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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

800 questions

ChemistryClass 112 markseasy

Some Basic Concepts of Chemistry

State the law of conservation of mass and the law of definite proportions.

Reveal model answer + marking points▾

Law of conservation of mass: matter can neither be created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products. Law of definite proportions: a given chemical compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source or method of preparation.

Marking-scheme points

  • ✓Conservation of mass: mass of reactants = mass of products
  • ✓Definite proportions: fixed ratio of elements by mass
  • ✓Example: water always 1:8 H:O by mass

The concept

Two founding laws: mass is conserved in a reaction, and a compound always has the same fixed ratio of elements.

Why it matters

They gave birth to the atomic theory and balanced equations - the start of modern chemistry.

💡 Memory trick

Mass stays (conservation); ratio stays (definite proportions) - 'both stay the same'.

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ChemistryClass 112 markseasy

Some Basic Concepts of Chemistry

Calculate the number of moles present in 11 g of carbon dioxide (CO2).

Reveal model answer + marking points▾

Molar mass of CO2 = 12 + 2(16) = 44 g/mol. Number of moles = given mass / molar mass = 11 / 44 = 0.25 mol.

n = m / M

Marking-scheme points

  • ✓Molar mass of CO2 = 44 g/mol
  • ✓moles = mass / molar mass
  • ✓n = 11/44 = 0.25 mol
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ChemistryClass 113 marksmedium

Some Basic Concepts of Chemistry

How many molecules and how many atoms are present in 0.25 mol of CO2? (Avogadro number = 6.022 x 10^23)

Reveal model answer + marking points▾

Number of molecules = 0.25 x 6.022 x 10^23 = 1.506 x 10^23 molecules. Each CO2 molecule has 3 atoms (1 C + 2 O), so number of atoms = 3 x 1.506 x 10^23 = 4.52 x 10^23 atoms.

N = n x N_A

Marking-scheme points

  • ✓molecules = n x N_A = 0.25 x 6.022e23 = 1.506e23
  • ✓Atoms per CO2 molecule = 3
  • ✓atoms = 3 x 1.506e23 = 4.52e23
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