Class 11 Chemistry Ch 1 — Unit Test Notes

← Class 11 Notes / Ch 1 & 2 Full Notes / Unit Test Notes

🎯 Unit Test Focus — 25 Marks Paper
This page covers all key points + important questions that come in the 25-mark unit test for Chapter 1. Read key points carefully, then practice every question below.

⭐ Key Points MCQ (1 Mark) 2 Mark Q&A 3 Mark Q&A Numericals

⭐ Chapter 1 — Key Points to Remember

📌

Unit 1 — Key Points

All Important Definitions & Facts

1. Chemistry — Study of matter, its physical and chemical properties and the changes it undergoes under different conditions. It is a central science.

2. Matter — Anything that occupies space and has mass.
Matter → Pure substances (definite composition) + Mixtures (no fixed composition)
Pure substances → Elements (cannot be broken down) + Compounds (can be broken down)
Mixtures → Homogeneous (uniform) + Heterogeneous (non-uniform)

3. Metals — Lustre, conduct heat and electricity, ductile, malleable. Examples: gold, silver, copper, iron. Mercury is liquid metal at room temperature.
Non-metals — No lustre (exception: diamond, iodine), poor conductors (exception: graphite), brittle. Examples: nitrogen, carbon, iodine.
Metalloids — Properties intermediate between metals and non-metals. Examples: arsenic, silicon, germanium.

4. Branches of Chemistry — Organic, Inorganic, Physical, Bio, Analytical (5 branches).
Physical chemistry provides basic framework for all other branches.

5. Physical properties — Measured without changing composition: colour, odour, melting point, boiling point, density.
Chemical properties — Substance undergoes change in composition: burning coal → CO₂; Mg burns → MgO.

6. SI Units (7 base units)

Quantity	SI Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	Kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

7. Key Conversions
1 kg = 1000 g = 10³ g | 1 nm = 10⁻⁹ m | 1 pm = 10⁻¹² m
1 L = 1 dm³ = 1000 mL = 1000 cm³
Density SI unit: kg/m³ | CGS: g/mL
Temperature: K = °C + 273.15 | °F = (9/5)°C + 32

8. Mass vs Weight
Mass = quantity of matter, does NOT change with position, SI unit = kg.
Weight = mass × gravitational pull, changes with position.
∴ Mass is more fundamental than weight.

9. Five Laws of Chemical Combination
(i) Law of Conservation of Mass (Lavoisier) — Mass can neither be created nor destroyed. Total mass of reactants = Total mass of products.
(ii) Law of Definite Proportions (Proust) — A compound always contains same proportion of elements by weight regardless of source.
(iii) Law of Multiple Proportions (Dalton, 1803) — When two elements form more than one compound, masses of B combining with fixed mass of A are in simple whole number ratio.
(iv) Gay Lussac's Law of Gaseous Volumes (1808) — Gases combine or are produced in simple ratio by volume at same T and P.
(v) Avogadro's Law (1811) — Equal volumes of all gases at same T and P contain equal number of molecules.

10. Dalton's Atomic Theory (1808) — 4 postulates
1. Matter = tiny indivisible particles called atoms.
2. All atoms of given element have identical properties including mass.
3. Compounds formed when atoms combine in fixed ratio.
4. Chemical reactions = reorganisation of atoms; atoms neither created nor destroyed.

11. Atomic Mass — Relative mass of atom w.r.t. 1/12th of C-12 atom.
1 amu (u / Da) = 1/12 × mass of C-12 = 1.66056 × 10⁻²⁴ g
Average atomic mass = weighted average of isotope masses × % abundance
Periodic table shows average atomic masses.

12. Molecular Mass — Sum of average atomic masses of all atoms in a molecule.
Example: CO₂ = 12 + 2(16) = 44 u | H₂O = 2(1) + 16 = 18 u | H₂SO₄ = 2+32+64 = 98 u
Formula mass — Used for ionic compounds (e.g., NaCl = 23 + 35.5 = 58.5 u)

13. Mole Concept
1 mole = amount of substance containing as many entities as atoms in exactly 12 g of C-12.
Avogadro's Constant (Nₐ) = 6.022 × 10²³ mol⁻¹
Molar mass = mass of 1 mole of substance in grams (numerically = molecular mass in u)
n (moles) = mass (g) ÷ molar mass (g/mol)
Number of particles = n × Nₐ

14. Moles and Gases (STP)
STP = Standard Temperature (0°C) and Pressure (1 atm).
Molar volume = 22.4 dm³ mol⁻¹ at STP.
n = Volume at STP (dm³) ÷ 22.4 dm³ mol⁻¹
(New STP at 1 bar: molar volume = 22.71 L mol⁻¹)

📐 Important Formulas — Quick Recall
n = mass/molar mass | Particles = n × 6.022×10²³
n (gas) = V(STP)/22.4 | K = °C + 273.15 | °F = (9/5)°C + 32
Density = mass/volume | 1 amu = 1.66056×10⁻²⁴ g
Avg atomic mass = Σ(mass × %abundance)/100

MCQ — 1 Mark Questions (Choose Correct Option)

🎯 Frequently asked in Unit Tests

✅

MCQ

1 Mark — Choose the Correct Answer

Q1. A sample of pure water always contains __ % oxygen by mass.

Options: a. 88.9 b. 18 c. 80 d. 16 Answer: a. 88.9% Reason: H₂O molar mass = 18. Oxygen = 16/18 × 100 = 88.9%

Q2. Which compound CANNOT demonstrate law of multiple proportions?

Options: a. NO, NO₂ b. CO, CO₂ c. H₂O, H₂O₂ d. Na₂S, NaF Answer: d. Na₂S, NaF Reason: Na₂S and NaF contain different elements (S and F), so they are NOT two compounds of the same two elements. Law of multiple proportions applies only to two compounds formed by SAME two elements.

Q3. Temperature reading same on Celsius and Fahrenheit scale is:

Options: a. −40° b. +40° c. −80° d. −20° Answer: a. −40° Proof: °F = (9/5)°C + 32. Put °F = °C = x: x = (9/5)x + 32 → −(4/5)x = 32 → x = −40

Q4. SI unit of electric current is:

Answer: b. Ampere (A)

Q5. In N₂ + 3H₂ → 2NH₃, volume ratio 1:3:2 illustrates the law of:

Answer: d. Gaseous volumes (Gay Lussac's Law)

Q6. Which of the following has maximum number of molecules?

Options: a. 7g N₂ b. 2g H₂ c. 8g O₂ d. 20g NO₂ Answer: b. 2g H₂ Calculation: Moles = a)7/28=0.25 b)2/2=1 c)8/32=0.25 d)20/46=0.43. H₂ has most moles → most molecules.

Q7. Mass of H₂O in 0.25 mol of it is:

Options: a. 4.5g b. 18g c. 0.25g d. 5.4g Answer: a. 4.5 g Calculation: mass = n × molar mass = 0.25 × 18 = 4.5 g

Q8. Number of molecules in 22.4 cm³ of N₂ at STP is:

Options: a. 6.022×10²⁰ b. 6.022×10²³ c. 22.4×10²⁰ d. 22.4×10²³ Answer: a. 6.022 × 10²⁰ Reason: 22.4 cm³ = 22.4 mL = 0.0224 L = 0.001 mol. Molecules = 0.001 × 6.022×10²³ = 6.022×10²⁰

Q9. Which has the largest number of atoms?

Options: a. 1g Au b. 1g Na c. 1g Li d. 1g Cl₂ Answer: c. 1g Li Reason: Atoms ∝ 1/molar mass. Li has lowest molar mass (6.94) among options → most atoms in 1g.

2 Mark — Short Answer Questions

🎯 Most common in 25 mark unit test

✍️

2 Mark Q&A

Short Answer Questions with Answers

Q1. Define chemistry. Why is it called a central science? 2M

Definition: Chemistry is the study of matter, its physical and chemical properties and the changes it undergoes under different conditions. Central Science: Chemistry is called a central science because its knowledge is required in the studies of physics, biological sciences, applied sciences, and earth and space sciences. It touches every aspect of life — air, food, clothing, fuel, medicines.

Q2. Distinguish between pure substances and mixtures. Give examples. 2M IMP

Pure substances: Definite chemical composition. Same properties regardless of origin.
Examples: Distilled water, pure metal, diamond. Mixtures: No definite chemical composition, no fixed properties. Can be separated by physical methods.
Examples: Sea water, air, paint, concrete.

Q3. What are metalloids? Give two examples. 2M

Some elements have properties intermediate between metals and non-metals. They are called metalloids or semi-metals. Examples: Arsenic (As), Silicon (Si), Germanium (Ge). They are used as semiconductors in electronics.

Q4. Distinguish between mass and weight. 2M IMP

Mass: Measure of quantity of matter. Does NOT change with position. SI unit = kilogram (kg). It is an inherent property of matter. Weight: Result of mass × gravitational attraction. Varies with distance from earth's centre. Hence mass is more fundamental than weight.

Q5. State law of conservation of mass. 2M IMP

Proposed by: Antoine Lavoisier (1743–1794) — father of modern chemistry. Statement: "Mass can neither be created nor destroyed." Total mass of reactants = Total mass of products. Example: When hydrogen burns with oxygen to yield water: mass of water formed = mass of hydrogen + mass of oxygen consumed.

Q6. State law of definite proportions. 2M

Proposed by: Joseph Proust (French chemist). Statement: "A given compound always contains exactly the same proportion of elements by weight", irrespective of its source. Example: Cupric carbonate — both natural and synthetic samples contain Cu: 51.35%, O: 38.91%, C: 9.74%.

Q7. Define mole. What is Avogadro's constant? 2M IMP

Mole: One mole is the amount of a substance that contains as many entities (atoms/molecules/ions) as there are atoms in exactly 12 g of Carbon-12. Avogadro's Constant (Nₐ): 1 mole = 6.0221367 × 10²³ particles. Named in honour of Amedeo Avogadro. In SI system, mole (symbol: mol) is the seventh base unit.

Q8. What is molar mass? Find molar mass of CO₂ and H₂SO₄. 2M

Molar mass: Mass of one mole of a substance in grams. Numerically equal to molecular/formula mass in u. CO₂: = 12 + 2(16) = 44 g mol⁻¹ H₂SO₄: = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g mol⁻¹

Q9. What is molar volume of a gas? State its value at STP. 2M IMP

Molar volume: The volume occupied by one mole of any gas at STP (Standard Temperature 0°C and Pressure 1 atm). Molar volume = 22.4 dm³ mol⁻¹ at STP Formula: n = Volume at STP (dm³) / 22.4 dm³ mol⁻¹

Q10. Define: 1 amu. What is its value in grams? 2M

1 amu (atomic mass unit / u / dalton): Mass exactly equal to one-twelfth of the mass of one Carbon-12 atom. Calculation: 1 amu = (1/12) × 1.992648 × 10⁻²³ g 1 amu = 1.66056 × 10⁻²⁴ g Carbon-12 atom is assigned a mass of exactly 12.00000 amu (agreed by IUPAC, 1961).

3 Mark — Answer in Detail

🎯 High weightage in 25 mark paper

📋

3 Mark Q&A

Important 3 Mark Questions with Full Answers

Q1. State and explain the law of multiple proportions with example. 3M IMP

Proposed by: John Dalton (1803). Statement: When two elements A and B form more than one compound, the masses of element B that combine with a given mass of A are always in the ratio of small whole numbers. Example 1: H + O form two compounds: H₂O: 2g H + 16g O H₂O₂: 2g H + 32g O Ratio of oxygen = 16:32 = 1:2 ✓ Example 2: N + O form two compounds: NO: 14g N + 16g O NO₂: 14g N + 32g O Ratio of oxygen = 16:32 = 1:2 ✓ Both are simple whole number ratios, confirming the law.

Q2. State and explain Gay Lussac's Law of Gaseous Volumes with example. 3M IMP

Proposed by: Gay Lussac (1808). Statement: When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure. Example 1: H₂ + O₂ → H₂O 100 mL H₂ + 50 mL O₂ → 100 mL H₂O vapour Ratio = 2:1:2 (simple whole numbers) ✓ Example 2: N₂ + H₂ → NH₃ 1L N₂ + 3L H₂ → 2L NH₃ Ratio = 1:3:2 ✓

Q3. State and explain Avogadro's Law. 3M IMP

Proposed by: Avogadro (1811). Statement: Equal volumes of all gases at the same temperature and pressure contain equal number of molecules. Explanation: Consider H₂ + O₂ → H₂O: 100 mL H₂ + 50 mL O₂ → 100 mL H₂O (Gay Lussac) If 1 volume = n molecules: 2n molecules H₂ + n molecules O₂ → 2n molecules H₂O Simplify: 2 H₂ + O₂ → 2 H₂O (molecules) ✓ Note: Avogadro made a clear distinction between atoms and molecules.

Q4. State the postulates of Dalton's Atomic Theory. 3M IMP

Dalton published "A New System of Chemical Philosophy" in 1808. Postulates: 1. Matter consists of tiny, indivisible particles called atoms. 2. All atoms of a given element have identical properties including mass. Atoms of different elements differ in mass. 3. Compounds are formed when atoms of different elements combine in a fixed ratio. 4. Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed. Dalton's theory explained all the laws of chemical combination.

Q5. Explain classification of matter on basis of chemical composition (with diagram). 3M

Matter is classified based on chemical composition as: 1. Pure substances: Definite composition, same properties always. → Elements: Cannot be broken down (gold, oxygen, silicon) → Metals (gold, iron) | Non-metals (nitrogen) | Metalloids (silicon) → Compounds: Can be broken down, elements in fixed ratio (H₂O, NaCl, CO₂) 2. Mixtures: No fixed composition, separated by physical methods. → Homogeneous: Uniform throughout (solution, air, vinegar) → Heterogeneous: Non-uniform (soil, suspension, salad)

Q6. What is average atomic mass? Explain with example of Carbon. 3M

Definition: Many elements exist as a mixture of isotopes. The atomic mass of such an element is the weighted average of atomic masses of all its isotopes (considering % abundance). Example — Carbon has 3 isotopes: ¹²C: atomic mass = 12.00000 u, abundance = 98.892% ¹³C: atomic mass = 13.00335 u, abundance = 1.108% ¹⁴C: atomic mass = 14.00317 u, abundance = 2×10⁻¹⁰% Calculation: Avg = (12 × 98.892/100) + (13.00335 × 1.108/100) + (14.00317 × 2×10⁻¹⁰/100) Average atomic mass of Carbon = 12.011 u Periodic table shows average atomic masses.

Q7. Explain the mole concept. 3M IMP

Need for Mole: Even a small amount of substance contains very large numbers of atoms/molecules. Just as "dozen" = 12 items, "mole" is used to express a large number of submicroscopic entities. Definition: One mole is the amount of a substance that contains as many entities as there are atoms in exactly 12 g of Carbon-12. Calculation: No. of atoms in 12g of C-12 = 12g / (1.992648 × 10⁻²³ g/atom) = 6.0221367 × 10²³ atoms = Avogadro's Constant (Nₐ) Molar mass: Mass of 1 mole of substance in grams = molecular mass in u (numerically). Formula: n = mass(g) / molar mass(g mol⁻¹); Particles = n × Nₐ

🔢 Numericals — Step-by-Step Solutions

🎯 Numericals carry 5–8 marks in unit test

🔢

Numericals

All Important Problems with Step-by-Step Solutions

Prob 1. Calculate average atomic mass of Neon. 2M IMP

Given: ²⁰Ne: 19.9924 u (90.92%) | ²¹Ne: 20.9940 u (0.26%) | ²²Ne: 21.9914 u (8.82%) Step 1: Formula: Avg = Σ(atomic mass × % abundance)/100 Step 2: = (19.9924 × 90.92 + 20.9940 × 0.26 + 21.9914 × 8.82) / 100 Step 3: = (1817.40 + 5.46 + 193.97) / 100 = 2016.83 / 100 Average atomic mass of Ne = 20.17 u

Prob 2. Find moles and molecules in 5.6 g of urea (NH₂CONH₂). 2M IMP

Step 1: Molecular mass of urea (NH₂CONH₂): = 2(14) + 4(1) + 12 + 16 = 28 + 4 + 12 + 16 = 60 u ∴ Molar mass = 60 g mol⁻¹ Step 2: n = mass / molar mass = 5.6 / 60 = 0.0933 mol Step 3: Molecules = n × Nₐ = 0.0933 × 6.022 × 10²³ = 5.618 × 10²² molecules

Prob 3. Calculate number of atoms in each: (i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He. 3M IMP

(i) 52 moles of Argon: 1 mol Ar = 6.022 × 10²³ atoms 52 mol Ar = 52 × 6.022 × 10²³ = 313.144 × 10²³ atoms of Ar (ii) 52 u of Helium: Atomic mass of He = 4.0 u = mass of 1 atom 4.0 u = 1 atom → 52 u = 52/4.0 = 13 atoms of He (iii) 52 g of Helium: Molar mass of He = 4.0 g mol⁻¹ n = 52/4.0 = 13 mol Atoms = 13 × 6.022 × 10²³ = 78.286 × 10²³ atoms of He

Prob 4. Calculate moles and molecules of NH₃ in 67.2 dm³ at STP. 2M

Step 1: n = Volume at STP / Molar volume = 67.2 / 22.4 Step 2: n = 3.0 mol Step 3: Molecules = 3.0 × 6.022 × 10²³ = 18.066 × 10²³ molecules of NH₃

Prob 5. Convert: (a) 40°C to °F and K (b) 30°C to °F 2M

(a) 40°C to °F: °F = (9/5)(40) + 32 = 72 + 32 = 104°F K = 40 + 273.15 = 313.15 K (b) 30°C to °F: °F = (9/5)(30) + 32 = 54 + 32 = 86°F

Prob 6. Calculate number of moles of H₂ in 0.448 L at STP. 1M

Step 1: n = Volume / 22.4 = 0.448 / 22.4 n = 0.02 mol of H₂

Prob 7. Calculate average atomic mass of Boron. (¹⁰B: 19.60%, mass=10.13 u; ¹¹B: 80.40%, mass=11.009 u) 2M IMP

Step 1: Avg = (10.13 × 19.60 + 11.009 × 80.40) / 100 Step 2: = (198.55 + 885.12) / 100 = 1083.67 / 100 Average atomic mass of Boron = 10.81 u

Prob 8. Find number of moles of MgO in (i) 80g (ii) 10g. (Mg=24, O=16) 2M

Step 1: Molar mass of MgO = 24 + 16 = 40 g mol⁻¹ (i) n = 80/40 = 2 mol (ii) n = 10/40 = 0.25 mol

Prob 9. Find mass of oxygen used: 24g C + O₂ → 88g CO₂. 1M

By Law of Conservation of Mass: Mass of reactants = Mass of products 24g (C) + mass of O₂ = 88g (CO₂) Mass of O₂ = 88 − 24 = 64 g

Prob 10. Calculate volume occupied by 5 moles of CO₂ at STP. 1M

Formula: V = n × 22.4 dm³ mol⁻¹ = 5 × 22.4 Volume = 112 dm³

📋 Suggested 25 Mark Unit Test Paper Pattern

Question Type	Marks	Topics
MCQ (5 questions × 1 mark)	5 marks	Laws, SI units, mole, molecules, temperature
Short Answer — 2 mark (3 questions)	6 marks	Definitions: mole, mass/weight, molar mass, amu
Short Answer — 2 mark (2 questions)	4 marks	Laws: conservation of mass, definite proportions
Long Answer — 3 mark (2 questions)	6 marks	Dalton's theory, mole concept, multiple proportions
Numericals (2 problems)	4 marks	Moles, molecules, avg atomic mass, temp conversion
Total	25 marks	—

Unit Test coming up? 🎯

Send your doubts to Shashikant sir on WhatsApp before the test

💬 Ask on WhatsApp

📖 Full Ch 1 & 2 Notes ← All Class 11 Notes

Class 11 Chemistry ⚗️