Chemical Laws and Mole Concept – NDA Chemistry PYQs

Solve NDA Chemistry previous-year questions on Chemical Laws and Mole Concept with detailed solutions and explanations.

Chapter-wise PYQs • Concept-based explanations • Exam insights

NDA 2026-II

Q. 1. What are the approximate percentages by mass of nitrogen and hydrogen respectively in urea ?

(a) 43% and 5%

(b) 47% and 5%

(d) 47% and 7%

Answer: (b) 47% and 7%

Explanation: Urea formula: CO(NH₂)₂. Molar mass = 12 + 16 + 2(14 + 2) = 12 + 16 + 32 = 60 g/mol. N: 2 × 14 = 28; %N = 28/60 × 100 = 46.67% ≈ 47%. H: 4 × 1 = 4; %H = 4/60 × 100 = 6.67% ≈ 7%. Wait – 4H gives ~6.67% ≈ 7%. So answer is (b) 47% and 7%. Rechecking: N=28/60=46.7%≈47%; H=4/60=6.7%≈7%. Answer is (b). Concept Tested: Percentage composition by mass – urea

NDA 2026-I

Q. 2. 1·7 g of silver nitrate is dissolved in 100 g of water, and 0·585 g of sodium chloride is dissolved in 100 g of water. Both are mixed and a chemical reaction occurs. 1·435 g of silver chloride and 0·85 g of sodium nitrate are formed. Which law of chemical combination is shown by the above reaction?

(a) Law of multiple proportions

(b) Law of conservation of mass

(d) Law of constant proportions

Answer: (b) Law of conservation of mass

Explanation: Reactants: 1.7 + 100 + 0.585 + 100 = 202.285 g. Products: 1.435 + 0.85 + remaining water + remaining dissolved salts. The key calculation: reactant mass of dissolved solutes = 1.7 + 0.585 = 2.285 g. Product mass = 1.435 + 0.85 = 2.285 g. Total mass is conserved – this demonstrates the law of conservation of mass. Concept Tested: Law of conservation of mass – mass balance verification

NDA 2025-I

Q. 3. Which one of the following statements is NOT correct?

(a) Half mole of nitrogen gas is measured 11·2 litre at STP

(b) 17 gram of ammonia gas contains 6·022 × 10²³ molecules at STP

(d) 4 gram of hydrogen gas contains 6·022 × 10²³ molecules

Answer: (d) 4 gram of hydrogen gas contains 6·022 × 10²³ molecules

Explanation: At STP, 1 mole of any gas occupies 22.4 L and contains 6.022 × 10²³ molecules. (a) Half mole N₂ = 11.2 L ✓. (b) 17 g NH₃ = 17/17 = 1 mole = 6.022×10²³ molecules ✓. (c) 22.4 L CO₂ at STP = 1 mole = 44 g ✓. (d) H₂ molar mass = 2 g/mol; 1 mole = 2 g, not 4 g. 4 g of H₂ = 2 moles = 2 × 6.022×10²³ molecules, not 6.022×10²³. So (d) is NOT correct. Concept Tested: Mole concept – STP volume, Avogadro’s number, molar mass calculations

NDA 2024-II

Q. 4. The mass of 0·5 mole of N₂ gas is

(a) 7 g

(b) 14 g

(d) 28 g

Answer: (b) 14 g

Explanation: Molar mass of N₂ = 2 × 14 = 28 g/mol. Mass of 0.5 mol = 0.5 × 28 = 14 g. Concept Tested: Mole concept – mass from moles of N₂

NDA 2024-I

Q. 5. Boron (which exists in two isotopic forms ¹⁰B and ¹¹B) has a relative atomic mass (combined) of 10·81. What will be the abundance of ¹⁰B and ¹¹B, respectively (consider a sample of 100 atoms)?

(a) 19% and 81%

(b) 81% and 19%

(d) 62% and 38%

Answer: (a) 19% and 81%

Explanation: Let x = fraction of ¹⁰B. Then (1−x) = fraction of ¹¹B. Average = 10x + 11(1−x) = 10.81. 10x + 11 − 11x = 10.81 → −x = −0.19 → x = 0.19. So ¹⁰B = 19%, ¹¹B = 81%. Concept Tested: Isotopic abundance calculation from average atomic mass

NDA 2023-I

Q. 6. Equal volume of all gases, when measured at the same temperature and pressure, contain an equal number of particles. Who proposed the above law ?

(a) Charles

(b) Boyle

(d) Lussac

Answer: (c) Avogadro

Explanation: This is Avogadro’s hypothesis (1811). It was proposed by Amedeo Avogadro. Charles proposed the law relating gas volume and temperature. Boyle proposed the law relating pressure and volume. Lussac proposed the law of combining volumes. Concept Tested: Avogadro’s hypothesis – attribution

NDA 2022-I

Q. 7. Which one of the following is the correct molecular formula of ammonium carbonate if the valency of ammonium ion is (+1) and carbonate anion is (−2)?

(a) (NH₄)₂CO₃

(b) NH₄(CO₃)₂

(d) NH₄CO₃

Answer: (a) (NH₄)₂CO₃

Explanation: NH₄⁺ has charge +1; CO₃²⁻ has charge −2. To balance charges: 2 × NH₄⁺ provides +2, balancing 1 × CO₃²⁻ (−2). Formula = (NH₄)₂CO₃. Option (b) uses incorrect parentheses placement. Options (c) and (d) are wrong in formula or charge balance. Concept Tested: Writing molecular formula from ionic charges – ammonium carbonate

NDA 2019-I

Q. 8. The equivalent weight of oxalic acid in C₂H₂O₄·2H₂O is

(a) 45

(b) 63

(d) 126

Answer: (b) 63

Explanation: Oxalic acid (C₂H₂O₄, anhydrous) has molar mass = 90. As a dibasic acid, n-factor = 2, so equivalent weight (anhydrous) = 90/2 = 45. But the question specifies C₂H₂O₄·2H₂O (the dihydrate). Molar mass of dihydrate = 90 + 2(18) = 90 + 36 = 126. Equivalent weight = 126/2 = 63. Concept Tested: Equivalent weight of dihydrate of oxalic acid

NDA 2018-II

Q. 9. What is the formula mass of anhydrous sodium carbonate? (Given that the atomic masses of sodium, carbon and oxygen are 23 u, 12 u and 16 u respectively)

(a) 286 u

(b) 106 u

(d) 53 u

Answer: (b) 106 u

Explanation: Anhydrous sodium carbonate is Na₂CO₃. Formula mass = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 u. Note: 286 u corresponds to Na₂CO₃·10H₂O (washing soda). Concept Tested: Formula mass of Na₂CO₃

Q. 10. Which one of the following statements about the law of conservation of mass is correct?

(a) A given compound always contains exactly same proportion of elements.

(b) When gases combine in a reaction, they do so in a simple ratio by volume, provided all gases are at room temperature.

(d) Equal volumes of all gases at same temperature and pressure contain equal number of molecules.

Answer: (c) Matter can neither be created nor destroyed.

Explanation: The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction – the total mass of reactants equals the total mass of products. (a) describes the law of definite proportions. (b) describes Gay-Lussac’s law of combining volumes. (d) is Avogadro’s hypothesis. Concept Tested: Law of conservation of mass – correct statement

★ JOVIK Exam Insight: NDA regularly tests identification of the correct law from a set of statements. Learn each law by its core statement: Conservation of mass = mass unchanged; Definite proportions = fixed ratio; Combining volumes = simple volume ratios.

NDA 2017-II

Q. 11. The proposition ‘equal volumes of different gases contain equal numbers of molecules at the same temperature and pressure’ is known as

(a) Avogadro’s hypothesis

(b) Gay-Lussac’s hypothesis

(d) Kirchhoff’s theory

Answer: (a) Avogadro’s hypothesis

Explanation: Avogadro’s hypothesis (1811) states that equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules. This is the direct statement described in the question. Concept Tested: Avogadro’s hypothesis – statement and attribution

Q. 12. How much CO₂ is produced on heating of 1 kg of carbon?

(a) 11/3 kg

(b) 3/11 kg

(d) 3/4 kg

Answer: (a) 11/3 kg

Explanation: C + O₂ → CO₂. Molar mass: C = 12, CO₂ = 44. 1 kg = 1000 g of C. Moles of C = 1000/12. Moles of CO₂ = same = 1000/12. Mass of CO₂ = (1000/12) × 44 = 44000/12 = 11000/3 g = 11/3 kg. Concept Tested: Mass-to-mass stoichiometry – combustion of carbon

★ JOVIK Exam Insight: Show the ratio: CO₂/C mass ratio = 44/12 = 11/3. Multiply mass of C by 11/3 to get CO₂ mass.

NDA 2016-II

Q. 13. Which one of the following has different number of molecules? (All are kept at normal temperature and pressure)

(a) 3 gram of Hydrogen

(b) 48 gram of Oxygen

(d) 2 gram of Carbon

Answer: (d) 2 gram of Carbon

Explanation: By Avogadro’s law, equal moles of any substance contain equal numbers of molecules. Calculate moles: (a) H₂: 3/2 = 1.5 mol. (b) O₂: 48/32 = 1.5 mol. (c) N₂: 42/28 = 1.5 mol. (d) C (carbon): 2/12 = 0.167 mol. Carbon is a solid element (monatomic, not a gas molecule). Its mole count is different from the others. So carbon has a different number of molecules (atoms). Concept Tested: Avogadro’s law – mole calculations and molecule counts

Q. 14. How many moles of hydrogen atom are present in one mole of Aluminium hydroxide?

(a) One mole

(b) Two moles

(d) Four moles

Answer: (c) Three moles

Explanation: The formula of aluminium hydroxide is Al(OH)₃. It contains 3 OH groups, each with one hydrogen atom. So one mole of Al(OH)₃ contains 3 moles of hydrogen atoms. Concept Tested: Mole concept – hydrogen atoms in Al(OH)₃

NDA 2016-I

Q. 15. What is the number of mole(s) of H₂(g) required to saturate one mole benzene ?

(a) 1

(b) 2

(d) 4

Answer: (c) 3

Explanation: Benzene (C₆H₆) has a ring structure with 3 double bonds. Hydrogenation (saturation) requires adding H₂ across each double bond: C₆H₆ + 3H₂ → C₆H₁₂ (cyclohexane). Three moles of H₂ are required per mole of benzene. Concept Tested: Stoichiometry of benzene hydrogenation

NDA 2015-II

Q. 16. Ammonia (NH₃) obtained from different sources always has same proportion of Nitrogen and Hydrogen. It proves the validity of law of:

(a) Reciprocal proportion

(b) Constant proportion

(d) None of the above

Answer: (b) Constant proportion

Explanation: The Law of Constant (Definite) Proportions states that a compound always contains the same elements in the same fixed mass ratios. Since NH₃ from different sources always has the same N:H ratio, this confirms the law of constant proportion. Concept Tested: Law of constant proportions – verification through examples

NDA 2014-II

Q. 17. Consider the following reaction: xAs₂S₃ + yO₂ → zAs₂O₃ + wSO₂ What is y (the coefficient for O₂) when this equation is balanced using whole number coefficients?

(a) 5

(b) 7

(d) 11

Answer: (c) 9

Explanation: Balance: 2As₂S₃ + 9O₂ → 2As₂O₃ + 6SO₂. Check: As: 4=4 ✓; S: 6=6 ✓; O: 18 left = 6+12=18 right ✓. So y = 9. Concept Tested: Balancing complex chemical equations – stoichiometric coefficients

Q. 18. How many grams of MgCO₃ contain 24·00 g of oxygen? (The molar mass of MgCO₃ is 84·30 g mol⁻¹)

(a) 42·15 g

(b) 84·30 g

(d) 154·00 g

Answer: (a) 42·15 g

Explanation: MgCO₃ has formula mass 84.30. It contains 3 oxygen atoms: mass of O = 3 × 16 = 48 g per mole. To get 24 g of O: fraction = 24/48 = 0.5 mol of MgCO₃. Mass = 0.5 × 84.30 = 42.15 g. Concept Tested: Mass calculations from molar composition

Q. 19. A compound X₂O₃ contains 31·58% oxygen by weight. The atomic mass of X is

(a) 34·66 g mol⁻¹

(b) 45·01 g mol⁻¹

(d) 104·00 g mol⁻¹

Answer: (c) 52·00 g mol⁻¹

Explanation: In X₂O₃: oxygen mass = 3 × 16 = 48. % O = 48/(2M+48) × 100 = 31.58. Solving: 4800 = 31.58(2M+48) → 2M+48 = 151.99 → 2M = 103.99 → M = 52. This is chromium (Cr). Concept Tested: Percentage composition – back-calculation of atomic mass

Q. 20. If the reaction of 1·0 mol NH₃(g) and 1·0 mol O₂(g) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l) is carried to completion, then

(a) all the O₂(g) is consumed

(b) 4·0 mol NO(g) is produced

(d) all the NH₃(g) is consumed

Answer: Editorial Review Required

Explanation: Limiting reagent analysis: 4NH₃ + 5O₂ → 4NO + 6H₂O. For 1 mol NH₃, O₂ needed = 1.25 mol; only 1 mol available, so O₂ is limiting. H₂O produced = (6/5) × 1.0 = 1.2 mol. NO produced = (4/5) × 1.0 = 0.8 mol. NH₃ consumed = (4/5) × 1.0 = 0.8 mol, so NH₃ is not fully consumed. None of the four options correctly reflect 1.2 mol H₂O or 0.8 mol NO. This question requires verification against the official NDA 2014-II key. Concept Tested: Limiting reagent problems – stoichiometry calculation

NDA 2014-I

Q. 21. The number of aluminium ions present in 54 g of aluminium (atomic weight 27) is

(a) 2

(b) 18

(d) 1·2 × 10²⁴

Answer: (d) 1·2 × 10²⁴

Explanation: Moles of Al = 54 ÷ 27 = 2 mol. Number of atoms = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ ≈ 1.2 × 10²⁴. Concept Tested: Mole concept – number of atoms from mass

Q. 22. Note the following balanced chemical equation: 2CO + O₂ ⇌ 2CO₂ Which one of the following statements is significant in relation to the above chemical equation?

(a) One can add to a vessel only 2 mol of CO for each mol of O₂ added

(b) No matter how much of these two reagents are added to a vessel, 1 mol of O₂ is consumed

(d) When 2 mol of CO and 1 mol of O₂ are placed in a vessel, they will react to give 1 mol of CO₂

Answer: (c) When they react, CO reacts with O₂ in a 2 : 1 mol ratio

Explanation: The equation 2CO + O₂ → 2CO₂ shows CO and O₂ react in a 2:1 molar ratio. (a) is incorrect – you can add any amount; the ratio applies only during reaction. (b) is incorrect – only as much O₂ as needed by the stoichiometry is consumed. (d) is incorrect – 2 mol CO + 1 mol O₂ gives 2 mol CO₂, not 1. Concept Tested: Interpretation of balanced equations – molar ratios

NDA 2012-II

Q. 23. Which one among the following equations is correctly balanced?

(a) NaOH + Al + H₂O → 2H₂ + NaAlO₂

(b) 2NaOH + 2Al + 2H₂O → 3H₂ + 2NaAlO₂

(d) 2NaOH + 2Al + H₂O → H₂ + 2NaAlO₂

Answer: (b) 2NaOH + 2Al + 2H₂O → 3H₂ + 2NaAlO₂

Explanation: The balanced reaction of aluminium with NaOH and water is: 2NaOH + 2Al + 2H₂O → 3H₂ + 2NaAlO₂. Check atoms: Left: Na=2, O=2+2=4, H=2+4=6, Al=2. Right: H=6 (3H₂), Na=2, Al=2, O=2+2=4 (NaAlO₂ has 2 O each × 2 = 4 plus no extra). Option (b) balances. Concept Tested: Balancing chemical equations – Al with NaOH

Q. 24. Three elements Na, S and O combine to form a compound Na₂SO₃. What is the valency of S in this compound?

(a) 2+

(b) 4+

(d) 8+

Answer: (b) 4+

Explanation: Na has valency +1; two Na atoms give +2 total. O has valency −2; three O atoms give −6 total. For Na₂SO₃ to be neutral: +2 + S + (−6) = 0 → S = +4. So the valency of S in Na₂SO₃ is 4+. Concept Tested: Determination of oxidation state/valency of sulphur in Na₂SO₃

Q. 25. Which one among the following is the equivalent weight of sulphuric acid? (Atomic weight: H = 1, S = 32, O = 16)

(a) 98

(b) 60

(d) 49

Answer: (d) 49

Explanation: Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol. H₂SO₄ is a dibasic acid (n-factor = 2). Equivalent weight = molar mass ÷ n-factor = 98 ÷ 2 = 49. Concept Tested: Equivalent weight of sulphuric acid

Q. 26. Given below is an approximate composition of a substance: CaO 60–70% SiO₂ 20–25% Al₂O₃ 5–10% Fe₂O₃ 2–3% The substance is

(a) Plaster of Paris

(b) Cement

(d) Quartz

Answer: (b) Cement

Explanation: Portland cement has the approximate composition: CaO (60–70%), SiO₂ (20–25%), Al₂O₃ (5–10%), Fe₂O₃ (2–3%). This matches exactly. Plaster of Paris is CaSO₄·½H₂O. Marble is CaCO₃. Quartz is SiO₂. Concept Tested: Composition of cement – CaO, SiO₂, Al₂O₃, Fe₂O₃

NDA 2010-II

Q. 27. Calcium carbonate is naturally available as limestone and can also be synthesized from quick lime. It is seen that the composition of the elements in both the natural and synthetic calcium carbonate are same. The validity of which one among the following laws is confirmed by this observation?

(a) Law of conservation of mass

(b) Law of definite proportion

(d) Avogadro’s law

Answer: (b) Law of definite proportion

Explanation: The Law of Definite Proportions (Law of Constant Composition) states that a pure chemical compound always contains the same elements in the same fixed proportions by mass, regardless of its source or method of preparation. Since CaCO₃ from both natural and synthetic sources has the same elemental composition, this confirms the law of definite proportion. Concept Tested: Law of definite proportions – same compound from different sources

NDA 2009-II

Q. 28. According to which one of the following laws it is indicated that when two or more gases react with one another, their volumes bear a simple ratio?

(a) Law of mass action

(b) Law of multiple proportions

(d) Law of combining volumes

Answer: (d) Law of combining volumes

Explanation: Gay-Lussac’s Law of Combining Volumes states that when gases react, their volumes bear a simple whole-number ratio to each other and to the products (at the same temperature and pressure). This is the law described in the question. Concept Tested: Gay-Lussac’s Law of Combining Volumes

Q. 29. Which of the following is the best example of the law of conservation of mass?

(a) When 12 gm of carbon is heated in vacuum, there is no change in mass

(b) Weight of platinum wire is the same before and after heating

(d) 12 gm of carbon combines with 32 gm of oxygen to give 44 gm of carbon dioxide

Answer: (d) 12 gm of carbon combines with 32 gm of oxygen to give 44 gm of carbon dioxide

Explanation: The law of conservation of mass states that in a chemical reaction, total mass of reactants = total mass of products. Option (d) shows 12 + 32 = 44 g, which directly demonstrates this law during a chemical change. Options (a), (b) and (c) show no chemical reaction – mass is unchanged for other reasons. Concept Tested: Law of conservation of mass – best example

NDA 2009-I

Q. 30. Consider the following equation for the formation of ammonia from nitrogen and hydrogen: N₂ + 3H₂ = 2NH₃ How many hydrogen molecules are required to react with 100 molecules of nitrogen?

(a) 100

(b) 200

(d) 400

Answer: (c) 300

Explanation: From the equation, 1 molecule of N₂ reacts with 3 molecules of H₂. For 100 N₂ molecules: H₂ required = 100 × 3 = 300 molecules. Concept Tested: Stoichiometry – mole/molecule ratios from balanced equations

Q. 31. Two elements gallium and oxygen combine to form a compound Ga₂O₃. Which among the following is the valency of gallium?

(a) 1

(b) 2

(d) 4

Answer: (c) 3

Explanation: In Ga₂O₃, oxygen has valency 2. Total valency of oxygen = 3 × 2 = 6. Total valency of gallium = 6 (must balance). Two gallium atoms share this: valency per Ga = 6 ÷ 2 = 3. Concept Tested: Valency determination from chemical formula