14 g/mol .
Q11. How many atoms of oxygen are there in the molecule Al₂(SO₄)₃?
A. 4
B. 8
C. 12
D. 16
Answer: C
Rationale: There are 3 sulfate groups, each containing 4 oxygen atoms: 3 × 4 = 12 oxygen
atoms .
Q12. A solution in which the pOH is 12.5 would be described as:
A. Distinctly basic
B. Distinctly acidic
C. Neutral
D. Slightly basic
Answer: B
Rationale: pOH = 12.5 → pH = 14 − 12.5 = 1.5, which is distinctly acidic .
Q13. Which of the following is true about a 0.10 M solution of a weak acid, HX?
A. [HX] = [H⁺]
B. [HX] > [H⁺]
C. [HX] < [H⁺]
D. [HX] = 0.10 M
CHEM 120 Week 3–6 Practice Exam: 100 Questions with Answers
Section 1: Solutions, Molarity & Concentration (Questions 1–20)
Q1. What is the molarity of a solution containing 2.5 moles of NaCl dissolved in enough
water to make 0.500 L of solution?
A. 0.20 M
B. 1.25 M
C. 5.0 M
D. 12.5 M
Answer: C
Rationale: Molarity (M) = moles of solute / liters of solution = 2.5 mol / 0.500 L = 5.0 M .
Q2. How many grams of glucose (C₆H₁₂O₆, molar mass 180.16 g/mol) are needed to prepare
250.0 mL of a 0.300 M solution?
A. 5.40 g
B. 13.5 g
C. 54.0 g
D. 135 g
Answer: B
Rationale: Moles = M × L = 0.300 mol/L × 0.250 L = 0.0750 mol. Grams = 0.0750 mol × 180.16
g/mol = 13.51 g ≈ 13.5 g .
Q3. What formula would you use to prepare a dilute solution from a concentrated solution?
A. M = mol/L
B. C₁V₁ = C₂V₂
C. P₁V₁ = P₂V₂
D. PV = nRT
Answer: B
Rationale: The dilution equation C₁V₁ = C₂V₂ relates the concentration and volume of the
concentrated and diluted solutions .
Q4. You are preparing a solution from 750 mL of a 3.0 M NaCl solution. How many mL of a
1.5 M NaCl solution could you produce by diluting the original solution?
A. 750 mL
B. 1500 mL
C. 2250 mL
D. 3000 mL
Answer: B
Rationale: C₁V₁ = C₂V₂ → (3.0 M)(750 mL) = (1.5 M)(V₂) → V₂ = 1500 mL .
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Q5. How many ions would you expect MgBr₂ to break down into in water?
A. 1
B. 2
C. 3
D. 4
Answer: C
Rationale: MgBr₂ dissociates into one Mg²⁺ ion and two Br⁻ ions, for a total of 3 ions .
Q6. A chemist dissolves FeBr₃ in water to make a 2.0 M solution. What is the osmolarity of
this solution?
A. 2.0 Osm/L
B. 4.0 Osm/L
C. 6.0 Osm/L
D. 8.0 Osm/L
Answer: D
Rationale: FeBr₃ dissociates into 1 Fe³⁺ and 3 Br⁻ ions (4 particles total). Osmolarity = 2.0 M ×
4 = 8.0 Osm/L .
Q7. Which of the following is a strong base?
A. NH₃ (ammonia)
B. KOH (potassium hydroxide)
C. CH₃OH (methanol)
D. H₂CO₃ (carbonic acid)
Answer: B
Rationale: KOH is a strong base that completely dissociates into K⁺ and OH⁻ in water. NH₃ is a
weak base .
Q8. A solution has a pH of 3.5. What is the hydronium ion concentration [H₃O⁺]?
A. 3.5 × 10⁻¹ M
B. 3.2 × 10⁻⁴ M
C. 3.2 × 10⁻¹¹ M
D. 1.0 × 10⁻⁷ M
Answer: B
Rationale: [H₃O⁺] = 10⁻ᵖᴴ = 10⁻³·⁵ = 3.16 × 10⁻⁴ M .
Q9. A 0.500-g sample of magnesium is burned completely in oxygen to form magnesium
oxide. The balanced equation is 2Mg + O₂ → 2MgO. If 0.825 g of MgO is collected, calculate
the percent yield (Mg = 24.3 g/mol, MgO = 40.3 g/mol).
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A. Approximately 50%
B. Approximately 75%
C. Approximately 83%
D. Approximately 99%
Answer: D
Rationale: Theoretical yield = 0.500 g Mg × (1 mol/24.3 g) × (1 mol MgO/1 mol Mg) × (40.3
g/1 mol) ≈ 0.830 g. Percent yield = (0.825/0.830) × 100 ≈ 99% .
Q10. What is the molar mass of CaSO₄?
A. 100.09 g/mol
B. 136.14 g/mol
C. 172.17 g/mol
D. 88.06 g/mol
Answer: B
Rationale: Ca (40.08) + S (32.07) + 4 O (4 × 16.00 = 64.00) = 136.15 g/mol ≈ 136.14 g/mol .
Q11. How many atoms of oxygen are there in the molecule Al₂(SO₄)₃?
A. 4
B. 8
C. 12
D. 16
Answer: C
Rationale: There are 3 sulfate groups, each containing 4 oxygen atoms: 3 × 4 = 12 oxygen
atoms .
Q12. A solution in which the pOH is 12.5 would be described as:
A. Distinctly basic
B. Distinctly acidic
C. Neutral
D. Slightly basic
Answer: B
Rationale: pOH = 12.5 → pH = 14 − 12.5 = 1.5, which is distinctly acidic .
Q13. Which of the following is true about a 0.10 M solution of a weak acid, HX?
A. [HX] = [H⁺]
B. [HX] > [H⁺]
C. [HX] < [H⁺]
D. [HX] = 0.10 M
