Some Basic Concepts of Chemistry is the first unit of Class 11 Chemistry and builds the foundation for numerical problem solving. Important questions class 11 chemistry chapter 1 cover matter, SI units, significant figures, laws of chemical combination, mole concept, stoichiometry, limiting reagent, empirical formula, molarity, and molality.
A small mistake in mole conversion can change the whole answer in Chemistry. That is why Chapter 1 matters more than its position in the textbook suggests.
Class 11 Chemistry Chapter 1 teaches students how to measure substances, write quantities correctly, and use balanced equations for calculations. Once students understand moles, molar mass, concentration terms, and stoichiometric ratios, later units like Equilibrium, Thermodynamics, and Redox Reactions become much easier.
Key Takeaways
| Topic |
What to Know |
| Matter |
Exists as solid, liquid, or gas and can be classified as pure substances or mixtures |
| Pure Substances |
Classified into elements and compounds |
| SI Units |
Seven base units: metre, kilogram, second, ampere, kelvin, mole, candela |
| Significant Figures |
Rules for counting and rounding measured values |
| Laws of Chemical Combination |
Conservation of mass, definite proportions, multiple proportions, Gay Lussac’s law, Avogadro’s law |
| Dalton’s Atomic Theory |
Atoms combine in fixed ratios to form compounds |
| Atomic Mass Unit |
1 u = 1/12th mass of one carbon-12 atom |
| Mole |
1 mol = 6.022 × 10²³ entities |
| Molar Mass |
Mass of one mole of a substance in grams |
| Molarity |
Moles of solute per litre of solution |
| Molality |
Moles of solute per kg of solvent |
| Limiting Reagent |
Reactant consumed first in a reaction |
Class 11 Chemistry Units at a Glance
Important Topics of Some Basic Concepts of Chemistry Class 11
Important topics of some basic concepts of chemistry class 11 are mostly calculation-based. Start with units and significant figures before moving to mole concept and stoichiometry.
- Nature and classification of matter
- Physical and chemical properties
- SI units and derived units
- Scientific notation and significant figures
- Laws of chemical combination
- Dalton’s atomic theory
- Atomic mass and molecular mass
- Mole concept and Avogadro number
- Percentage composition
- Empirical and molecular formula
- Stoichiometry and balanced equations
- Limiting reagent
- Molarity, molality, mole fraction, and mass percent
Important Questions of Some Basic Concepts of Chemistry Class 11
Important questions of some basic concepts of chemistry class 11 should be practised by mark level. Definitions help in 1-mark questions, but most marks come from numericals.
These some basic concepts of chemistry class 11 questions and answers cover direct definitions, formula-based questions, MCQs, assertion-reason, PYQ-style numericals, and common mistakes.
1 Mark Questions from Some Basic Concepts of Chemistry
These questions test direct definitions, formulas, and recall. Write short answers with exact terms.
Q1. What is chemistry?
Answer: Chemistry is the branch of science that studies the composition, structure, properties, and reactions of matter.
Q2. What is the SI unit of amount of substance?
Answer: The SI unit of amount of substance is mole.
Q3. Define one atomic mass unit.
Answer: One atomic mass unit is exactly equal to one-twelfth of the mass of one carbon-12 atom.
1 u = 1.66056 × 10⁻²⁴ g
Q4. What is the SI unit of density?
Answer: The SI unit of density is kg m⁻³.
In chemistry, density is often expressed as g cm⁻³.
Q5. State the law of conservation of mass.
Answer: The law of conservation of mass states that matter is neither created nor destroyed in a chemical reaction.
The total mass of reactants equals the total mass of products.
Q6. What is the value of Avogadro’s number?
Answer: Avogadro’s number is 6.022 × 10²³ mol⁻¹.
Q7. What is formula mass?
Answer: Formula mass is the sum of atomic masses of all atoms present in one formula unit of an ionic compound.
Example: NaCl has formula mass because it does not exist as a discrete molecule.
Q8. State Avogadro’s law.
Answer: Avogadro’s law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
Q9. Identify the limiting reagent in A + B₂ → AB₂ when 300 atoms of A react with 200 molecules of B₂.
Answer: B₂ is the limiting reagent.
200 molecules of B₂ react with 200 atoms of A. So, 100 atoms of A remain unreacted.
Q10. How many significant figures are present in 0.200?
Answer: 0.200 has three significant figures.
Zeros after the decimal point and after a non-zero digit are significant.
2 Mark Questions from Some Basic Concepts of Chemistry
These questions usually test laws, comparisons, and short explanations.
Q11. State the law of definite proportions with one example.
Answer: The law of definite proportions states that a given compound always contains the same elements in the same proportion by mass.
Example: Water always contains hydrogen and oxygen in the mass ratio 1:8.
This ratio remains the same whether water is natural or prepared in a laboratory.
Q12. Differentiate between molarity and molality.
Answer:
| Basis |
Molarity |
Molality |
| Meaning |
Moles of solute per litre of solution |
Moles of solute per kg of solvent |
| Unit |
mol L⁻¹ |
mol kg⁻¹ |
| Depends on temperature |
Yes |
No |
| Uses |
Solution concentration |
Temperature-independent concentration |
Q13. What is the difference between precision and accuracy?
Answer: Precision means how close repeated measurements are to each other.
Accuracy means how close a measurement is to the true value.
A measurement can be precise but not accurate if repeated values are close together but far from the correct value.
Q14. State the law of multiple proportions with an example.
Answer: The law of multiple proportions states that when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a simple whole-number ratio.
Example: Hydrogen and oxygen form water and hydrogen peroxide.
In water, 2 g hydrogen combines with 16 g oxygen.
In hydrogen peroxide, 2 g hydrogen combines with 32 g oxygen.
The oxygen ratio is 16:32 = 1:2.
Q15. Convert 35°C to Fahrenheit and Kelvin.
Answer:
°F = (9/5 × °C) + 32
= (9/5 × 35) + 32
= 63 + 32
= 95°F
K = °C + 273.15
= 35 + 273.15
= 308.15 K
3 Mark Questions from Some Basic Concepts of Chemistry
These questions usually need formulas, steps, and correct units.
Q16. Calculate the molecular mass of H₂O, CO₂, and CH₄.
Answer:
H₂O = 2(1.008) + 16.00
= 18.016 u
CO₂ = 12.011 + 2(16.00)
= 44.011 u
CH₄ = 12.011 + 4(1.008)
= 16.043 u
Q17. Calculate the mass percent of Ca, P, and O in calcium phosphate, Ca₃(PO₄)₂.
Answer:
Molar mass of Ca₃(PO₄)₂:
= 3(40) + 2(31) + 8(16)
= 120 + 62 + 128
= 310 g mol⁻¹
Mass % of Ca = 120/310 × 100
= 38.71%
Mass % of P = 62/310 × 100
= 20.00%
Mass % of O = 128/310 × 100
= 41.29%
Q18. How many significant figures are present in 0.003500, 8.256, and 100?
Answer:
0.003500 has 4 significant figures.
Leading zeros are not significant. The digits 3, 5, 0, and 0 are significant.
8.256 has 4 significant figures.
All non-zero digits are significant.
100 usually has 1 significant figure unless written with a decimal or scientific notation.
Q19. Write four postulates of Dalton’s atomic theory. State one limitation.
Answer:
Postulates:
- Matter is made up of indivisible atoms.
- All atoms of an element have identical properties and mass.
- Atoms of different elements have different masses.
- Compounds form when atoms combine in fixed ratios.
- Chemical reactions involve rearrangement of atoms.
Limitation:
Dalton’s atomic theory could not explain the law of gaseous volumes. It also could not explain why atoms combine.
MCQs from Some Basic Concepts of Chemistry with Answers
MCQs from this chapter often test molarity, mole concept, empirical formula, significant figures, and law-based reasoning.
Always check units before choosing the answer.
Q20. What is the molarity of a solution containing 5.85 g NaCl in 500 mL solution?
(a) 4 mol L⁻¹
(b) 20 mol L⁻¹
(c) 0.2 mol L⁻¹
(d) 2 mol L⁻¹
Answer: (c) 0.2 mol L⁻¹
Moles of NaCl = 5.85/58.5
= 0.1 mol
Volume = 500 mL = 0.5 L
Molarity = 0.1/0.5
= 0.2 mol L⁻¹
Q21. A compound has empirical formula CH₂O and molecular mass 180 g. What is its molecular formula?
(a) C₉H₁₈O₉
(b) CH₂O
(c) C₆H₁₂O₆
(d) C₂H₄O₂
Answer: (c) C₆H₁₂O₆
Empirical formula mass = 12 + 2 + 16
= 30
n = 180/30
= 6
Molecular formula = C₆H₁₂O₆
Q22. Which sample contains the greatest number of atoms?
(a) 4 g He
(b) 0.40 g Ca
(c) 46 g Na
(d) 12 g He
Answer: (d) 12 g He
Moles of 12 g He = 12/4 = 3 mol
Moles of 46 g Na = 46/23 = 2 mol
Moles of 4 g He = 1 mol
Moles of 0.40 g Ca = 0.01 mol
The greatest number of atoms is in 12 g He.
Q23. Which of the following are unitless quantities?
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Mass percent
Answer: (c) and (d)
Mole fraction and mass percent are dimensionless quantities.
Q24. When 500 mL of 5 M solution is diluted to 1500 mL, what is the new molarity?
(a) 1.5 M
(b) 1.66 M
(c) 0.017 M
(d) 1.59 M
Answer: (b) 1.66 M
M₁V₁ = M₂V₂
5 × 500 = M₂ × 1500
M₂ = 2500/1500
= 1.66 M
Q25. Which reaction does not follow the law of conservation of mass as written?
(a) 2Mg + O₂ → 2MgO
(b) C₃H₈ + O₂ → CO₂ + H₂O
(c) P₄ + 5O₂ → P₄O₁₀
(d) CH₄ + 2O₂ → CO₂ + 2H₂O
Answer: (b) C₃H₈ + O₂ → CO₂ + H₂O
This equation is not balanced.
Correct equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Q26. One mole of oxygen gas at STP is equal to which of the following?
(a) 6.022 × 10²³ molecules of oxygen
(b) 6.022 × 10²³ atoms of oxygen
(c) 16 g of oxygen
(d) 32 g of oxygen
Answer: (a) and (d)
One mole of O₂ contains 6.022 × 10²³ molecules and has mass 32 g.
Assertion-Reason Questions from Some Basic Concepts of Chemistry
Assertion-reason questions from this chapter usually test significant figures, mole concept, empirical formula, and balanced equations.
Read the assertion and reason separately before choosing the answer.
Q27. Assertion (A): Significant figures for 0.200 are 3, while for 200 it is usually 1. Reason (R): Zeros at the end of a number are significant only when they are on the right side of the decimal point.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
(iii) A is true, but R is false.
(iv) Both A and R are false.
Answer: (iii) A is true, but R is false.
Zeros after a decimal point are significant.
Zeros at the end of a whole number may or may not be significant, depending on measurement accuracy.
Q28. Assertion (A): One atomic mass unit is defined as one-twelfth of the mass of one carbon-12 atom. Reason (R): Carbon-12 is used as the standard atomic mass reference.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) Both A and R are true, but R is not the correct explanation of A.
Answer: (i) Both A and R are true, and R is the correct explanation of A.
Atomic masses are compared with carbon-12 as the standard.
Q29. Assertion (A): Combustion of 16 g methane gives 18 g water. Reason (R): Water is one of the products in methane combustion.
(i) Both A and R are true, but R is not the correct explanation of A.
(ii) A is true, but R is false.
(iii) A is false, but R is true.
(iv) Both A and R are false.
Answer: (iii) A is false, but R is true.
Balanced equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
16 g methane produces 36 g water, not 18 g.
Q30. Assertion (A): Ethene’s empirical formula mass is half its molecular formula mass. Reason (R): Empirical formula represents the simplest whole-number ratio of atoms in a compound.
(i) Both A and R are true, and R is the correct explanation of A.
(ii) A is true, but R is false.
(iii) A is false, but R is true.
(iv) Both A and R are false.
Answer: (i) Both A and R are true, and R is the correct explanation of A.
Ethene has molecular formula C₂H₄ and empirical formula CH₂.
Empirical formula mass = 14
Molecular formula mass = 28
Numerical Questions on Mole Concept, Molarity and Stoichiometry
Some basic concepts of chemistry class 11 extra questions are mainly numerical. Write the balanced equation first for all stoichiometry questions.
Mole Concept and Molarity Questions
Q31. Calculate the number of molecules in 0.5 mol H₂O.
Answer:
1 mol contains 6.022 × 10²³ molecules.
0.5 mol contains:
0.5 × 6.022 × 10²³
= 3.011 × 10²³ molecules
Q32. Calculate the molarity of NaOH in a solution prepared by dissolving 4 g NaOH in enough water to form 250 mL solution.
Answer:
Molar mass of NaOH = 40 g mol⁻¹
Moles of NaOH = 4/40
= 0.1 mol
Volume = 250 mL = 0.250 L
Molarity = 0.1/0.250
= 0.4 M
Q33. Four litres of water is added to 2 L of 6 M HCl solution. What is the molarity of the resulting solution?
Answer:
Initial molarity, M₁ = 6 M
Initial volume, V₁ = 2 L
Final volume = 2 + 4 = 6 L
Using M₁V₁ = M₂V₂:
6 × 2 = M₂ × 6
M₂ = 12/6
= 2 M
Q34. What volume of 10 M HCl and 3 M HCl should be mixed to obtain 1 L of 6 M HCl?
Answer:
Let volume of 10 M HCl = V L.
Volume of 3 M HCl = (1 - V) L.
10V + 3(1 - V) = 6
10V + 3 - 3V = 6
7V = 3
V = 3/7 L
= 0.428 L
= 428 mL
Volume of 3 M HCl = 1000 - 428
= 572 mL
Stoichiometry and Limiting Reagent Questions
Q35. Calculate the amount of water produced by combustion of 16 g methane.
Answer:
Balanced equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
16 g CH₄ = 1 mol CH₄
1 mol CH₄ gives 2 mol H₂O.
Mass of water = 2 × 18
= 36 g
Q36. How much potassium chlorate should be heated to produce 2.24 L oxygen at NTP?
Answer:
Balanced equation:
2KClO₃ → 2KCl + 3O₂
At NTP:
1 mol O₂ = 22.4 L
2.24 L O₂ = 2.24/22.4
= 0.1 mol O₂
From the equation:
3 mol O₂ is produced from 2 mol KClO₃.
0.1 mol O₂ is produced from:
2/3 × 0.1 mol KClO₃
Molar mass of KClO₃ = 122.5 g mol⁻¹
Mass of KClO₃ = 2/3 × 0.1 × 122.5
= 8.17 g
Q37. 50.0 kg N₂ and 10.0 kg H₂ are mixed to produce NH₃. Identify the limiting reagent and calculate mass of NH₃ formed.
Answer:
Balanced equation:
N₂ + 3H₂ → 2NH₃
Moles of N₂ = 50,000/28
= 1786 mol
Moles of H₂ = 10,000/2
= 5000 mol
For 1786 mol N₂, H₂ required:
= 1786 × 3
= 5358 mol
Available H₂ = 5000 mol
So, H₂ is the limiting reagent.
From equation:
3 mol H₂ gives 2 mol NH₃.
5000 mol H₂ gives:
2/3 × 5000
= 3333 mol NH₃
Mass of NH₃ = 3333 × 17
= 56,661 g
= 56.7 kg approximately
Q38. Calculate the mass percent of carbon in carbon dioxide.
Answer:
Molecular mass of CO₂ = 12 + 32
= 44 g mol⁻¹
Mass percent of carbon = 12/44 × 100
= 27.27%
Important Questions on Significant Figures, Units and Measurements
Significant figures and unit conversion questions are scoring if students apply the rule correctly.
Do not round off in the middle of a numerical unless needed.
Q39. Express 0.00016 in scientific notation.
Answer:
0.00016 = 1.6 × 10⁻⁴
The decimal point moves four places to the right, so the exponent is -4.
Q40. Round off 34.216, 10.4107, 0.04597, and 2808 to three significant figures.
Answer:
34.216 → 34.2
10.4107 → 10.4
0.04597 → 0.0460
2808 → 2810 or 2.81 × 10³
Q41. Fill in the blanks: 1 km = ___ mm = ___ pm; 1 mg = ___ kg = ___ ng; 1 mL = ___ L = ___ dm³.
Answer:
1 km = 10⁶ mm = 10¹⁵ pm
1 mg = 10⁻⁶ kg = 10⁶ ng
1 mL = 10⁻³ L = 10⁻³ dm³
Q42. If the speed of light is 3.0 × 10⁸ m s⁻¹, calculate the distance covered by light in 2.00 ns.
Answer:
Distance = Speed × Time
Time = 2.00 ns
= 2.00 × 10⁻⁹ s
Distance = 3.0 × 10⁸ × 2.00 × 10⁻⁹
= 6.0 × 10⁻¹ m
= 0.600 m
Some Basic Concepts of Chemistry Extra Questions with Answers
Some basic concepts of chemistry extra questions often combine mole concept, percentage composition, molarity, molality, and limiting reagent.
These questions are useful for 3-mark and 5-mark practice.
Q43. Calculate the average atomic mass of hydrogen using ¹H = 99.985%, mass = 1 u and ²H = 0.015%, mass = 2 u.
Answer:
Average atomic mass:
= (99.985 × 1 + 0.015 × 2)/100
= (99.985 + 0.030)/100
= 100.015/100
= 1.00015 u
Q44. The density of 3 molal NaOH solution is 1.110 g mL⁻¹. Calculate molarity.
Answer:
3 molal NaOH means 3 mol NaOH in 1000 g water.
Mass of 3 mol NaOH:
= 3 × 40
= 120 g
Total mass of solution:
= 1000 + 120
= 1120 g
Volume of solution:
= 1120/1.110
= 1009 mL
= 1.009 L
Molarity = 3/1.009
= 2.97 M
Q45. In the reaction 2A + 4B → 3C + 4D, if 5 mol A reacts with 6 mol B, identify the limiting reagent and calculate moles of C formed.
Answer:
From equation:
2 mol A reacts with 4 mol B.
So, 5 mol A requires:
10 mol B
But only 6 mol B is available.
Therefore, B is the limiting reagent.
From equation:
4 mol B gives 3 mol C.
6 mol B gives:
3/4 × 6
= 4.5 mol C
Q46. Vitamin C combustion gives 0.2998 g CO₂ and 0.0819 g H₂O from 0.2000 g Vitamin C. Find the empirical formula.
Answer:
Mass of carbon:
= 12/44 × 0.2998
= 0.08177 g
Percentage of carbon:
= 0.08177/0.2000 × 100
= 40.88%
Mass of hydrogen:
= 2/18 × 0.0819
= 0.00910 g
Percentage of hydrogen:
= 0.00910/0.2000 × 100
= 4.55%
Percentage of oxygen:
= 100 - 40.88 - 4.55
= 54.57%
Moles:
C = 40.88/12 = 3.41
H = 4.55/1 = 4.55
O = 54.57/16 = 3.41
Ratio:
C:H:O = 3.41:4.55:3.41
= 1:1.33:1
Multiplying by 3:
= 3:4:3
Empirical formula = C₃H₄O₃
PYQ-Style Questions from Some Basic Concepts of Chemistry
These questions follow common school exam patterns for Class 11 Chemistry Chapter 1.
Use full steps because numericals can carry 3 to 5 marks.
Q47. The following data are obtained when dinitrogen and dioxygen react to form different compounds. Which law of chemical combination does this obey?
| Mass of N₂ |
Mass of O₂ |
| 14 g |
16 g |
| 14 g |
32 g |
| 28 g |
32 g |
| 28 g |
80 g |
Answer: This obeys the law of multiple proportions.
For a fixed mass of nitrogen, oxygen combines in simple whole-number ratios.
Taking 28 g nitrogen, the oxygen masses are 32 g, 64 g, 32 g, and 80 g.
The ratio is 2:4:2:5.
This is a simple whole-number ratio.
Q48. If 4 g NaOH is dissolved in 36 g H₂O, find the mole fraction of each component.
Answer:
Moles of NaOH = 4/40
= 0.1 mol
Moles of H₂O = 36/18
= 2 mol
Total moles = 2.1 mol
Mole fraction of NaOH:
= 0.1/2.1
= 0.048
Mole fraction of H₂O:
= 2/2.1
= 0.952
Q49. CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. 250 mL of 0.76 M HCl reacts with 1000 g CaCO₃. Name the limiting reagent and calculate moles of CaCl₂ formed.
Answer:
Moles of HCl:
= 0.76 × 0.250
= 0.19 mol
Moles of CaCO₃:
= 1000/100
= 10 mol
From equation:
2 mol HCl gives 1 mol CaCl₂.
0.19 mol HCl gives:
0.19/2
= 0.095 mol CaCl₂
To consume 10 mol CaCO₃, 20 mol HCl would be needed.
Only 0.19 mol HCl is available.
So, HCl is the limiting reagent.
Moles of CaCl₂ formed = 0.095 mol
Most Important Questions from Some Basic Concepts of Chemistry for 2026 Exams
These are high-probability questions from class 11 some basic concepts of chemistry important questions.
Revise them after completing all formulas.
- Calculate molecular mass of H₂O, CO₂, and CH₄.
- State and illustrate the law of multiple proportions.
- Calculate number of moles in 7.85 g Fe and 7.9 mg Ca.
- Determine empirical formula of a compound containing 4.07% H, 24.27% C, and 71.65% Cl.
- Calculate molarity of 5.85 g NaCl in 500 mL solution.
- Solve a limiting reagent numerical using CaCO₃ and HCl.
- Identify the limiting reagent and mass of NH₃ formed from N₂ and H₂.
- Calculate molality from molarity and density.
- Find the mass of one carbon-12 atom in grams.
- Solve assertion-reason questions on significant figures and methane combustion.
Common Mistakes in Some Basic Concepts of Chemistry Questions
Most mistakes in this chapter happen because students skip units or use the wrong base quantity.
Read every numerical twice before starting calculation.
Mole Conversion Mistakes
Students often divide mass by the wrong molar mass.
For diatomic gases:
O₂ = 32 g mol⁻¹
H₂ = 2 g mol⁻¹
N₂ = 28 g mol⁻¹
Do not use 16, 1, and 14 for these gases.
Molarity and Molality Mistakes
Molarity uses volume of solution in litres.
Molality uses mass of solvent in kilograms.
Molarity changes with temperature. Molality does not change with temperature.
Significant Figure Mistakes
In multiplication and division, keep the same number of significant figures as the measurement with the fewest significant figures.
In addition and subtraction, keep the fewest decimal places.
Limiting Reagent Mistakes
Do not compare only given moles.
Divide available moles by the stoichiometric coefficient. The smallest result gives the limiting reagent.
Unit Conversion Mistakes
In M₁V₁ = M₂V₂, both volumes must be in the same unit.
In molality, solvent mass must be in kilograms.
Important Formulas from Some Basic Concepts of Chemistry
| Concept |
Formula |
| Moles |
n = Given mass / Molar mass |
| Number of particles |
N = n × Nₐ |
| Molarity |
M = Moles of solute / Volume of solution in litres |
| Molality |
m = Moles of solute / Mass of solvent in kg |
| Mole fraction |
X = Moles of component / Total moles |
| Mass percent |
Mass of solute / Mass of solution × 100 |
| Percentage composition |
Mass of element / Molar mass of compound × 100 |
| Empirical formula factor |
n = Molecular mass / Empirical formula mass |
| Dilution formula |
M₁V₁ = M₂V₂ |
| Density |
d = Mass / Volume |