,CHEM 121 Final Module Exam Foundations of
General Chemistry w/Lab | Portage | 26/27 Actual
(PDF)
1. A metal sample weighing 30.9232 g was added to a graduated cylinder containing 23.20 mL of
water. The volume of the water plus the sample was 24.80 mL. What is the density of the metal,
expressed to the correct number of significant figures?
A) 19.3 g/mL
B) 19.33 g/mL
C) 19.327 g/mL
D) 2.0 × 10¹ g/mL
Correct Answer: 19.3 g/mL
Rationale: Volume of metal = 24.80 mL − 23.20 mL = 1.60 mL. Density = 30.9232 g / 1.60 mL = 19.327
g/mL. Since the volume has only two significant figures, the density is rounded to 19.3 g/mL. Options
B and C overstate precision, and option D is an incorrect calculation.
2. Rutherford's gold foil experiment led to which conclusion about atomic structure?
A) Electrons are embedded in a positively charged sphere.
B) The nucleus is dense, positively charged, and occupies a small fraction of the atom's volume.
C) Atoms are indivisible and cannot be broken down further.
D) Electrons orbit the nucleus in fixed circular paths.
Correct Answer: The nucleus is dense, positively charged, and occupies a small fraction of the atom's
volume.
,Rationale: The deflection of alpha particles indicated a dense, positively charged center. Option A
describes the plum pudding model, option C is Dalton's atomic theory, and option D is the Bohr
model. Options A, C, and D are incorrect.
3. Chlorine has two stable isotopes: ³⁵Cl (34.96885 amu) and ³⁷Cl (36.96590 amu). If the atomic mass of
chlorine is 35.45 amu, what is the approximate natural abundance of ³⁵Cl?
A) 24.33%
B) 50.00%
C) 75.77%
D) 64.55%
Correct Answer: 75.77%
Rationale: Let x be the abundance of ³⁵Cl. 34.96885x + 36.96590(1 − x) = 35.45. Solving gives x ≈
0.7577, or 75.77%. Options A, B, and D do not yield the correct average atomic mass.
4. How many grams of CH₄ contain the same number of molecules as 2.50 g of O₂?
A) 0.0781 g
B) 1.25 g
C) 0.156 g
D) 4.88 × 10⁻³ g
Correct Answer: 1.25 g
Rationale: Moles O₂ = 2.50 g / 32.00 g/mol = 0.0781 mol. Mass CH₄ = 0.0781 mol × 16.04 g/mol = 1.25
g. Option A is the number of moles, option C is half the correct mass, and option D is an incorrect
calculation.
5. Consider the reaction: 2 NaI(aq) + Cl₂(g) → I₂(aq) + 2 NaCl(aq). Which element undergoes reduction?
, A) Sodium
B) Iodide
C) Chlorine
D) Iodine
Correct Answer: Chlorine
Rationale: Chlorine goes from oxidation state 0 in Cl₂ to −1 in NaCl, gaining electrons and undergoing
reduction. Iodide is oxidized from −1 to 0. Sodium and iodine are not reduced. Options A, B, and D are
incorrect.
6. What is the oxidation number of chlorine in KClO₃?
A) +6
B) +5
C) −1
D) +2
Correct Answer: +5
Rationale: K is +1, O is −2. Let x be Cl: +1 + x + 3(−2) = 0, so x = +5. Options A, C, and D result from
incorrect algebraic setups or ignoring the polyatomic ion's charge.
7. If 25.0 g of Al₂O₃ and 75.0 g of carbon react according to Al₂O₃ + 3 C → 2 Al + 3 CO, what is the
maximum mass of Al that can be produced? (Molar masses: Al₂O₃ = 101.96 g/mol, C = 12.01 g/mol, Al
= 26.98 g/mol)
A) 13.2 g
B) 112 g
General Chemistry w/Lab | Portage | 26/27 Actual
(PDF)
1. A metal sample weighing 30.9232 g was added to a graduated cylinder containing 23.20 mL of
water. The volume of the water plus the sample was 24.80 mL. What is the density of the metal,
expressed to the correct number of significant figures?
A) 19.3 g/mL
B) 19.33 g/mL
C) 19.327 g/mL
D) 2.0 × 10¹ g/mL
Correct Answer: 19.3 g/mL
Rationale: Volume of metal = 24.80 mL − 23.20 mL = 1.60 mL. Density = 30.9232 g / 1.60 mL = 19.327
g/mL. Since the volume has only two significant figures, the density is rounded to 19.3 g/mL. Options
B and C overstate precision, and option D is an incorrect calculation.
2. Rutherford's gold foil experiment led to which conclusion about atomic structure?
A) Electrons are embedded in a positively charged sphere.
B) The nucleus is dense, positively charged, and occupies a small fraction of the atom's volume.
C) Atoms are indivisible and cannot be broken down further.
D) Electrons orbit the nucleus in fixed circular paths.
Correct Answer: The nucleus is dense, positively charged, and occupies a small fraction of the atom's
volume.
,Rationale: The deflection of alpha particles indicated a dense, positively charged center. Option A
describes the plum pudding model, option C is Dalton's atomic theory, and option D is the Bohr
model. Options A, C, and D are incorrect.
3. Chlorine has two stable isotopes: ³⁵Cl (34.96885 amu) and ³⁷Cl (36.96590 amu). If the atomic mass of
chlorine is 35.45 amu, what is the approximate natural abundance of ³⁵Cl?
A) 24.33%
B) 50.00%
C) 75.77%
D) 64.55%
Correct Answer: 75.77%
Rationale: Let x be the abundance of ³⁵Cl. 34.96885x + 36.96590(1 − x) = 35.45. Solving gives x ≈
0.7577, or 75.77%. Options A, B, and D do not yield the correct average atomic mass.
4. How many grams of CH₄ contain the same number of molecules as 2.50 g of O₂?
A) 0.0781 g
B) 1.25 g
C) 0.156 g
D) 4.88 × 10⁻³ g
Correct Answer: 1.25 g
Rationale: Moles O₂ = 2.50 g / 32.00 g/mol = 0.0781 mol. Mass CH₄ = 0.0781 mol × 16.04 g/mol = 1.25
g. Option A is the number of moles, option C is half the correct mass, and option D is an incorrect
calculation.
5. Consider the reaction: 2 NaI(aq) + Cl₂(g) → I₂(aq) + 2 NaCl(aq). Which element undergoes reduction?
, A) Sodium
B) Iodide
C) Chlorine
D) Iodine
Correct Answer: Chlorine
Rationale: Chlorine goes from oxidation state 0 in Cl₂ to −1 in NaCl, gaining electrons and undergoing
reduction. Iodide is oxidized from −1 to 0. Sodium and iodine are not reduced. Options A, B, and D are
incorrect.
6. What is the oxidation number of chlorine in KClO₃?
A) +6
B) +5
C) −1
D) +2
Correct Answer: +5
Rationale: K is +1, O is −2. Let x be Cl: +1 + x + 3(−2) = 0, so x = +5. Options A, C, and D result from
incorrect algebraic setups or ignoring the polyatomic ion's charge.
7. If 25.0 g of Al₂O₃ and 75.0 g of carbon react according to Al₂O₃ + 3 C → 2 Al + 3 CO, what is the
maximum mass of Al that can be produced? (Molar masses: Al₂O₃ = 101.96 g/mol, C = 12.01 g/mol, Al
= 26.98 g/mol)
A) 13.2 g
B) 112 g
