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Some Basic Concepts of Chemistry – Class 11 Chemistry notes, important questions and MCQs

NCERT Class 11 Chemistry, chapter 1. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Chemistry measures matter in moles. Laws of chemical combination, molar mass, empirical formulae and stoichiometry let us calculate how much reacts and forms.

Mole concept

1 mole = 6.022 × 10²³ particles (Avogadro number); molar mass in g/mol equals the atomic or molecular mass.
At STP, 1 mole of an ideal gas occupies 22.4 L (older convention; 22.7 L at 1 bar).
Laws: conservation of mass, definite proportions, multiple proportions, Gay-Lussac's law of gaseous volumes, Avogadro's law.

Formulae and concentration

Empirical formula is the simplest ratio; molecular formula = n × empirical formula.
Limiting reagent is consumed first and decides the amount of product.
Molarity M = moles of solute/litre of solution; molality m = moles of solute/kg of solvent; mole fraction.

Key terms

Mole: amount containing 6.022 × 10²³ particles
Limiting reagent: reactant that is used up first
Mass percent: (mass of a component / total mass) × 100

Formulas

n = mass/molar mass
M = n/V(L)

Important questions with answers

Calculate the number of atoms in 0.5 mol of carbon.2 marks

0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms.

A compound has 40% C, 6.7% H and 53.3% O. Find its empirical formula.3 marks

Moles: C 40/12 = 3.33, H 6.7/1 = 6.7, O 53.3/16 = 3.33. Ratio 1 : 2 : 1, so the empirical formula is CH₂O.

3 g of H₂ reacts with 29 g of O₂ to form water. Which is the limiting reagent? How much water forms and how much of the excess reactant is left?5 marks

2H₂ + O₂ → 2H₂O. Moles: H₂ = 3/2 = 1.5 mol; O₂ = 29/32 ≈ 0.906 mol. 1.5 mol H₂ needs 0.75 mol O₂, which is available, so H₂ is limiting. Water = 1.5 mol × 18 = 27 g. O₂ left = 0.906 − 0.75 = 0.156 mol ≈ 5 g.

Quick revision

Avogadro number 6.022 × 10²³.
Empirical formula = simplest ratio.
Limiting reagent decides product.

Exam tips

Molality does not change with temperature; molarity does.
In stoichiometry, always convert given masses to moles first, then use the mole ratio from the balanced equation.

Practice MCQs

Number of particles in 1 mole is
1. 6.022 × 10²²
2. 1.6 × 10⁻¹⁹
3. 3.01 × 10²³
4. 6.022 × 10²³
Show answer
Answer: (d) 6.022 × 10²³. Avogadro number.
Molar mass of water (H₂O) is
1. 16 g/mol
2. 10 g/mol
3. 20 g/mol
4. 18 g/mol
Show answer
Answer: (d) 18 g/mol. 2 + 16.
Number of moles in 88 g of CO₂ is
1. 1
2. 2
3. 4
4. 0.5
Show answer
Answer: (b) 2. 88/44.
Molarity of a solution with 4 g NaOH in 500 mL is
1. 0.1 M
2. 0.2 M
3. 0.4 M
4. 1 M
Show answer
Answer: (b) 0.2 M. 0.1 mol/0.5 L.
Mass percent of carbon in CO₂ is about
1. 12%
2. 44%
3. 72.7%
4. 27.3%
Show answer
Answer: (d) 27.3%. 12/44 × 100.
A compound has empirical formula CH₂O and molar mass 180 g/mol. Its molecular formula is
1. C₃H₆O₃
2. C₆H₁₂O₆
3. C₂H₄O₂
4. CH₂O
Show answer
Answer: (b) C₆H₁₂O₆. n = 180/30 = 6.
Which concentration term is independent of temperature
1. molality
2. normality
3. molarity
4. mass per volume
Show answer
Answer: (a) molality. Based on mass.
4 g of H₂ reacts with 32 g of O₂. The mass of water formed is
1. 4 g
2. 18 g
3. 32 g
4. 36 g
Show answer
Answer: (d) 36 g. 2 mol H₂ + 1 mol O₂ → 2 mol H₂O.

All chapters of Class 11 Chemistry

Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.