Case Study 1: Law of Conservation of Mass
In a laboratory experiment, Sneha heated 5 g of calcium carbonate in a test tube. On heating, it decomposed to give 2.8 g of calcium oxide and 2.2 g of carbon dioxide gas. Her teacher asked her to verify whether the law of conservation of mass is followed in this reaction. Sneha carefully measured the mass before and after the reaction and found that the total mass of products was equal to the mass of the reactant, which proved the law.
Questions:
- The reaction is an example of:
(a) combination (b) decomposition (c) displacement (d) neutralization
Answer: (b) - The law verified here is:
(a) Multiple proportions (b) Constant proportions (c) Conservation of mass (d) Avogadro’s law
Answer: (c) - Mass of reactant taken = ?
(a) 2.2 g (b) 2.8 g (c) 5.0 g (d) 10 g
Answer: (c) - Total mass of products formed = ?
(a) 2.2 g (b) 2.8 g (c) 5.0 g (d) 7.8 g
Answer: (c) - Which gas is released in this reaction?
(a) O₂ (b) CO₂ (c) H₂ (d) N₂
Answer: (b)
Case Study 2: Law of Constant Proportion
Hydrogen combines with oxygen to form water. In the laboratory, Ravi performed electrolysis of water and obtained hydrogen and oxygen in the ratio of 2:1 by volume. His teacher explained that no matter what sample of pure water is taken, the ratio of hydrogen to oxygen by mass is always 1:8. This verified the law of constant proportion.
Questions:
- Ratio of hydrogen to oxygen by mass in water = ?
(a) 2:1 (b) 1:2 (c) 1:8 (d) 8:1
Answer: (c) - Ratio of hydrogen to oxygen by volume in water = ?
(a) 1:1 (b) 2:1 (c) 1:2 (d) 8:1
Answer: (b) - This proves which law?
(a) Conservation of mass (b) Constant proportion (c) Multiple proportion (d) Avogadro’s law
Answer: (b) - Electrolysis of water gives:
(a) only oxygen (b) only hydrogen (c) hydrogen and oxygen (d) hydrogen and nitrogen
Answer: (c) - Water is a:
(a) mixture (b) compound (c) element (d) colloid
Answer: (b)
Case Study 3: Dalton’s Atomic Theory
Dalton proposed that matter is made up of small indivisible particles called atoms. For example, a sample of hydrogen contains identical atoms of hydrogen. Dalton also said that atoms of different elements differ in mass and combine in simple whole-number ratios to form compounds. Later, it was found that atoms are divisible into subatomic particles, but Dalton’s theory was important to explain chemical reactions.
Questions:
- Who gave the first atomic theory?
(a) Thomson (b) Rutherford (c) Dalton (d) Bohr
Answer: (c) - Atoms of the same element are:
(a) identical in mass and properties (b) different in properties (c) random (d) none
Answer: (a) - Atoms combine to form:
(a) mixtures (b) compounds (c) elements (d) none
Answer: (b) - Limitation of Dalton’s theory:
(a) atoms are indivisible (b) atoms combine in fixed ratios (c) atoms can neither be created nor destroyed (d) all of these
Answer: (a) - Which particles were later discovered inside atoms?
(a) protons (b) neutrons (c) electrons (d) all
Answer: (d)
Case Study 4: Writing Chemical Formulae
Teacher asked students to write formulae for compounds. Sneha was told to write the formula of aluminium oxide. She recalled that aluminium has valency +3 and oxygen has valency –2. By cross-multiplying, she wrote Al₂O₃. Similarly, for calcium hydroxide, she wrote Ca(OH)₂ using valencies of Ca (+2) and OH (–1).
Questions:
- Valency of aluminium = ?
(a) 1 (b) 2 (c) 3 (d) 4
Answer: (c) - Formula of aluminium oxide = ?
(a) AlO (b) AlO₂ (c) Al₂O₃ (d) Al₃O₂
Answer: (c) - Valency of hydroxide ion (OH⁻) = ?
(a) 1 (b) 2 (c) 3 (d) 0
Answer: (a) - Formula of calcium hydroxide = ?
(a) CaOH (b) Ca(OH)₂ (c) Ca₂OH (d) none
Answer: (b) - Writing chemical formula depends on:
(a) symbols only (b) valency (c) atomic mass (d) both (a) and (b)
Answer: (b)
Case Study 5: Balancing Chemical Equations
In a classroom, students were asked to balance the equation for the combustion of methane. Initially, the reaction was written as CH₄ + O₂ → CO₂ + H₂O, which was unbalanced. After balancing, the final equation became CH₄ + 2O₂ → CO₂ + 2H₂O. This exercise helped students understand that atoms are neither created nor destroyed in a chemical reaction.
Questions:
1. Combustion of methane is an example of:
(a) displacement (b) double displacement (c) combustion (d) decomposition
Answer: (c)
2. In CH₄ + 2O₂ → CO₂ + 2H₂O, number of oxygen atoms on each side = ?
(a) 1 (b) 2 (c) 4 (d) 3
Answer: (c)
3. Number of hydrogen atoms on each side = ?
(a) 2 (b) 4 (c) 6 (d) 8
Answer: (b)
4. The balancing proves which law?
(a) Law of constant proportion (b) Law of conservation of mass (c) Law of multiple proportion (d) Dalton’s law
Answer: (b)
5. Product gases formed are:
(a) CO and H₂ (b) CO₂ and H₂O (c) C and O₂ (d) CH₄ and O₂
Answer: (b)
Case Study 6: Concept of Mole
During a classroom demonstration, the teacher burned 2 g of hydrogen completely with oxygen to form water. She explained that 2 g of hydrogen contains 1 mole of molecules, i.e., 6.022 × 10²³ molecules. This example helped students understand Avogadro’s number and the mole concept.
Questions:
1. Mass of 1 mole of H₂ = ?
(a) 1 g (b) 2 g (c) 4 g (d) 18 g
Answer: (b)
2.Avogadro’s number = ?
(a) 6.022 × 10²³ (b) 3.011 × 10²³ (c) 9.022 × 10²³ (d) 6.022 × 10²²
Answer: (a)
3.Number of molecules in 2 g of hydrogen = ?
(a) 3.011 × 10²³ (b) 6.022 × 10²³ (c) 9.022 × 10²³ (d) 12.044 × 10²³
Answer: (b)
4.Hydrogen reacts with oxygen to form:
(a) CO₂ (b) H₂O (c) H₂O₂ (d) HOCl
Answer: (b)
5. 1 mole of any gas at STP occupies:
(a) 11.2 L (b) 22.4 L (c) 44.8 L (d) 1 L
Answer: (b)
Case Study 7: Molar Mass Calculations
In an experiment, students calculated the number of moles in 180 g of glucose (C₆H₁₂O₆). The molar mass was found as 12×6 + 1×12 + 16×6 = 180 g/mol. Therefore, 180 g of glucose = 1 mole = 6.022 × 10²³ molecules.
Questions:
- Molar mass of glucose = ?
(a) 60 g/mol (b) 120 g/mol (c) 180 g/mol (d) 12 g/mol
Answer: (c) - Number of moles in 180 g glucose = ?
(a) 1 (b) 2 (c) 6 (d) 12
Answer: (a) - Number of molecules in 1 mole glucose = ?
(a) 6.022 × 10²³ (b) 3.011 × 10²³ (c) 12.044 × 10²³ (d) none
Answer: (a) - Number of atoms in 1 molecule of glucose = ?
(a) 24 (b) 18 (c) 20 (d) none
Answer: (a) (6C + 12H + 6O = 24) - Number of atoms in 1 mole glucose = ?
(a) 24 × 6.022 × 10²³ (b) 12 × 6.022 × 10²³ (c) 6 × 6.022 × 10²³ (d) none
Answer: (a)
Case Study 8: Numerical Problem on Moles
A factory produced 22 g of CO₂. The molar mass of CO₂ is 44 g/mol. Number of moles = 22 ÷ 44 = 0.5 mole. Therefore, molecules present = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules.
Questions:
- Molar mass of CO₂ = ?
(a) 22 g (b) 32 g (c) 44 g (d) 28 g
Answer: (c) - Number of moles in 22 g CO₂ = ?
(a) 0.25 (b) 0.5 (c) 1 (d) 2
Answer: (b) - Number of molecules = ?
(a) 6.022 × 10²³ (b) 3.011 × 10²³ (c) 1.204 × 10²⁴ (d) 2.204 × 10²³
Answer: (b) - Number of oxygen atoms = ?
(a) 3.011 × 10²³ (b) 6.022 × 10²³ (c) 1.204 × 10²⁴ (d) none
Answer: (c) - Number of carbon atoms = ?
(a) 3.011 × 10²³ (b) 6.022 × 10²³ (c) 1.204 × 10²⁴ (d) none
Answer: (a)
Case Study 9: Molecular vs Atomic Mass
During a test, Aman was asked to calculate the molecular mass of H₂SO₄. He added the atomic masses: 2(1) + 32 + 4(16) = 98. Similarly, atomic mass of oxygen is 16 u. This helped him differentiate between atomic and molecular masses.
Questions:
- Atomic mass of oxygen = ?
(a) 8 u (b) 16 u (c) 32 u (d) 12 u
Answer: (b) - Molecular mass of H₂SO₄ = ?
(a) 98 u (b) 50 u (c) 32 u (d) 16 u
Answer: (a) - Unit of atomic mass = ?
(a) g/mol (b) kg (c) amu (u) (d) none
Answer: (c) - Sulphur in H₂SO₄ has atomic mass = ?
(a) 16 (b) 12 (c) 32 (d) 14
Answer: (c) - Molecular mass of O₂ = ?
(a) 16 u (b) 32 u (c) 18 u (d) none
Answer: (b)
Case Study 10: Gay-Lussac’s Law of Combining Volumes
During an experiment, Gay-Lussac observed that when hydrogen and oxygen gases react, 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of water vapour. This illustrates that gases combine in simple whole-number ratios by volume.
Questions:
- Ratio of hydrogen:oxygen by volume = ?
(a) 1:1 (b) 2:1 (c) 1:2 (d) 1:8
Answer: (b) - Ratio of products:reactants by volume = ?
(a) 2:3 (b) 1:2 (c) 3:2 (d) none
Answer: (a) - This law is given by:
(a) Dalton (b) Gay-Lussac (c) Avogadro (d) Rutherford
Answer: (b) - Water vapour formed is in ratio:
(a) 2 volumes (b) 1 volume (c) 3 volumes (d) none
Answer: (a) - This law supports:
(a) atomic theory (b) conservation of mass (c) mole concept (d) combining volumes of gases
Answer: (d)