Class 11 Chemistry Basic Concepts of Chemistry Exam Questions

Find the Class 11 Chemistry Basic Concepts of Chemistry Exam Questions right below. We provide advanced study materials for Class 11 Chemistry learners for the 2026-27 term, including expert notes and solved practice questions. Every section matches the latest guidelines from CBSE, NCERT, and KVS.

Advanced Study Material for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

Every Class 11 student studying Chemistry can use this Chapter 1 Some Basic Concepts of Chemistry study material to look past basic textbook lessons. It features helpful chapter summaries and solved questions to make learning much easier.

Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry Advanced Resources

LEVEL WISE QUESTIONS
 
LEVEL- 1

1. State the number of significant figures in (i) 0.436 and (ii) 802.42
Ans. (i) 0.436 has three significant figures
       (ii) 802.42 has five significant figures
 
2. What do you mean by Mole fraction?
Ans. Mole fraction is the ratio of number of moles of one component to the total number of moles (solute & solvents) present in the solution. It is expressed as ‘x’.
 
3. Calculate the molecular mass of H2SO4.
Ans. Molecular Mass of H2SO4 = 2 × 1 + 32 + 4 × 16 = 2 + 32 + 64 = 98u
 
4. What is the SI unit of mass? How is it defined?
Ans. The S.I. unit of mass is Kilogram.
Mass of a substance is the amount of matter present in it.
 
5. What is limiting reagent?
Ans. The reactant which gets consumed first or limits the amount of product formed is known as limiting reagent.
 
LEVEL- 2

1. What is the value of Avogadro number?
Ans. 6.022 × 1023
 
2. What is the law called which deals with the ratios of the volumes of the gaseous reactants and products of the volumes of the gaseous reactants and products?
Ans. Gay Lussac’s law of gaseous volumes.
 
3. How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?
Ans. Molar mass of Na2CO3 = 2 × 23 + 12 × 3 × 16 = 106 g/mol.
0.50 mol Na2CO3 means 0.50 × 106 = 53 g.
0.50 mol Na2CO3 means 0.50 mol i.e. 53 g of Na2CO3 are present in 1 L of the solution.
 
4. What is the percentage of carbon, hydrogen and oxygen in ethanol?
Ans. Molecular formula of ethanol is C2H5OH
Molar mass of ethanol is:
                                            = 2 × 12 + 5 × 1 + 16 + 1
                                           = 24 + 5 + 17 = 46 g
 
Class 11 Chemistry Basic Concepts of Chemistry Exam Questions
 
5. Classify the following as pure substances or mixture.
(a) ethyl alcohol
(b) oxygen
(c) carbon
(d) distilled water
(e) blood
(f) steel
Ans. Pure substance – ethyl alcohol, oxygen, carbon, distilled water mixture – blood, steel
 
 
LEVEL – 3

1. What is the difference between 160 cm and 160.0 cm?
Ans. 160 has three significant figures while 160.0 has four significant figures. Hence160.0 represents greater accuracy.
 
2. In the combustion of methane in air, what is the limiting reagent & why?
Ans. Methane is the limiting reagent because the other reactant is oxygen of air which is always present in excess. Thus the amounts of carbon dioxide & water formed will depend upon the amount of methane burnt.
 
3. Which aqueous solution has higher concentration, 1 molar or 1 molal solution of the same solute? Give reason.
Ans. 1 molar aqueous solution has higher concentration than 1 molal solution.
A molar solution contains one mole of solute in one litre of solution while one molal solution contains one mole of solute in 1000 g of solvent. If density of water is 1, then one mole of solute is present in 1000 mL of water in 1 molal solution while one mole of solute is present in less than 1000mL of water in 1 molar solution (1000 mL sol = amount of solute + amount of solvent). Thus 1 molar solution is more concentrated.
 
4. Calculate the molarity of water if its density is 1000 kg/m3.
Ans. Molarity of water means the number of moles of water in 1 litre of water
 
Class 11 Chemistry Basic Concepts of Chemistry Exam Questions-1
 
5. A compound made up of two elements A and B has = 70%, B = 30%. Their relative number of moles in the compound is 1.25 and 1.88. Calculate
a. Atomic masses of the elements A and B
b. Molecular formula of the compound, if its molecular mass is found to be 160.
Ans. 
Class 11 Chemistry Basic Concepts of Chemistry Exam Questions-2


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CBSE Class 11 Chemistry Study Material: Chapter 1 Some Basic Concepts of Chemistry

Comprehensive Study Resources for Chapter 1 Some Basic Concepts of Chemistry

Gather all necessary academic resources for Chapter 1 Some Basic Concepts of Chemistry on this dedicated page. Formulated in direct alignment with the active 2026 standards for Class 11 Chemistry, the pack offers thorough notes, handy Mind Maps, and exam-focused Sure Shot Questions for CBSE exams. Instructors advise leveraging these assets daily to optimize study speed.

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