CHM 2046 General Chemistry 2 Exam 1
Practice Questions And Correct Answers
(Verified Answers) Plus Rationale 2027
Q&A| Instant Download Pdf.
1. A 0.250 mol sample of NaCl is dissolved in enough water to produce
exactly 500.0 mL of solution. What is the molarity of the NaCl
solution?
A. 0.125 M
B. 0.250 M
C. 0.500 M
D. 2.00 M
Rationale: Molarity is moles of solute divided by liters of solution. Thus,
M=0.250 mol/0.5000 L=0.500 M.
2. A student dilutes 25.0 mL of 2.00 M HCl to a final volume of 250.0 mL.
What is the concentration of HCl after dilution?
A. 0.0200 M
B. 0.200 M
C. 0.500 M
D. 20.0 M
Rationale: During dilution, the number of moles of solute remains
constant, so M1V1=M2V2. Therefore, M2=(2.00)(25.0)/250.0=0.200 M.
, 3. Which statement correctly describes a strong electrolyte when
dissolved in water?
A. It remains almost entirely as intact molecules.
B. It dissolves without producing ions.
C. It dissociates essentially completely into ions.
D. It reacts only with water molecules but produces no ions.
Rationale: Strong electrolytes, including soluble ionic compounds and
strong acids and bases, produce essentially complete ionization or
dissociation in aqueous solution.
4. Which compound is expected to behave as a nonelectrolyte in
aqueous solution?
A. NaNO₃
B. HCl
C. KOH
D. C₂H₅OH
Rationale: Ethanol is a molecular compound that dissolves as intact neutral
molecules and therefore does not produce a significant concentration of
ions in water.
5. What is the net ionic equation for the reaction between aqueous
AgNO₃ and aqueous NaCl?
A. AgNO₃ + NaCl → AgCl + NaNO₃
B. Ag⁺ + NO₃⁻ → AgNO₃
C. Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
D. Na⁺(aq) + NO₃⁻(aq) → NaNO₃(s)
Rationale: AgCl is an insoluble precipitate, while Na⁺ and NO₃⁻ remain
spectator ions. The net ionic equation therefore contains only Ag⁺ and Cl⁻.
6. A solution contains 0.150 mol of CaCl₂ in 750.0 mL of solution. What is
the molarity of Ca²⁺ ions?
,A. 0.100 M
B. 0.400 M
C. 0.200 M
D. 0.600 M
Rationale: CaCl₂ produces one Ca²⁺ ion per formula unit, so the Ca²⁺
concentration equals the CaCl₂ concentration: 0.150/0.750=0.200 M, not
0.400 M. Therefore the correct choice is C.
7. A solution is prepared by dissolving 5.85 g of NaCl, molar mass 58.44
g/mol, in enough water to make 500.0 mL of solution. What is the
molarity?
A. 0.0500 M
B. 0.200 M
C. 0.500 M
D. 1.00 M
Rationale: The number of moles is 5.85/58.44=0.1001 mol. Dividing by
0.5000 L gives approximately 0.200 M.
8. Which reaction represents a Brønsted–Lowry acid-base reaction?
A. NaCl(s) → Na⁺(aq) + Cl⁻(aq)
B. HCl + H₂O → H₃O⁺ + Cl⁻
C. Ag⁺ + Cl⁻ → AgCl(s)
D. CaCO₃ → CaO + CO₂
Rationale: A Brønsted–Lowry acid donates a proton and a Brønsted–Lowry
base accepts one. In this reaction, HCl donates H⁺ to H₂O, producing H₃O⁺.
9. According to the Brønsted–Lowry definition, which species acts as a
base in the reaction NH3+H2O⇌NH4++OH−?
A. H₂O only
B. NH₃
, C. NH₄⁺
D. OH⁻
Rationale: NH₃ accepts a proton from H₂O to form NH₄⁺, so NH₃ is the
Brønsted–Lowry base.
10. Which species is the conjugate base of H₂SO₄ after loss of one
proton?
A. SO₄²⁻
B. HSO₄⁻
C. H₃SO₄⁺
D. OH⁻
Rationale: A conjugate base is formed when an acid loses one proton.
Removing one H⁺ from H₂SO₄ gives HSO₄⁻.
11. What is the pH of a 1.0 × 10⁻³ M HCl solution at 25 °C?
A. 1.00
B. 2.00
C. 3.00
D. 11.00
Rationale: HCl is a strong acid and dissociates completely, giving
[H+]=1.0×10−3 M. Therefore, pH=−log(1.0×10−3)=3.00.
12. What is the pOH of a solution having [OH−]=1.0×10−5 M?
A. 5.00
B. 7.00
C. 5.00
D. 9.00
Rationale: The definition of pOH is −log[OH−]. Thus,
pOH=−log(1.0×10−5)=5.00.
Practice Questions And Correct Answers
(Verified Answers) Plus Rationale 2027
Q&A| Instant Download Pdf.
1. A 0.250 mol sample of NaCl is dissolved in enough water to produce
exactly 500.0 mL of solution. What is the molarity of the NaCl
solution?
A. 0.125 M
B. 0.250 M
C. 0.500 M
D. 2.00 M
Rationale: Molarity is moles of solute divided by liters of solution. Thus,
M=0.250 mol/0.5000 L=0.500 M.
2. A student dilutes 25.0 mL of 2.00 M HCl to a final volume of 250.0 mL.
What is the concentration of HCl after dilution?
A. 0.0200 M
B. 0.200 M
C. 0.500 M
D. 20.0 M
Rationale: During dilution, the number of moles of solute remains
constant, so M1V1=M2V2. Therefore, M2=(2.00)(25.0)/250.0=0.200 M.
, 3. Which statement correctly describes a strong electrolyte when
dissolved in water?
A. It remains almost entirely as intact molecules.
B. It dissolves without producing ions.
C. It dissociates essentially completely into ions.
D. It reacts only with water molecules but produces no ions.
Rationale: Strong electrolytes, including soluble ionic compounds and
strong acids and bases, produce essentially complete ionization or
dissociation in aqueous solution.
4. Which compound is expected to behave as a nonelectrolyte in
aqueous solution?
A. NaNO₃
B. HCl
C. KOH
D. C₂H₅OH
Rationale: Ethanol is a molecular compound that dissolves as intact neutral
molecules and therefore does not produce a significant concentration of
ions in water.
5. What is the net ionic equation for the reaction between aqueous
AgNO₃ and aqueous NaCl?
A. AgNO₃ + NaCl → AgCl + NaNO₃
B. Ag⁺ + NO₃⁻ → AgNO₃
C. Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
D. Na⁺(aq) + NO₃⁻(aq) → NaNO₃(s)
Rationale: AgCl is an insoluble precipitate, while Na⁺ and NO₃⁻ remain
spectator ions. The net ionic equation therefore contains only Ag⁺ and Cl⁻.
6. A solution contains 0.150 mol of CaCl₂ in 750.0 mL of solution. What is
the molarity of Ca²⁺ ions?
,A. 0.100 M
B. 0.400 M
C. 0.200 M
D. 0.600 M
Rationale: CaCl₂ produces one Ca²⁺ ion per formula unit, so the Ca²⁺
concentration equals the CaCl₂ concentration: 0.150/0.750=0.200 M, not
0.400 M. Therefore the correct choice is C.
7. A solution is prepared by dissolving 5.85 g of NaCl, molar mass 58.44
g/mol, in enough water to make 500.0 mL of solution. What is the
molarity?
A. 0.0500 M
B. 0.200 M
C. 0.500 M
D. 1.00 M
Rationale: The number of moles is 5.85/58.44=0.1001 mol. Dividing by
0.5000 L gives approximately 0.200 M.
8. Which reaction represents a Brønsted–Lowry acid-base reaction?
A. NaCl(s) → Na⁺(aq) + Cl⁻(aq)
B. HCl + H₂O → H₃O⁺ + Cl⁻
C. Ag⁺ + Cl⁻ → AgCl(s)
D. CaCO₃ → CaO + CO₂
Rationale: A Brønsted–Lowry acid donates a proton and a Brønsted–Lowry
base accepts one. In this reaction, HCl donates H⁺ to H₂O, producing H₃O⁺.
9. According to the Brønsted–Lowry definition, which species acts as a
base in the reaction NH3+H2O⇌NH4++OH−?
A. H₂O only
B. NH₃
, C. NH₄⁺
D. OH⁻
Rationale: NH₃ accepts a proton from H₂O to form NH₄⁺, so NH₃ is the
Brønsted–Lowry base.
10. Which species is the conjugate base of H₂SO₄ after loss of one
proton?
A. SO₄²⁻
B. HSO₄⁻
C. H₃SO₄⁺
D. OH⁻
Rationale: A conjugate base is formed when an acid loses one proton.
Removing one H⁺ from H₂SO₄ gives HSO₄⁻.
11. What is the pH of a 1.0 × 10⁻³ M HCl solution at 25 °C?
A. 1.00
B. 2.00
C. 3.00
D. 11.00
Rationale: HCl is a strong acid and dissociates completely, giving
[H+]=1.0×10−3 M. Therefore, pH=−log(1.0×10−3)=3.00.
12. What is the pOH of a solution having [OH−]=1.0×10−5 M?
A. 5.00
B. 7.00
C. 5.00
D. 9.00
Rationale: The definition of pOH is −log[OH−]. Thus,
pOH=−log(1.0×10−5)=5.00.
