Matter can neither be created nor destroyed in a chemical reaction. Total mass of reactants = total mass of products. Lavoisier (1789).
-- NCERT Class 11 Chemistry, Ch. 1, p. 4Laws Chemical Combination
8 MCQs9-step worked example
Source: NCERT Some Basic Concepts of ChemistryPYQ coverage: NEET 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
The laws of chemical combination are the experimental foundation on which modern chemistry rests. NEET questions on this topic test whether you can state, distinguish, and apply these laws — and the common confusion is mixing up which law applies to which observation.
Law of Conservation of Mass (Lavoisier, 1789). In a chemical reaction, total mass of reactants equals total mass of products. No mass is created or destroyed. NCERT Class 11 Chemistry Chapter 1, page 4 treats this as the first law. The trap: students forget this applies to the entire closed system — if gas escapes an open vessel, the apparent mass change does not violate the law.
Law of Definite Proportions (Proust, 1799). A given compound always contains the same elements in the same ratio by mass, regardless of source or method of preparation. Water from a river or a lab synthesis is always 1:8 by mass (H:O). Do not confuse this with the law of multiple proportions — definite proportions is about ONE compound, not two.
Law of Multiple Proportions (Dalton, 1803). When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. Example: CO and CO₂ — for a fixed mass of carbon, the oxygen masses are in the ratio 1:2. NCERT Class 11 Chemistry Chapter 1, page 4.
Gay-Lussac's Law of Gaseous Volumes (1808). Gases react in simple ratios by volume (at the same T and P). Two volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapour.
The high-frequency confusion in NEET: misidentifying which law a given experimental observation illustrates. When a stem describes two different compounds of the same two elements, the answer is multiple proportions — not definite proportions. Read the stem for "one compound, fixed ratio" versus "two compounds, ratio of ratios."
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which law states that in a chemical reaction, the total mass of reactants equals the total mass of products?
MCQ 2Easy RecallPractice
The Law of Definite Proportions was proposed by:
MCQ 3Easy RecallPractice
Gay-Lussac's Law of Gaseous Volumes states that gases react in:
MCQ 4Direct ApplicationPractice
Carbon forms two oxides: CO and CO₂. In CO, 12 g of carbon combines with 16 g of oxygen. In CO₂, 12 g of carbon combines with 32 g of oxygen. The ratio of oxygen masses combining with a fixed mass of carbon is 16:32 = 1:2. This observation illustrates:
MCQ 5Direct ApplicationPractice
A sample of pure water is decomposed and found to contain hydrogen and oxygen in the mass ratio 1:8. A second sample of water from a completely different source is also decomposed and gives the same 1:8 ratio. This observation is an illustration of:
MCQ 6Direct ApplicationPractice
Hydrogen and oxygen react as: 2H₂(g) + O₂(g) → 2H₂O(g). At constant temperature and pressure, 100 mL of hydrogen reacts completely with 50 mL of oxygen. Which law does this volume ratio (2:1) directly illustrate?
MCQ 7Concept TrapPractice
Nitrogen forms several oxides: N₂O, NO, NO₂, N₂O₃, and N₂O₅. For a fixed mass of nitrogen, the masses of oxygen in these compounds are in the ratio 1:2:4:3:5. Which law is illustrated?
MCQ 8Concept TrapPractice
In a closed container, 10.0 g of calcium carbonate is heated and decomposes completely: CaCO₃ → CaO + CO₂. If the calcium oxide produced has a mass of 5.6 g, what is the mass of CO₂ produced?
Worked Example
- 1
Given
- Oxide I: 32 g S combines with 32 g O - Oxide II: 32 g S combines with 48 g O - Mass of sulphur is fixed at 32 g in both compounds
- 2
Required
Verify that the masses of oxygen combining with a fixed mass of sulphur form a ratio of small whole numbers.
- 3
Concept
The Law of Multiple Proportions (Dalton) states: when two elements form two or more compounds, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. NCERT Class 11 Chemistry Chapter 1, page 4.
- 4
Formula
No algebraic formula needed. The procedure is: fix the mass of one element, take the ratio of the other element's masses across the two compounds, and simplify.
- 5
Substitution
Ratio of oxygen masses = mass of O in Oxide I : mass of O in Oxide II = 32 : 48
- 6
Calculation
32 : 48 = 32/16 : 48/16 = 2 : 3 Note on exact values: 32 g and 48 g are given values in this problem context. The ratio 2:3 follows by simple division and involves no significant-figure ambiguity.
- 7
Final answer
The oxygen masses are in the ratio **2:3**, which is a ratio of small whole numbers. The data obey the Law of Multiple Proportions.
- 8
Common trap
Confusing this with the Law of Definite Proportions. Definite proportions applies to the fixed composition of ONE compound (e.g., Oxide I is always 50% S and 50% O by mass). Multiple proportions compares the composition ACROSS two or more different compounds of the same two elements. When a stem gives you data for two compounds, the answer is multiple proportions.
- 9
Similar NEET-style question
Carbon forms two chlorides. In one, 12 g of carbon combines with 71 g of chlorine. In the other, 12 g of carbon combines with 142 g of chlorine. What is the ratio of chlorine masses for a fixed mass of carbon, and which law does this illustrate? *(Answer: 71:142 = 1:2, Law of Multiple Proportions.)* ---
Before solving, remember these
A given chemical compound always contains the same elements combined in fixed proportions by mass. Proust (1799).
-- NCERT Class 11 Chemistry, Ch. 1, p. 4When 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 integer ratios. Dalton (1803).
-- NCERT Class 11 Chemistry, Ch. 1, p. 4Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
-- NCERT Class 11 Chemistry, Ch. 1, p. 5Formulas
Molality
Molal concentration: moles of solute per kg of solvent. Temperature-independent.
| Symbol | Quantity | SI Unit |
|---|---|---|
| m | molality | mol/kg |
| n | moles solute | mol |
Valid when
- Mass of SOLVENT (not solution)
Molarity
Molar concentration: moles of solute per litre of solution.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | molarity | mol/L |
| n | moles solute | mol |
| V | solution volume | L |
Valid when
- Volume of SOLUTION not solvent
- Temperature dependent (volume changes with T)
Number of moles
Three equivalent ways to compute moles: from mass and molar mass, from number of particles, from gas volume at STP.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | moles | mol |
| m | mass | g |
| M | molar mass | g/mol |
| N | particle count | - |
| NA | Avogadro 6.022e23 | 1/mol |
| V | gas volume | L |
Valid when
- Use g/(g/mol) for mass route
- STP for gas volume route
- Pure substance
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student picks reagent with smaller absolute moles as limiting, ignoring stoichiometric coefficients.
When it triggers
Stoichiometry problem with two reactants.
How to avoid
Compute (moles available / coefficient) for each. Smaller value = limiting reagent. Compare ratios, not raw moles.
Category: Overthinking
Student computes from total mole amounts without checking stoichiometric ratios. Limiting reagent is the one that runs out first; rest is excess.
When it triggers
Stoichiometry problem with two reactants given.
How to avoid
Compare moles available to stoichiometric requirement. Compute moles_actual / coefficient for each; smaller value = limiting reagent.
Category: Unit Conversion
Student uses 22.4 L/mol at non-STP conditions or assumes ideal-gas-only molar volume.
When it triggers
Question gives gas at non-STP T or P.
How to avoid
22.4 L/mol applies ONLY at STP (273.15 K, 100 kPa). For other conditions, use PV=nRT.
Category: Overthinking
Student forgets to multiply by purity% before computing moles from impure sample mass.
When it triggers
Question states sample is X% pure (e.g. limestone, ore).
How to avoid
Effective mass = sample mass × (purity/100). Then compute moles using effective mass / molar mass.
Root cause: concept gap
Correction
Compute (moles_actual / coefficient) for each reactant. Smaller value = limiting.
Root cause: concept gap
Correction
22.4 L/mol applies ONLY at STP. Use PV=nRT for other conditions.
Root cause: concept gap
Correction
Effective mass = stated mass × (purity/100). Then convert to moles using molar mass.
Past Year Questions
4 questions from NEET 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
Dalton’s Atomic theory could not explain which of the following?
1Law of gaseous volume
2Law of conservation of mass
3Law of constant proportion
4Law of multiple proportion
NTA Answer: Option 1(final)
NEET 2024Revised key
1750 mg
2250 mg
3Zero mg
4200 mg
NTA Answer: Option 2(revised_final)
NEET 2024Revised key
1A BC 2 2
2ABC 3
3AB C 2 2
4ABC 4
NTA Answer: Option 2(revised_final)
NEET 2023
11.76 g
22.64 g
31.32 g
41.12 g
NTA Answer: Option 1(final)
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Mole concept and stoichiometry calculations: mole-mass conversion, limiting reagent, percentage purity, % composition.
Direct ApplicationMedium
Common distractors
swapped classes
Tempts surface-level recall.
Sources
NCERT refs: Class 11 Chemistry Chapter 1, p.4
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