Some Basic Concepts of Chemistry Class 11 Notes

Some Basic Concepts of Chemistry

MOLE CONCEPT

1. What is a Mole?

    1 mole = 6.022 × 10²³ particles

    This number is called Avogadro's Number (Nₐ)

    Nₐ = 6.02214076 × 10²³ mol⁻¹ (exact value since 2019)

Just like:

1 dozen = 12 items
1 pair = 2 items
1 mole = 6.022 × 10²³ items (atoms, molecules, ions, electrons, etc.)

Molar Mass (M)

The mass of 1 mole of a substance (in grams) = numerical value of relative atomic/molecular mass (A_r or M_r).

Molar mass of a substance = A_r or M_r in g/mol

Substance	Formula	Molar Mass
Carbon	C	12.01 g/mol
Oxygen gas	O₂	32.00 g/mol
Water	H₂O	18.02 g/mol
Sodium chloride	NaCl	58.44 g/mol
Glucose	C₆H₁₂O₆	180.16 g/mol

2. Core Mole Calculations

    Number of moles (n)

    n = Mass (g)⁄Molar Mass (g/mol)    →    n = m / M

    n = Number of particles⁄6.022×10²³    →    n = N / Nₐ

    n = Volume of gas (dm³)⁄24 dm³/mol (at RTP)    →    n = V / 24

    n = Concentration (mol/dm³) × Volume (dm³)    →    n = C × V

 n = Pressure x Volume/Gas Constant x Temperature    →    PV/RT

Summary Table

Quantity	Symbol	Unit	Formula
Amount of substance	n	mol	—
Mass	m	g	m = n × M
Number of particles	N	—	N = n × 6.022×10²³
Volume of gas (RTP)	V	dm³ or L	V = n × 24
Concentration	C	mol/dm³	C = n / V(dm³)

3. Moles and Gases

RTP (Room Temperature & Pressure): 25°C (298 K) and 1 atm

At RTP: 1 mole of any gas occupies 24 dm³ (24 liters)

STP (Standard Temperature & Pressure): 0°C (273 K) and 1 atm

    At STP: 1 mole occupies 22.7 dm³ (older syllabus) or 22.4 dm³ (most boards now use 24 dm³)

Ideal Gas Equation:

PV = nRT

R = 8.314 J mol⁻¹ K⁻¹ (or 0.0821 L atm mol⁻¹ K⁻¹)

4. Moles in Solutions

    Concentration (molarity) = moles of solute⁄volume of solution (dm³)

    C = n / V    (V in dm³ or liters)

Conversions:

1 dm³ = 1000 cm³ = 1 liter
500 cm³ = 0.5 dm³
25.0 cm³ = 0.025 dm³

5. Stoichiometry – The Heart of Mole Calculations

Mole ratios from balanced equations enable mass-mass, mass-volume, or mole-mole calculations in reactions.
For instance, in 2H₂ + O₂ → 2H₂O, 2 moles H₂ react with 1 mole O₂ to produce 2 moles H₂O

Steps for any stoichiometry problem:

Write the balanced equation
Convert given data → moles
Use mole ratio from equation
Convert moles → required unit (mass, volume, particles, etc.)

Example: How many grams of oxygen are needed to completely burn 24 g of carbon?
C + O₂ → CO₂
Moles of C = 24 / 12 = 2 mol
Mole ratio C:O₂ = 1:1 → 2 mol O₂ needed
Mass of O₂ = 2 × 32 = 64 g

6. Fully Worked Examples

Q1. Calculate the number of molecules in 9.0 g of water.
Molar mass H₂O = 18 g/mol
n = 9.0 / 18 = 0.50 mol
Number of molecules = 0.50 × 6.022×10²³ = 3.01 × 10²³ molecules

Q2. What volume will 0.80 mol of CO₂ occupy at RTP?
V = n × 24 = 0.80 × 24 = 19.2 dm³

Q3. 25.0 cm³ of 0.200 mol/dm³ NaOH neutralizes how many grams of HCl?
HCl + NaOH → NaCl + H₂O
Moles NaOH = 0.200 × (25/1000) = 0.005 mol
Mole ratio 1:1 → 0.005 mol HCl
Mass HCl = 0.005 × 36.5 = 0.1825 g

Q4. Calculate the empirical formula of a compound containing 40% C, 6.67% H, 53.33% O.
Assume 100 g → C:40g, H:6.67g, O:53.33g
Moles: C=40/12=3.33, H=6.67/1=6.67, O=53.33/16=3.33
Divide by smallest: C:1, H:2, O:1
Empirical formula = CH₂O

Q5. 20 cm³ of a gas weighs 0.90 g at RTP. Find its molecular mass.
V = 0.020 dm³ → n = 0.020 / 24 = 0.000833 mol
M = mass/n = 0.90 / 0.000833 ≈ 108 g/mol

7. Previous Year Questions & Exam-Specific Tricks

NEET 2024 9.0 g water contains how many molecules?
→ 9/18 = 0.5 mol → 0.5 × 6.022×10²³ = 3.011 × 10²³

JEE Main 2023 Volume of 0.4 mol CO₂ at RTP?
→ 0.4 × 24 = 9.6 L

JEE Adv 2022 A gas occupies 5.6 L at 1 atm and 273 K. Find molecules.
→ 5.6/22.4 = 0.25 mol → 0.25 × 6.022×10²³ = 1.5055 × 10²³

CSIR-NET Empirical formula from combustion: 0.1 mol CO₂ and 0.2 mol H₂O from 0.1 mol compound → EF = CH₂O

Quick Revision Checklist (Before Exam)

Avogadro number: 6.022 × 10²³
Molar volume RTP: 24 L
1 dm³ = 1 L = 1000 cm³
Molarity × volume(in L) = moles
Limiting reagent gives least product
% yield = (actual/theoretical) × 100

Some Basic Concepts of Chemistry Class 11 Notes