Mole Concept Explained for NEET & CBSE | Easy Notes + Examples

  

Simple Explanation of Mole Concept with Formulas & Examples for NEET Preparation

Mole Concept Explained for NEET & CBSE | Easy Notes + Examples 

Brief Overview

This note covers mole concept and was created from the MOLE Concept : CBSE / ICSE :. It explains the basic definition, Avogadro's number, and how to calculate moles from mass or volume, with practical worked examples.

Key Points

• Avogadro's number and its role in counting particles.

• Formula for moles from mass and from gas volume at STP.

• Step‑by‑step worked examples covering atoms, molecules, and gases.

• Summary table indicating when to use each formula.

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🔬 The Mole Concept

Mole: A counting unit equal to units of any substance. Just as one dozen equals 12 items, one mole equals items.

🎯 Avogadro's Number

Term	Symbol	Value
Avogadro's Number

• 1 mole of atoms = atoms

• 1 mole of molecules = molecules

• 1 mole of bicycles = bicycles

The mole is just a number — a very large number used to count microscopic particles.

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⚖️ Calculating Moles: Two Essential Formulas

Formula 1: From Mass (Universal — works for all substances)

Molar mass = Atomic mass or molecular mass expressed in grams (add "g" to the numerical value)

Formula 2: From Volume (Ideal Gases only — at STP)

Critical condition: This formula applies only at STP (Standard Temperature and Pressure) and only for ideal gases

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📝 Worked Examples

Example 1: Moles and Atoms from Mass

Problem: Find number of moles and number of atoms in 120 g of calcium

Solution:

• Mass given = 120 g

• Molar mass of Ca = 40 g/mol

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Example 2: Moles of Molecules from Mass

Problem: Find moles of molecules in 34 g of ammonia (NH₃)

Solution:

• Mass given = 34 g

• Molecular mass of NH₃ = 14 + (1 × 3) = 17

• Molar mass = 17 g/mol

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Example 3: Moles from Volume (Gas at STP)

Problem: Find moles of molecules in 56 L of CO₂ at STP

Solution:

• Volume given = 56 L (at STP)

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Example 4: Mass from Volume (Gas at STP)

Problem: Find mass of 11.2 L of nitrogen (N₂) at STP

Solution

Example3.

📘 Step 1: Use Standard Concept

At STP:
👉 1 mole of any gas = 22.4 L

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📘 Step 2: Apply Formula

$$\text{Moles}=\frac{\text{Given Volume}}{22.4}$$

$n=\frac{V}{22.4}$

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📘 Step 3: Substitute Values

$$n=\frac{56}{22.4}$$

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📘 Step 4: Solve

$$n=2.5$$

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✅ Final Answer

👉 Number of moles of CO₂ = 2.5 mol

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🎯 Quick NEET Trick

• Just remember: 22.4 × 2.5 = 56

• So directly → 2.5 mol

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🧪 Problem

Find the mass of 11.2 L of nitrogen (N₂) at STP

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📘 Step 1: Find moles from volume

At STP:
👉 1 mol gas = 22.4 L

$$n=\frac{V}{22.4}$$

$n=\frac{V}{22.4}$

Substitute values:

$$n=\frac{11.2}{22.4}=0.5\text{ mol}$$

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📘 Step 2: Find mass from moles

👉 Molar mass of N₂ = 28 g/mol

$$\text{Mass}=n\times \text{Molar Mass}$$

$m=n\times M$

Substitute values:

$$\text{Mass}=0.5\times 28=14\text{ g}$$

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✅ Final Answer

👉 Mass of N₂ = 14 g

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🎯 Quick NEET Trick

• 11.2 L = half of 22.4 L

• So moles = 0.5

• Half of 28 = 14 g

Example 5: Volume from Mass (Gas at STP)

Problem: Find volume of 32 g of sulfur dioxide (SO₂) at STP

Solution:Step 1: 

📘 Step 1: Calculate moles from mass

👉 Molar mass of SO₂
= 32 (S) + 16×2 (O) = 64 g/mol

$$n = \frac{m}{M}$$

$n = \frac{m}{M}$

$$n = \frac{32}{64} = 0.5 \text{ mol}$$

Step 2: 

📘 Step 2: Convert moles to volume

At STP:

👉 1 mol gas = 22.4 L

$$V = n \times 22.4$$

$V = n \times 22.4$

$$V = 0.5 \times 22.4 = 11.2 \text{ L}$$$$\text{✅ Final Answer 👉 Volume = 11.2 L 🎯 Quick Trick (Exam Booster) 32 g is half of 64 g (molar mass) So directly → half mole → half volume = 11.2 L}$$

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🔗  Internal Links

1. Atomic Structure Notes (Class 11)
2. Stoichiometry Complete Guide
3. Units & Dimensions Physics Notes
4. NEET Chemistry Syllabus Breakdown
5. 100 Mole Concept Numericals Practice
6. Gas Laws Explained (Boyle, Charles Law)

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🎯 Summary Table: When to Use Each Formula

Given Information	Formula to Use	Conditions
Mass		Universal — works for all substances
Volume of gas		Only for ideal gases at STP