Laws of Chemical Combination — NEET Chemistry MCQs with Solutions
Free NEET Chemistry Laws of Chemical Combination MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
▶ Practise Laws of Chemical Combination online (free)
Questions with solutions
Q1 — Laws of Chemical Combination · easy · theory
Which law states that a given compound always contains exactly the same proportion of elements by weight, irrespective of the source of the compound?
A. Avogadro’s law
B. Law of conservation of mass
C. Law of multiple proportions
D. Law of definite proportions ✓ Correct
Solution: The law of definite (or constant) proportions was given by Joseph Proust. Working with natural and synthetic cupric carbonate he found the composition was identical (51.35% Cu, 9.74% C, 38.91% O in both), showing a pure compound always contains the same elements combined in the same proportion by mass.
Q2 — Laws of Chemical Combination · easy · theory
Avogadro’s law states that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain equal numbers of:
A. electrons
B. atoms
C. protons
D. molecules ✓ Correct
Solution: Avogadro proposed in 1811 that equal volumes of all gases at the same temperature and pressure contain equal numbers of MOLECULES (not atoms). This distinction between atoms and molecules explained why 2 volumes of hydrogen react with 1 volume of oxygen to give 2 volumes of water vapour.
Q3 — Laws of Chemical Combination · easy · numerical
When 12 g of carbon is burnt completely in 32 g of oxygen, the whole of it is converted into carbon dioxide. By the law of conservation of mass, the mass of carbon dioxide formed is:
A. 20 g
B. 384 g
C. 44 g ✓ Correct
D. 22 g
Solution: C + O₂ → CO₂. Mass of products = mass of reactants = 12 g + 32 g = 44 g of CO₂. Subtracting the masses (32 − 12) wrongly gives 20 g; multiplying them (12 × 32) gives 384 g.
Q4 — Laws of Chemical Combination · medium · numerical
5.3 g of sodium carbonate reacts completely with 6 g of ethanoic acid to give 8.2 g of sodium ethanoate, 2.2 g of carbon dioxide and water. By the law of conservation of mass, the mass of water formed is:
A. 0.9 g ✓ Correct
B. 0.45 g
C. 1.8 g
D. 3.1 g
Solution: Total mass of reactants = 5.3 + 6 = 11.3 g. This must equal the total mass of products. Water = 11.3 − (8.2 + 2.2) = 11.3 − 10.4 = 0.9 g. Ignoring the CO₂ (11.3 − 8.2) wrongly gives 3.1 g.
Q5 — Laws of Chemical Combination · medium · numerical
Pure water from any source always contains hydrogen and oxygen combined in the fixed mass ratio 1 : 8. The mass of oxygen present in 27 g of pure water is:
A. 3 g
B. 21.6 g
C. 13.5 g
D. 24 g ✓ Correct
Solution: By the law of definite proportions H : O = 1 : 8, so water is made of 1 + 8 = 9 mass parts. Each part = 27 ÷ 9 = 3 g. Oxygen = 8 parts = 8 × 3 = 24 g (and hydrogen = 3 g). Taking oxygen as 80% of the mass wrongly gives 0.8 × 27 = 21.6 g.
Q6 — Laws of Chemical Combination · medium · numerical
Carbon combines with oxygen to form two compounds: in carbon monoxide 12 g of carbon combines with 16 g of oxygen, while in carbon dioxide 12 g of carbon combines with 32 g of oxygen. The ratio of the masses of oxygen combining with the fixed mass of carbon is:
A. 1 : 2 ✓ Correct
B. 2 : 3
C. 1 : 1
D. 2 : 1
Solution: This illustrates the law of multiple proportions. For a fixed 12 g of carbon, oxygen is 16 g in CO and 32 g in CO₂, giving 16 : 32 = 1 : 2, a simple whole-number ratio.
Q7 — Laws of Chemical Combination · medium · numerical
In the reaction N₂ + 3H₂ → 2NH₃, all volumes measured at the same temperature and pressure, the volume of hydrogen that reacts completely with 20 mL of nitrogen is:
A. 30 mL
B. 60 mL ✓ Correct
C. 20 mL
D. 40 mL
Solution: By Gay-Lussac’s law of gaseous volumes the combining volumes are in the ratio of the coefficients, N₂ : H₂ = 1 : 3. Hydrogen needed = 3 × 20 = 60 mL. Using a 1 : 1 ratio wrongly gives 20 mL.