NEET 2027
HIGH-WEIGHTAGE CHEMISTRY
PART 1
MOLE CONCEPT • ATOMIC STRUCTURE CHEMICAL BONDING • THERMODYNAMICS •
EQUILIBRIUM
100 Most Important Practice Questions with Full Explanations
Accurate • High-Weightage • Professional
Special Features
✓ Only the most important topics that repeatedly appear in NEET
✓ 100 carefully selected accurate questions
✓ Short theory before every chapter
✓ Detailed reason why the correct option is right
✓ Clear reason why each wrong option is wrong
✓ Perfect for NEET 2027 scoring
Physical Chemistry Core + Chemical Bonding
, CONTENTS
1. Some Basic Concepts of Chemistry (Mole Concept) — Theory + Q1 to Q20
2. Atomic Structure — Theory + Q21 to Q40
3. Chemical Bonding and Molecular Structure — Theory + Q41 to Q65
4. Thermodynamics — Theory + Q66 to Q85
5. Chemical Equilibrium — Theory + Q86 to Q100
Quick Revision Formula Sheet — Last page
How to Use This Book
1. Read the short theory of each chapter carefully.
2. Solve the questions without looking at answers.
3. Then read why the correct option is right and why others are wrong.
4. Revise the formula sheet before the exam.
NEET 2027 | High-Weightage Chemistry | Part 1 Page 2
, 1. SOME BASIC CONCEPTS OF CHEMISTRY (MOLE CONCEPT)
One mole = 6.022 × 10²³ particles (Avogadro’s number, N■). Molar mass = mass of 1 mole in grams. Number of
moles n = Given mass/Molar mass = Number of particles/N■ = Volume at STP (L)/22.4. Empirical formula =
simplest whole-number ratio of atoms. Molecular formula = n × empirical formula (n = Molar mass/Empirical
formula mass). Limiting reagent is completely consumed and decides the amount of product formed. Percentage
composition is calculated from the molecular formula.
Q1. The number of molecules in 22.4 L of N■ gas at STP is:
A. 6.022 × 10²³
B. 3.011 × 10²³
C. 1.204 × 10²■
D. 6.022 × 10²²
✓ Correct Answer: A
Why this is correct: At STP, 1 mole of any ideal gas occupies 22.4 L and contains Avogadro’s number of molecules (6.022 ×
10²³).
✗ Why the other options are wrong:
• B corresponds to 0.5 mole.
• C corresponds to 2 moles.
• D corresponds to 0.1 mole.
Q2. Mass of one molecule of water is:
A. 18 g
B. 18 / N■ g
C. 18 × N■ g
D. 1/18 g
✓ Correct Answer: B
Why this is correct: Molar mass of H■O = 18 g mol■¹. Mass of one molecule = molar mass / N■ = .022 × 10²³ g.
✗ Why the other options are wrong:
• A is the mass of one mole.
• C is incorrectly multiplied.
• D has no physical meaning here.
Q3. Number of moles in 4.4 g of CO■ is:
A. 0.1
B. 0.2
C. 1.0
D. 0.01
✓ Correct Answer: A
Why this is correct: Molar mass of CO■ = 44 g mol■¹. n = given mass / molar mass = 4. = 0.1 mol.
✗ Why the other options are wrong:
• B would require 8.8 g.
• C would require 44 g.
• D would require 0.44 g.
Q4. A compound has empirical formula CH■O and molar mass 180 g mol■¹. Its molecular formula is:
A. CH■O
B. C■H■O■
C. C■H■■O■
D. C■H■O■
✓ Correct Answer: C
Why this is correct: Empirical formula mass = 12+2+16 = 30. n = 180/30 = 6. Molecular formula = C■H■■O■.
NEET 2027 | High-Weightage Chemistry | Part 1 Page 3
HIGH-WEIGHTAGE CHEMISTRY
PART 1
MOLE CONCEPT • ATOMIC STRUCTURE CHEMICAL BONDING • THERMODYNAMICS •
EQUILIBRIUM
100 Most Important Practice Questions with Full Explanations
Accurate • High-Weightage • Professional
Special Features
✓ Only the most important topics that repeatedly appear in NEET
✓ 100 carefully selected accurate questions
✓ Short theory before every chapter
✓ Detailed reason why the correct option is right
✓ Clear reason why each wrong option is wrong
✓ Perfect for NEET 2027 scoring
Physical Chemistry Core + Chemical Bonding
, CONTENTS
1. Some Basic Concepts of Chemistry (Mole Concept) — Theory + Q1 to Q20
2. Atomic Structure — Theory + Q21 to Q40
3. Chemical Bonding and Molecular Structure — Theory + Q41 to Q65
4. Thermodynamics — Theory + Q66 to Q85
5. Chemical Equilibrium — Theory + Q86 to Q100
Quick Revision Formula Sheet — Last page
How to Use This Book
1. Read the short theory of each chapter carefully.
2. Solve the questions without looking at answers.
3. Then read why the correct option is right and why others are wrong.
4. Revise the formula sheet before the exam.
NEET 2027 | High-Weightage Chemistry | Part 1 Page 2
, 1. SOME BASIC CONCEPTS OF CHEMISTRY (MOLE CONCEPT)
One mole = 6.022 × 10²³ particles (Avogadro’s number, N■). Molar mass = mass of 1 mole in grams. Number of
moles n = Given mass/Molar mass = Number of particles/N■ = Volume at STP (L)/22.4. Empirical formula =
simplest whole-number ratio of atoms. Molecular formula = n × empirical formula (n = Molar mass/Empirical
formula mass). Limiting reagent is completely consumed and decides the amount of product formed. Percentage
composition is calculated from the molecular formula.
Q1. The number of molecules in 22.4 L of N■ gas at STP is:
A. 6.022 × 10²³
B. 3.011 × 10²³
C. 1.204 × 10²■
D. 6.022 × 10²²
✓ Correct Answer: A
Why this is correct: At STP, 1 mole of any ideal gas occupies 22.4 L and contains Avogadro’s number of molecules (6.022 ×
10²³).
✗ Why the other options are wrong:
• B corresponds to 0.5 mole.
• C corresponds to 2 moles.
• D corresponds to 0.1 mole.
Q2. Mass of one molecule of water is:
A. 18 g
B. 18 / N■ g
C. 18 × N■ g
D. 1/18 g
✓ Correct Answer: B
Why this is correct: Molar mass of H■O = 18 g mol■¹. Mass of one molecule = molar mass / N■ = .022 × 10²³ g.
✗ Why the other options are wrong:
• A is the mass of one mole.
• C is incorrectly multiplied.
• D has no physical meaning here.
Q3. Number of moles in 4.4 g of CO■ is:
A. 0.1
B. 0.2
C. 1.0
D. 0.01
✓ Correct Answer: A
Why this is correct: Molar mass of CO■ = 44 g mol■¹. n = given mass / molar mass = 4. = 0.1 mol.
✗ Why the other options are wrong:
• B would require 8.8 g.
• C would require 44 g.
• D would require 0.44 g.
Q4. A compound has empirical formula CH■O and molar mass 180 g mol■¹. Its molecular formula is:
A. CH■O
B. C■H■O■
C. C■H■■O■
D. C■H■O■
✓ Correct Answer: C
Why this is correct: Empirical formula mass = 12+2+16 = 30. n = 180/30 = 6. Molecular formula = C■H■■O■.
NEET 2027 | High-Weightage Chemistry | Part 1 Page 3
