Some Basic Concepts of Chemistry — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Some Basic Concepts of Chemistry" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Some Basic Concepts of Chemistry" — 8 important questions with detailed answers for…
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NCERT Solutions for Class 11 Chemistry — Some Basic Concepts of Chemistry. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Some Basic Concepts of Chemistry — All 8 Questions Solved
Q1. Calculate the number of moles in 27 g of Al (Atomic mass of Al = 27 u).
Number of moles = mass (g) / molar mass (g/mol)
Given:
mass = 27 g
Molar mass of Al = 27 g/mol
Number of moles = 27 g / 27 g/mol
Number of moles = 1 mol
Answer: 1 mole of Al is present in 27 g.
Q2. Calculate the mass of 3.2 mol of oxygen atoms. (Atomic mass of O = 16 u)
mass (g) = number of moles × molar mass (g/mol)
Given:
Number of moles = 3.2 mol
Molar mass of O = 16 g/mol
mass = 3.2 mol × 16 g/mol
mass = 51.2 g
Answer: The mass of 3.2 mol of oxygen atoms is 51.2 g.
Q3. Calculate the number of molecules in 11.2 L of CO2 at STP (Molar mass of CO2 = 44 g/mol).
Step 1: Find number of moles using molar volume at STP.
At STP, 1 mole of gas = 22.4 L
Number of moles = Volume (L) / Molar volume (L/mol)
Number of moles = 11.2 L / 22.4 L/mol
Number of moles = 0.5 mol
Step 2: Find number of molecules.
Number of molecules = number of moles × Avogadro's number
Number of molecules = 0.5 mol × 6.022 × 10²³ /mol
Number of molecules = 3.011 × 10²³
Answer: Number of molecules in 11.2 L of CO2 at STP is 3.011 × 10²³.
Q4. What is the empirical formula of a compound containing 75% C and 25% H by mass? (Atomic mass: C = 12, H = 1)
Step 1: Convert percentage to grams (assume 100 g sample).
Mass of C = 75 g
Mass of H = 25 g
Step 2: Convert to moles.
Moles of C = 75 g / 12 g/mol = 6.25 mol
Moles of H = 25 g / 1 g/mol = 25 mol
Step 3: Find mole ratio (divide by smallest).
Mole ratio C : H = 6.25 : 25
Dividing by 6.25: C : H = 1 : 4
Answer: The empirical formula is CH₄.
Q5. Define atomic mass unit (amu). What is the relationship between amu and gram?
Atomic Mass Unit (amu) definition:
Atomic mass unit is defined as 1/12th of the mass of a carbon-12 atom.
Relationship with gram:
1 amu = 1.66054 × 10⁻²⁴ g
Or, in reverse:
1 g = 6.02214 × 10²³ amu
Note: This relationship shows why Avogadro's number (6.02214 × 10²³) is defined as it is — it converts between atomic mass units and grams.
Q6. A compound has the molecular formula C₂H₆O. What are the possible structures? (This relates to isomerism.)
Molecular formula: C₂H₆O
Molecular mass = (2 × 12) + (6 × 1) + (1 × 16) = 24 + 6 + 16 = 46 u
Possible structures (isomers):
1. Ethanol (CH₃CH₂OH)
Structure: H₃C-CH₂-OH
Functional group: Alcohol
2. Dimethyl ether (CH₃OCH₃)
Structure: H₃C-O-CH₃
Functional group: Ether
Note: Both have the same molecular formula but different structures, properties, and functional groups. This phenomenon is called structural isomerism.
Q7. Calculate the molarity of a solution prepared by dissolving 5.3 g of Na2CO3 in water to make 250 mL of solution. (Molar mass of Na2CO3 = 106 g/mol)
Molarity (M) = number of moles / volume in liters
Step 1: Calculate number of moles of Na2CO3.
Number of moles = mass / molar mass
Number of moles = 5.3 g / 106 g/mol
Number of moles = 0.05 mol
Step 2: Convert volume to liters.
Volume = 250 mL = 0.25 L
Step 3: Calculate molarity.
Molarity = 0.05 mol / 0.25 L
Molarity = 0.2 M
Answer: The molarity of the solution is 0.2 M.
Q8. State the law of conservation of mass and explain its significance in chemistry.
Law of Conservation of Mass:
Matter can neither be created nor destroyed during a chemical reaction. The total mass of reactants is equal to the total mass of products.
Mathematical form:
Total mass of reactants = Total mass of products
Example:
For the combustion of hydrogen:
2H₂ + O₂ → 2H₂O
Reactants: 2 × 2 = 4 g (H₂) + 32 g (O₂) = 36 g total
Products: 2 × 18 = 36 g (H₂O)
Mass is conserved.
Significance:
1. Forms the basis for stoichiometric calculations in chemistry
2. Helps in balancing chemical equations
3. Confirms that all matter involved in a reaction is accounted for
4. Important for understanding mass relationships in reactions
Some Basic Concepts of Chemistry — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following is a homogeneous mixture?
- A. Air
- B. Soil
- C. Milk
- D. Sand and water
Answer: A. Air — A homogeneous mixture has a uniform composition throughout. Air is a mixture of gases that is uniform in composition. Soil is heterogeneous, milk is a colloid (heterogeneous at a microscopic level), and sand and water clearly separate.
MCQ 2. The SI unit of temperature is:
- A. Celsius
- B. Fahrenheit
- C. Kelvin
- D. Joule
Answer: C. Kelvin — Kelvin (K) is the SI unit for temperature. Celsius and Fahrenheit are common non-SI units, while Joule is the SI unit for energy.
MCQ 3. How many significant figures are present in 0.00250?
- A. 2
- B. 3
- C. 4
- D. 5
Answer: B. 3 — Leading zeros (0.00) before the first non-zero digit are not significant. Trailing zeros after a decimal point and after a non-zero digit (like the '0' in 250) are significant. Thus, 2, 5, and 0 are significant, giving 3 significant figures.
MCQ 4. How many moles of water (H₂O) are present in 90 grams of water? (Atomic mass: H=1 u, O=16 u)
- A. 2 moles
- B. 5 moles
- C. 10 moles
- D. 18 moles
Answer: B. 5 moles — The molar mass of H₂O = (2 × 1) + 16 = 18 g/mol. Moles = Mass / Molar mass = 90 g / 18 g/mol = 5 moles.
MCQ 5. If 10 g of H₂ reacts with 64 g of O₂ to form H₂O, what is the limiting reagent? (Atomic mass: H=1 u, O=16 u)
- A. H₂
- B. O₂
- C. H₂O
- D. Neither
Answer: B. O₂ — The balanced equation is 2H₂ + O₂ → 2H₂O. Moles of H₂ = 10 g / 2 g/mol = 5 mol. Moles of O₂ = 64 g / 32 g/mol = 2 mol. From stoichiometry, 2 moles of H₂ react with 1 mole of O₂. For 5 moles of H₂, (5/2) = 2.5 moles of O₂ are required. Since only 2 moles of O₂ are available, O₂ is the limiting reagent.
MCQ 6. According to the Law of Conservation of Mass, the total mass of reactants in a chemical reaction is always equal to the:
- A. Mass of products
- B. Mass of the limiting reagent
- C. Mass of the excess reagent
- D. Mass of the catalyst
Answer: A. Mass of products — The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. Therefore, the total mass of the reactants must equal the total mass of the products.
MCQ 7. What is the percentage of carbon in methane (CH₄)? (Atomic mass: C=12 u, H=1 u)
- A. 25%
- B. 75%
- C. 12%
- D. 16%
Answer: B. 75% — Molar mass of CH₄ = 12 + (4 × 1) = 16 g/mol. The mass of carbon in one mole of CH₄ is 12 g. Percentage of Carbon = (12 g / 16 g) × 100 = 75%.
MCQ 8. An organic compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen. Its molecular mass is 180 g/mol. What is its molecular formula? (Atomic mass: C=12, H=1, O=16)
- A. CH₂O
- B. C₂H₄O₂
- C. C₆H₁₂O₆
- D. C₃H₆O₃
Answer: C. C₆H₁₂O₆ — Moles ratio: C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. Dividing by the smallest (3.33), the ratio is C:H:O = 1:2:1. Thus, the empirical formula is CH₂O. Empirical formula mass = 12 + (2×1) + 16 = 30 g/mol. The n factor = Molecular mass / Empirical formula mass = 180 / 30 = 6. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
MCQ 9. Which state of matter has definite volume but indefinite shape?
- A. Solid
- B. Liquid
- C. Gas
- D. Plasma
Answer: B. Liquid — Liquids have a definite volume but take the shape of their container, hence an indefinite shape. Solids have both definite shape and volume, while gases have indefinite shape and volume.
MCQ 10. A student measures the mass of a substance three times and gets 15.5 g, 15.6 g, and 15.5 g. The actual mass is 18.0 g. The measurements are:
- A. Precise and accurate
- B. Precise but not accurate
- C. Accurate but not precise
- D. Neither precise nor accurate
Answer: B. Precise but not accurate — The measurements (15.5 g, 15.6 g, 15.5 g) are very close to each other, indicating high precision. However, they are significantly different from the actual mass of 18.0 g, meaning they are not accurate.
MCQ 11. The Law of Definite Proportions states that:
- A. Elements always combine in simple whole-number ratios by mass
- B. Compounds always contain the same elements in the same proportion by mass
- C. The total mass of reactants equals the total mass of products
- D. When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios.
Answer: B. Compounds always contain the same elements in the same proportion by mass — The Law of Definite Proportions (also known as Proust's Law) states that a given chemical compound always contains its component elements in fixed ratio by mass, regardless of its source or method of preparation.
MCQ 12. Calculate the area of a rectangle with length 2.34 cm and width 1.2 cm, expressing the answer with correct significant figures.
- A. 2.808 cm²
- B. 2.81 cm²
- C. 2.8 cm²
- D. 3 cm²
Answer: C. 2.8 cm² — Area = Length × Width = 2.34 cm × 1.2 cm = 2.808 cm². When multiplying or dividing, the result must have the same number of significant figures as the measurement with the fewest significant figures. 2.34 has 3 significant figures, and 1.2 has 2 significant figures. Thus, the answer must be rounded to 2 significant figures, giving 2.8 cm².
MCQ 13. Avogadro's Law states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of:
- A. Atoms
- B. Molecules
- C. Ions
- D. Moles of mass
Answer: B. Molecules — Avogadro's Law specifically states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.
MCQ 14. How many moles of oxygen are required for the complete combustion of 2 moles of methane (CH₄)? Reaction: CH₄ + 2O₂ → CO₂ + 2H₂O
- A. 1 mole
- B. 2 moles
- C. 4 moles
- D. 8 moles
Answer: C. 4 moles — From the balanced chemical equation, 1 mole of CH₄ reacts with 2 moles of O₂. Therefore, 2 moles of CH₄ will require (2 × 2) = 4 moles of O₂ for complete combustion.
MCQ 15. Which of the following is a compound?
- A. Oxygen
- B. Gold
- C. Water
- D. Nitrogen
Answer: C. Water — A compound is formed when two or more different elements chemically combine in a fixed ratio. Water (H₂O) is a compound formed from hydrogen and oxygen. Oxygen, Gold, and Nitrogen are elements.
MCQ 16. What is the molality of a solution containing 10 g of NaOH in 500 mL of water? (Density of water = 1 g/mL, Atomic mass: Na=23, O=16, H=1)
- A. 0.5 m
- B. 1.0 m
- C. 0.25 m
- D. 2.0 m
Answer: A. 0.5 m — Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = 10 g / 40 g/mol = 0.25 mol. Mass of water (solvent) = Volume × Density = 500 mL × 1 g/mL = 500 g = 0.5 kg. Molality = Moles of solute / Mass of solvent (in kg) = 0.25 mol / 0.5 kg = 0.5 m.
MCQ 17. Which postulate of Dalton's Atomic Theory was disproved by the discovery of isotopes?
- A. Atoms are indivisible
- B. All atoms of a given element are identical in all respects
- C. Atoms of different elements have different properties
- D. Atoms combine in simple whole-number ratios.
Answer: B. All atoms of a given element are identical in all respects — Isotopes are atoms of the same element with the same atomic number but different mass numbers (due to different numbers of neutrons). This contradicts Dalton's postulate that all atoms of a given element are identical in all respects, especially mass.
MCQ 18. What is the molarity of a solution prepared by dissolving 49 g of H₂SO₄ in enough water to make 250 mL of solution? (Atomic mass: H=1, S=32, O=16)
- A. 0.5 M
- B. 1.0 M
- C. 2.0 M
- D. 4.0 M
Answer: C. 2.0 M — Molar mass of H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 98 g/mol. Moles of H₂SO₄ = 49 g / 98 g/mol = 0.5 mol. Volume of solution = 250 mL = 0.25 L. Molarity = Moles of solute / Volume of solution (L) = 0.5 mol / 0.25 L = 2.0 M.
MCQ 19. How many atoms are present in 0.1 mole of H₂SO₄? (Avogadro's number = 6.022 × 10²³)
- A. 6.022 × 10²² atoms
- B. 1.2044 × 10²³ atoms
- C. 4.2154 × 10²³ atoms
- D. 6.022 × 10²³ atoms
Answer: C. 4.2154 × 10²³ atoms — 0.1 mole of H₂SO₄ contains 0.1 × (6.022 × 10²³) molecules. Each H₂SO₄ molecule has 2 hydrogen atoms + 1 sulfur atom + 4 oxygen atoms = 7 atoms. Total number of atoms = (0.1 × 6.022 × 10²³) × 7 = 4.2154 × 10²³ atoms.
MCQ 20. One mole of any substance contains:
- A. Its molar mass in grams
- B. 6.022 × 10²³ grams of that substance
- C. 6.022 × 10²³ particles
- D. 22.4 L at STP
Answer: C. 6.022 × 10²³ particles — One mole is defined as the amount of substance that contains Avogadro's number (6.022 × 10²³) of elementary entities (atoms, molecules, ions, etc.). Option A describes molar mass, not the definition of a mole.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.