CHY-103 GENERAL CHEMISTRY I TERM
ASSESSMENT FALL 2025/26 SOLUTIONS -
TORONTO METROPOLITAN UNIVERSITY.
149 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
CHY-103 GENERAL CHEMISTRY I TERM ASSESSMENT FALL 2025/26 SOLUTIONS - TORONTO
METROPOLITAN UNIVERSITY.. It contains 149 carefully selected questions that reflect the most current exam
content and testing strategies. Each question is accompanied by a correct answer and a detailed rationale that
explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 149 Questions
Foundations - Application - Chy-103 General Chemistry I TERM Assessment FALL 2025/26 Solutions -
Toronto Metropolitan University General Chemistry I Chy-103 Atomic Structure Stoichiometry
Thermochemistry Bonding Gases AND Solution Chemistry Undergraduate YEAR 1 First-year General
Chemistry R1 University Standard
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Atomic Structure AND 1-25 Reaction, Temperature, Electron, Sample, Point
Periodicity
Chemical Bonding AND 26-50 Reaction, Standard, Sample, Energy, Polar
Molecular Geometry
Stoichiometry AND Chemical 51-75 Reaction, Correctly, Sample, Describes, Point
Reactions
Gases AND GAS LAWS 76-100 Electron, Reaction, Standard, Correctly, Galvanic CELL
Thermochemistry AND 101-125 Reaction, Sample, Titrated, Unknown, Molecules
Thermodynamics
Solutions AND Colligative 126-149 Reaction, Sample, RATE Constant, Electron, Equilibrium
Properties
TOTAL 149 All questions include answers and detailed rationales
,Section A - Atomic Structure AND Periodicity
Q1.
An electron in a hydrogen atom transitions from n = 4 to n = 2. Using the Rydberg
equation (R_H = 1.097 × 10^7 m¹), what is the wavelength of the emitted photon?
A. 486 nm B. 656 nm
C. 122 nm D. 103 nm
Correct: A - 486 nm
Rationale:1/» = R_H(1/2² " 1/4²) = 1.097×10^7 (1/4 " 1/16) = 1.097×10^7 × 0.1875 =
2.057×10^6 m¹, so 4.86×10 m = 486 nm (Balmer series, H line).
Why the other answers are wrong:
B. 656 nm corresponds to the n=3->2 transition (H), not n=4->2.
C. 122 nm is a Lyman-series (n=2->1) wavelength, not Balmer.
D. 103 nm arises from n=3->1 Lyman transition, not n=4->2.
Reference: Tro, N.J. (2023). Chemistry: A Molecular Approach, 6th Ed., Ch. 7 (Quantum Mechanics &
Atomic Structure)
Q2.
A 25.0 mL sample of 0.100 M HSO is titrated with 0.100 M NaOH. What volume of NaOH is
required to reach the second equivalence point?
A. 25.0 mL B. 50.0 mL
C. 75.0 mL D. 12.5 mL
Correct: B - 50.0 mL
Rationale:H ‚SO „ is diprotic, so moles of H z = 2 × (0.0250 L × 0.100 mol/L) = 5.00×10 {³ mol.
NaOH needed = 5.00×10³ mol ÷ 0.100 M = 0.0500 L = 50.0 mL.
Why the other answers are wrong:
A. 25.0 mL would only neutralize one proton (first equivalence point).
C. 75.0 mL over-titrates by 50%, exceeding the second equivalence point.
D. 12.5 mL is half the first equivalence volume; insufficient for even the first proton.
Reference: Brown, T.L. et al. (2022). Chemistry: The Central Science, 15th Ed., Ch. 4 (Aqueous
Reactions & Solution Stoichiometry)
Q3.
Which of the following sets of quantum numbers (n, , m, ms) is NOT permitted for an
electron in an atom?
Page 3
, Section A - Atomic Structure AND Periodicity
A. (3, 2, 1, +½) B. (2, 1, 0, ½)
C. (4, 0, 1, +½) D. (3, 1, 1, +½)
Correct: C - (4, 0, 1, +½)
Rationale:For ! = 0 (s orbital), m! must be 0; m! = 1 is invalid. All other sets satisfy the
constraints |m| and ms = ±½.
Why the other answers are wrong:
A. Valid: = 2 allows m = 1 and ms = +½.
B. Valid: = 1 allows m = 0 and ms = ½.
D. Valid: = 1 allows m = 1 and ms = +½.
Reference: Zumdahl, S.S. & Zumdahl, S.A. (2023). Chemistry, 11th Ed., Ch. 7 (Atomic Structure &
Periodicity)
Q4.
For the reaction N(g) + 3H(g) -> 2NH(g), H° = 92 kJ/mol. According to Le Chatelier's
principle, which change will increase the equilibrium yield of NH?
A. Increasing temperature B. Increasing volume at constant
temperature
C. Removing H from the system D. Increasing pressure by decreasing
volume
Correct: D - Increasing pressure by decreasing volume
Rationale:The reaction reduces moles of gas (4 !’ 2), so increasing pressure (decreasing
volume) shifts equilibrium toward products, increasing NH yield. The exothermic nature
means lower temperature also favors products, but among the options only pressure increase
works.
Why the other answers are wrong:
A. Higher temperature shifts an exothermic reaction toward reactants, decreasing NH.
B. Increasing volume lowers pressure and shifts toward the side with more gas moles
(reactants).
C. Removing a reactant (H) shifts equilibrium toward reactants, decreasing NH.
Reference: Atkins, P. & Jones, L. (2023). Chemical Principles, 8th Ed., Ch. 10 (Chemical Equilibrium)
Q5.
A 0.500 g sample of an unknown hydrocarbon is combusted in a bomb calorimeter with
heat capacity 4.20 kJ/°C. The temperature rises from 22.00 °C to 25.50 °C. What is the heat
of combustion per gram of the sample?
A. 14.7 kJ/g B. 29.4 kJ/g
Page 4
ASSESSMENT FALL 2025/26 SOLUTIONS -
TORONTO METROPOLITAN UNIVERSITY.
149 Questions with Answers and Detailed Rationales
100 PERCENT GUARANTEED PASS
INSTANT DOWNLOAD ANSWERS INCLUDED
IMPORTANCE OF THIS DOCUMENT
This comprehensive examination preparation guide has been meticulously developed to help you succeed in the
CHY-103 GENERAL CHEMISTRY I TERM ASSESSMENT FALL 2025/26 SOLUTIONS - TORONTO
METROPOLITAN UNIVERSITY.. It contains 149 carefully selected questions that reflect the most current exam
content and testing strategies. Each question is accompanied by a correct answer and a detailed rationale that
explains the underlying pathophysiology, pharmacology, or clinical reasoning.
Self-Assessment – Test your knowledge and Exam Preparation – Familiarize yourself with the
identify areas requiring further question format and content
study areas
Concept Reinforcement – Deepen your Confidence Building – Develop test-taking
understanding through strategies and reduce
evidence-based exam anxiety
rationales
Time Management – Practice answering
questions under simulated
exam conditions
Review Summary 149 Questions
Foundations - Application - Chy-103 General Chemistry I TERM Assessment FALL 2025/26 Solutions -
Toronto Metropolitan University General Chemistry I Chy-103 Atomic Structure Stoichiometry
Thermochemistry Bonding Gases AND Solution Chemistry Undergraduate YEAR 1 First-year General
Chemistry R1 University Standard
All answers with rationales
,Table of Contents
Content Area Questions Key Topics
Atomic Structure AND 1-25 Reaction, Temperature, Electron, Sample, Point
Periodicity
Chemical Bonding AND 26-50 Reaction, Standard, Sample, Energy, Polar
Molecular Geometry
Stoichiometry AND Chemical 51-75 Reaction, Correctly, Sample, Describes, Point
Reactions
Gases AND GAS LAWS 76-100 Electron, Reaction, Standard, Correctly, Galvanic CELL
Thermochemistry AND 101-125 Reaction, Sample, Titrated, Unknown, Molecules
Thermodynamics
Solutions AND Colligative 126-149 Reaction, Sample, RATE Constant, Electron, Equilibrium
Properties
TOTAL 149 All questions include answers and detailed rationales
,Section A - Atomic Structure AND Periodicity
Q1.
An electron in a hydrogen atom transitions from n = 4 to n = 2. Using the Rydberg
equation (R_H = 1.097 × 10^7 m¹), what is the wavelength of the emitted photon?
A. 486 nm B. 656 nm
C. 122 nm D. 103 nm
Correct: A - 486 nm
Rationale:1/» = R_H(1/2² " 1/4²) = 1.097×10^7 (1/4 " 1/16) = 1.097×10^7 × 0.1875 =
2.057×10^6 m¹, so 4.86×10 m = 486 nm (Balmer series, H line).
Why the other answers are wrong:
B. 656 nm corresponds to the n=3->2 transition (H), not n=4->2.
C. 122 nm is a Lyman-series (n=2->1) wavelength, not Balmer.
D. 103 nm arises from n=3->1 Lyman transition, not n=4->2.
Reference: Tro, N.J. (2023). Chemistry: A Molecular Approach, 6th Ed., Ch. 7 (Quantum Mechanics &
Atomic Structure)
Q2.
A 25.0 mL sample of 0.100 M HSO is titrated with 0.100 M NaOH. What volume of NaOH is
required to reach the second equivalence point?
A. 25.0 mL B. 50.0 mL
C. 75.0 mL D. 12.5 mL
Correct: B - 50.0 mL
Rationale:H ‚SO „ is diprotic, so moles of H z = 2 × (0.0250 L × 0.100 mol/L) = 5.00×10 {³ mol.
NaOH needed = 5.00×10³ mol ÷ 0.100 M = 0.0500 L = 50.0 mL.
Why the other answers are wrong:
A. 25.0 mL would only neutralize one proton (first equivalence point).
C. 75.0 mL over-titrates by 50%, exceeding the second equivalence point.
D. 12.5 mL is half the first equivalence volume; insufficient for even the first proton.
Reference: Brown, T.L. et al. (2022). Chemistry: The Central Science, 15th Ed., Ch. 4 (Aqueous
Reactions & Solution Stoichiometry)
Q3.
Which of the following sets of quantum numbers (n, , m, ms) is NOT permitted for an
electron in an atom?
Page 3
, Section A - Atomic Structure AND Periodicity
A. (3, 2, 1, +½) B. (2, 1, 0, ½)
C. (4, 0, 1, +½) D. (3, 1, 1, +½)
Correct: C - (4, 0, 1, +½)
Rationale:For ! = 0 (s orbital), m! must be 0; m! = 1 is invalid. All other sets satisfy the
constraints |m| and ms = ±½.
Why the other answers are wrong:
A. Valid: = 2 allows m = 1 and ms = +½.
B. Valid: = 1 allows m = 0 and ms = ½.
D. Valid: = 1 allows m = 1 and ms = +½.
Reference: Zumdahl, S.S. & Zumdahl, S.A. (2023). Chemistry, 11th Ed., Ch. 7 (Atomic Structure &
Periodicity)
Q4.
For the reaction N(g) + 3H(g) -> 2NH(g), H° = 92 kJ/mol. According to Le Chatelier's
principle, which change will increase the equilibrium yield of NH?
A. Increasing temperature B. Increasing volume at constant
temperature
C. Removing H from the system D. Increasing pressure by decreasing
volume
Correct: D - Increasing pressure by decreasing volume
Rationale:The reaction reduces moles of gas (4 !’ 2), so increasing pressure (decreasing
volume) shifts equilibrium toward products, increasing NH yield. The exothermic nature
means lower temperature also favors products, but among the options only pressure increase
works.
Why the other answers are wrong:
A. Higher temperature shifts an exothermic reaction toward reactants, decreasing NH.
B. Increasing volume lowers pressure and shifts toward the side with more gas moles
(reactants).
C. Removing a reactant (H) shifts equilibrium toward reactants, decreasing NH.
Reference: Atkins, P. & Jones, L. (2023). Chemical Principles, 8th Ed., Ch. 10 (Chemical Equilibrium)
Q5.
A 0.500 g sample of an unknown hydrocarbon is combusted in a bomb calorimeter with
heat capacity 4.20 kJ/°C. The temperature rises from 22.00 °C to 25.50 °C. What is the heat
of combustion per gram of the sample?
A. 14.7 kJ/g B. 29.4 kJ/g
Page 4
