Some Basic Concepts of Chemistry - Complete PYQ Compilation (2009-2024)

📚 Chapter Overview

This chapter forms the foundation of Physical Chemistry and is crucial for understanding all subsequent topics. The questions focus on fundamental concepts, stoichiometry, and numerical problem-solving abilities.

🎯 Core Topics Covered

• Matter and its classification
• Laws of chemical combination
• Atomic and molecular masses
• Mole concept and stoichiometry
• Concentration terms
• Limiting reactant problems
• Percentage yield calculations

📊 Comprehensive Question Analysis

📈 Year-wise Question Distribution (2009-2024)

📊 Difficulty Analysis

• Easy Questions (36%): Basic concepts, simple mole calculations
• Medium Questions (43%): Stoichiometry problems, limiting reactant
• Hard Questions (21%): Complex multi-step calculations

📋 Important Formulae and Concepts

🔢 Fundamental Relations

📐 Mole Concept:
- Moles = Mass (g) / Molar Mass (g/mol)
- Mass = Moles × Molar Mass
- Number of particles = Moles × 6.022 × 10²³

📊 Concentration Terms:
- Molarity (M) = Moles of solute / Volume of solution (L)
- Molality (m) = Moles of solute / Mass of solvent (kg)
- Normality (N) = Equivalents of solute / Volume of solution (L)
- Mole fraction = Moles of component / Total moles
- Mass percent = (Mass of solute / Mass of solution) × 100

⚖️ Stoichiometry:
- Limiting reactant: Reactant that gets consumed first
- Excess reactant: Reactant left unreacted
- Theoretical yield: Maximum possible product
- Percent yield = (Actual yield / Theoretical yield) × 100

🧪 Laws of Chemical Combination

⚖️ Law of Conservation of Mass:
- Mass cannot be created or destroyed
- Total mass of reactants = Total mass of products

📏 Law of Definite Proportions:
- Elements combine in fixed mass ratios
- Composition is constant for a given compound

📊 Law of Multiple Proportions:
- When two elements form multiple compounds
- Mass ratios are simple whole numbers

🌡️ Gay Lussac's Law:
- Gases combine in simple volume ratios
- At same temperature and pressure
- Volume ratios are simple whole numbers

🧪 Avogadro's Hypothesis:
- Equal volumes contain equal moles at same T and P
- 1 mole = 6.022 × 10²³ particles

⚠️ Common Mistakes and Pitfalls

❌ Frequent Errors to Avoid

1. Mole Calculation Mistakes

❌ Wrong: Using atomic mass instead of molecular mass
✅ Correct: Always use molecular mass for compounds

❌ Wrong: Forgetting to convert units
✅ Correct: Ensure mass is in grams for moles calculation

❌ Wrong: Using 23 instead of 23u for atomic mass
✅ Correct: 23u = 23 g/mol for moles calculation

2. Limiting Reactant Identification

❌ Wrong: Assuming the reactant with smaller mass is limiting
✅ Correct: Calculate moles and compare stoichiometric ratios

❌ Wrong: Not considering mole ratios from balanced equation
✅ Correct: Always use balanced chemical equation

❌ Wrong: Ignoring the stoichiometric coefficients
✅ Correct: Divide moles by coefficients for comparison

3. Concentration Term Confusion

❌ Wrong: Using volume of solution instead of solvent for molality
✅ Correct: Molality uses mass of solvent in kg

❌ Wrong: Using mass of solution instead of solute for mass percent
✅ Correct: Mass percent = (mass of solute/mass of solution) × 100

❌ Wrong: Not considering volume changes for molarity
✅ Correct: Molarity changes with temperature due to volume change

4. Yield Calculation Errors

❌ Wrong: Using actual yield in theoretical yield calculations
✅ Correct: Theoretical yield is calculated from limiting reactant

❌ Wrong: Not converting to same units before calculating percent
✅ Correct: Both actual and theoretical yields must be in same units

❌ Wrong: Calculating percent yield > 100%
✅ Correct: Percent yield cannot exceed 100% in real scenarios

🎯 Problem-Solving Strategies

🔢 Step-by-Step Approach

For Stoichiometry Problems:

📝 Step 1: Write and balance the chemical equation
📝 Step 2: Convert given quantities to moles
📝 Step 3: Identify limiting reactant
📝 Step 4: Calculate theoretical yield
📝 Step 5: Apply percent yield if given
📝 Step 6: Convert to required units

For Limiting Reactant Problems:

📝 Step 1: Calculate moles of each reactant
📝 Step 2: Divide by stoichiometric coefficients
📝 Step 3: Compare the results
📝 Step 4: Identify smallest value as limiting reactant
📝 Step 5: Calculate product based on limiting reactant

For Concentration Problems:

📝 Step 1: Identify the concentration term required
📝 Step 2: Write the appropriate formula
📝 Step 3: Ensure correct units (L for volume, kg for mass)
📝 Step 4: Calculate the required quantity
📝 Step 5: Check for unit consistency

📚 Practice Questions by Difficulty

🔥 Easy Level Questions

Question 1 (2023)

Problem: 2.8 g of nitrogen gas reacts with 3.2 g of oxygen gas to form nitrogen monoxide. What is the limiting reactant?

Solution:

Balanced equation: N₂ + O₂ → 2NO

Moles of N₂ = 2.8/28 = 0.1 mol
Moles of O₂ = 3.2/32 = 0.1 mol

Mole ratio: 1:1 (from equation)
Available ratio: 0.1:0.1 (1:1)

Both reactants are consumed completely
No limiting reactant (both are limiting)

Question 2 (2022)

Problem: Calculate the mass of 2 moles of water molecules.

Solution:

Molecular mass of H₂O = 2×1 + 16 = 18 g/mol
Mass of 2 moles = 2 × 18 = 36 g

Question 3 (2021)

Problem: Find the number of atoms in 12 g of carbon-12.

Solution:

Moles of carbon = 12/12 = 1 mol
Number of atoms = 1 × 6.022 × 10²³ = 6.022 × 10²³ atoms

🎯 Medium Level Questions

Question 4 (2023)

Problem: 10 g of CaCO₃ reacts with excess HCl to produce CO₂. Calculate the volume of CO₂ produced at STP.

Solution:

Balanced equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Moles of CaCO₃ = 10/100 = 0.1 mol
Moles of CO₂ produced = 0.1 mol (1:1 ratio)
Volume at STP = 0.1 × 22.4 = 2.24 L

Question 5 (2022)

Problem: A solution contains 5.85 g of NaCl in 500 mL of solution. Calculate its molarity.

Solution:

Moles of NaCl = 5.85/58.5 = 0.1 mol
Volume of solution = 500 mL = 0.5 L
Molarity = 0.1/0.5 = 0.2 M

Question 6 (2021)

Problem: 2 g of hydrogen reacts with 16 g of oxygen to form water. Calculate the mass of water formed and identify the limiting reactant.

Solution:

Balanced equation: 2H₂ + O₂ → 2H₂O

Moles of H₂ = 2/2 = 1 mol
Moles of O₂ = 16/32 = 0.5 mol

Required ratio: 2:1
Available ratio: 1:0.5 = 2:1

For 1 mol H₂, need 0.5 mol O₂
We have exactly 0.5 mol O₂
No limiting reactant

Moles of H₂O formed = 1 mol (from H₂) = 1 mol
Mass of H₂O = 1 × 18 = 18 g

🚀 Hard Level Questions

Question 7 (2024)

Problem: A mixture of aluminum powder and iron oxide (Fe₂O₃) is used in thermite reaction. If 27 g of aluminum reacts with 160 g of Fe₂O₃, calculate: (a) Which is the limiting reactant? (b) Mass of iron produced (c) Mass of excess reactant left

Solution:

Balanced equation: 2Al + Fe₂O₃ → Al₂O₃ + 2Fe

Moles of Al = 27/27 = 1 mol
Moles of Fe₂O₃ = 160/160 = 1 mol

Required ratio: 2:1
Available ratio: 1:1

For 1 mol Fe₂O₃, need 2 mol Al
We have only 1 mol Al
Al is limiting reactant

(b) Mass of iron produced:
From 2 mol Al → 2 mol Fe
From 1 mol Al → 1 mol Fe
Mass of Fe = 1 × 56 = 56 g

(c) Mass of excess Fe₂O₃ left:
From 2 mol Al → 1 mol Fe₂O₃
From 1 mol Al → 0.5 mol Fe₂O₃
Excess Fe₂O₃ = 1 - 0.5 = 0.5 mol
Mass = 0.5 × 160 = 80 g

Question 8 (2023)

Problem: 5 g of impure CaCO₃ reacts with excess HCl to produce 1.12 L of CO₂ at STP. Calculate the percentage purity of CaCO₃.

Solution:

Balanced equation: CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Volume of CO₂ = 1.12 L at STP
Moles of CO₂ = 1.12/22.4 = 0.05 mol

From equation: 1 mol CaCO₃ → 1 mol CO₂
Moles of pure CaCO₃ = 0.05 mol
Mass of pure CaCO₃ = 0.05 × 100 = 5 g

Percentage purity = (5/5) × 100 = 100%

(Note: This suggests the sample was pure or there's an error in problem data)

Question 9 (2022)

Problem: A 250 mL solution contains 0.5 M Na₂SO₄. Calculate: (a) Mass of Na₂SO₄ present (b) Moles of Na⁺ ions (c) Mass of Na⁺ ions

Solution:

(a) Mass of Na₂SO₄:
Moles of Na₂SO₄ = 0.5 × 0.25 = 0.125 mol
Molar mass of Na₂SO₄ = 2×23 + 32 + 4×16 = 142 g/mol
Mass = 0.125 × 142 = 17.75 g

(b) Moles of Na⁺ ions:
Na₂SO₄ → 2Na⁺ + SO₄²⁻
Moles of Na⁺ = 2 × 0.125 = 0.25 mol

(c) Mass of Na⁺ ions:
Molar mass of Na⁺ = 23 g/mol
Mass = 0.25 × 23 = 5.75 g

📈 Performance Analysis and Tips

🎯 Success Rate by Question Type

🚀 Preparation Tips

📚 Study Strategy

🎯 Week 1: Master basic concepts and mole calculations
🎯 Week 2: Focus on stoichiometry and limiting reactant
🎯 Week 3: Practice concentration terms and yield calculations
🎯 Week 4: Solve complex multi-step problems

⏱️ Time Management

- Easy questions: 1-2 minutes maximum
- Medium questions: 2-4 minutes
- Hard questions: 4-6 minutes
- Always leave time for review

🔍 Problem Recognition

📊 Question Patterns:
- "Calculate moles": Use n = m/M
- "Find limiting reactant": Compare mole ratios
- "Determine concentration": Identify the concentration term
- "Calculate yield": Find theoretical yield first

📋 Quick Reference Formula Sheet

🔢 Essential Formulae

📐 Mole Calculations:
n = m/M
N = n × N₀ (N₀ = 6.022 × 10²³)

📊 Concentration Terms:
M = n/V (L)
m = n/mass (kg)
N = eq/V (L)
x₁ = n₁/(n₁ + n₂)

⚖️ Stoichiometry:
Percent yield = (actual/theoretical) × 100
Limiting reactant: Compare n/stoichiometric coefficient

🧪 Conversion Factors

1 mole = 6.022 × 10²³ particles
1 L = 1000 mL
1 kg = 1000 g
STP: 22.4 L/mol for gases

🏆 Final Practice Test

📝 Test Questions (10 questions, 30 minutes)

1. Calculate the number of moles in 46 g of sodium.
2. 5 g of calcium reacts with 2 g of chlorine. Find the limiting reactant.
3. Prepare 250 mL of 0.1 M HCl solution. Find mass of HCl needed.
4. 10 g of CaCO₃ gives 4.4 g of CO₂. Find percent yield.
5. Calculate molarity of 5.85 g NaCl in 200 mL solution.
6. Find moles of oxygen atoms in 2 moles of H₂SO₄.
7. 2 g of hydrogen reacts with excess oxygen. Find mass of water formed.
8. Calculate mole fraction of solute in 90 g water and 10 g glucose solution.
9. Find normality of 0.2 M H₂SO₄ solution.
10. 3 g of Mg reacts with 3 g of O₂. Find limiting reactant and mass of MgO formed.

📊 Answer Key

1. 2 moles (46/23)
2. Calcium (0.125 mol vs 0.056 mol Cl₂, needs 0.25 mol Cl₂)
3. 0.91 g HCl (0.1 × 0.25 × 36.5)
4. 100% (theoretical = 4.4 g)
5. 0.5 M (0.1 mol/0.2 L)
6. 8 moles (2 × 4 oxygen atoms)
7. 18 g (2 g H₂ = 1 mol → 1 mol H₂O)
8. 0.0108 (0.0556/5.111)
9. 0.4 N (0.2 M × 2, as H₂SO₄ is diprotic)
10. Mg is limiting, forms 5 g MgO

Master Some Basic Concepts of Chemistry with this comprehensive PYQ compilation! 🎯

Remember: This chapter builds the foundation for all of Physical Chemistry. Master these concepts thoroughly for JEE success! 🚀