ICSE Class 10 Chemistry: Mole Concept and Stoichiometry — Important Questions with Answers 2026
Tushar Parik
Author
ICSE Class 10 Chemistry: Mole Concept and Stoichiometry — Important Questions with Answers 2026
This comprehensive guide from Bright Tutorials covers everything you need to know — with clear explanations, exam tips, and key points for board exam preparation.
In This Article
Short Answer Questions (2-3 Marks)
- Q: What is a mole? What is Avogadro's number?
Ans: A mole is the amount of substance that contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. Avogadro's number (NA) = 6.022 × 10²³. 1 mole of any substance contains 6.022 × 10²³ particles. It is also called Avogadro's constant. - Q: Define: (i) Atomic mass, (ii) Molecular mass, (iii) Molar mass.
Ans: (i) Atomic mass: Relative mass of one atom compared to 1/12th mass of C-12 atom. Unit: amu/u. (ii) Molecular mass: Sum of atomic masses of all atoms in a molecule. Example: H₂O = 2(1) + 16 = 18 u. (iii) Molar mass: Mass of 1 mole of a substance in grams. Numerically equal to molecular mass but unit is g/mol. - Q: Calculate the number of moles in 44 g of CO₂.
Ans: Molar mass of CO₂ = 12 + 2(16) = 44 g/mol. Number of moles = Given mass / Molar mass = 44/44 = 1 mole. Number of molecules = 1 × 6.022 × 10²³ = 6.022 × 10²³ molecules. - Q: State Gay-Lussac's Law of Gaseous Volumes. Give an example.
Ans: Gay-Lussac's Law states: When gases react, their volumes bear a simple whole number ratio to each other and to the volumes of gaseous products, measured at the same temperature and pressure. Example: N₂ + 3H₂ → 2NH₃. Volume ratio = 1:3:2. - Q: What is the molar volume of a gas at STP? Calculate the volume of 2 moles of O₂ at STP.
Ans: At STP (Standard Temperature and Pressure: 0°C, 1 atm), 1 mole of any gas occupies 22.4 litres. This is called molar volume. Volume of 2 moles of O₂ = 2 × 22.4 = 44.8 litres at STP.
Long Answer / Application Questions (4-6 Marks)
- Q: In the reaction: CaCO₃ → CaO + CO₂, calculate the mass of CaO formed from 200 g of CaCO₃.
Ans: Molar masses: CaCO₃ = 100 g/mol, CaO = 56 g/mol. From equation: 100 g CaCO₃ produces 56 g CaO. 200 g CaCO₃ produces (56/100) × 200 = 112 g CaO.
Exam Tips for This Chapter
- Revise all definitions and laws from Mole Concept and Stoichiometry — they are commonly asked as 1-2 mark questions
- Practice diagrams related to Mole Concept and Stoichiometry — neat labelled diagrams carry 2-3 marks
- For numericals, always show formula → substitution → answer with correct units
- Previous year analysis shows Mole Concept and Stoichiometry carries 8-12 marks in the board exam
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