Nature of Matter and Laws of Chemical Combination (Nature and classification of matter, Physical & chemical properties, Dalton's Atomic Theory, Laws of Chemical Combination, Significant Figures, Scientific Notation, Dimensional Analysis): NEET MCQs with Answers

15 free questions on Nature of Matter and Laws of Chemical Combination (Nature and classification of matter, Physical & chemical properties, Dalton's Atomic Theory, Laws of Chemical Combination, Significant Figures, Scientific Notation, Dimensional Analysis) for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
How many significant figures are present in the three measurements 161 cm, 0.161 cm and 0.0161 cm respectively? [AIPMT 1998]
  1. 3, 4 and 5
  2. 3, 4 and 4
  3. 3, 3 and 4
  4. 3, 3 and 3
Show answer
Answer: (D) 3, 3 and 3
Explanation: Rules for counting significant figures: (i) All non-zero digits are significant. (ii) Zeros to the right of the decimal point that lie after a non-zero digit are significant. (iii) Zeros placed to the left of the first non-zero digit are NOT significant — they only fix the position of the decimal point. Therefore 161 → 3; 0.161 → 3 (the leading zero is not counted); 0.0161 → 3 (both leading zeros are not counted). All three measurements have 3 significant figures.

Question 2

Previous-year question
Equal masses of H₂, O₂ and methane are enclosed in a container of volume V at 27 °C under identical conditions. What is the ratio of the volumes occupied by H₂ : O₂ : CH₄? [AIPMT 2014]
  1. 8 : 16 : 1
  2. 16 : 8 : 1
  3. 16 : 1 : 2
  4. 8 : 1 : 2
Show answer
Answer: (C) 16 : 1 : 2
Explanation: By Avogadro’s hypothesis, at the same temperature and pressure the volume of a gas is directly proportional to its number of moles. For an equal mass w of each gas: n(H₂) = w/2, n(O₂) = w/32, n(CH₄) = w/16. Ratio of volumes = w/2 : w/32 : w/16. Multiplying throughout by 32/w gives 16 : 1 : 2. Concept: the lighter the gas, the more moles (and hence the greater the volume) a fixed mass occupies.

Question 3

Previous-year question
The correction factor 'a' in the van der Waals modification of the ideal-gas equation corresponds to: [NEET, 2018]
  1. Electric field present between the gas molecules
  2. Volume of the gas molecules
  3. Density of the gas molecules
  4. Forces of attraction between the gas molecules
Show answer
Answer: (D) Forces of attraction between the gas molecules
Explanation: The constant a accounts for the intermolecular forces of attraction; b accounts for molecular volume.

Question 4

Previous-year question
Maximum deviation from ideal-gas behaviour is expected from: [NEET-UG, 2013]
  1. NH₃(g)
  2. H₂(g)
  3. N₂(g)
  4. CH₄(g)
Show answer
Answer: (A) NH₃(g)
Explanation: NH₃ molecules experience hydrogen bonding, giving the strongest intermolecular attractions among the options, hence the maximum deviation from ideal behaviour.

Question 5

Previous-year question
A certain gas takes three times as long to effuse out as helium. Its molecular mass will be: [AIPMT Main, 2012]
  1. 64 u
  2. 9 u
  3. 27 u
  4. 36 u
Show answer
Answer: (D) 36 u
Explanation: r(gas)/r(He) = t(He)/t(gas) = 1/3 = √(4/M). So 4/M = 1/9 → M = 36 u.

Question 6

Previous-year question
50 mL each of gas A and gas B take 150 and 200 seconds respectively for effusing through a pinhole under similar conditions. If the molecular mass of gas B is 36, the molecular mass of gas A will be: [AIPMT Pre., 2012]
  1. 20.25
  2. 64
  3. 96
  4. 128
Show answer
Answer: (A) 20.25
Explanation: r_A/r_B = t_B/t_A = 200/150 = 4/3 = √(M_B/M_A). So 36/M_A = 16/9 → M_A = 20.25.

Question 7

Previous-year question
Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. The molecular mass of B will be: [AIPMT Pre., 2011]
  1. 50.00 u
  2. 12.25 u
  3. 6.50 u
  4. 25.00 u
Show answer
Answer: (B) 12.25 u
Explanation: r_A/r_B = t_B/t_A = 10/20 = 1/2 = √(M_B/M_A). So M_B/49 = 1/4 → M_B = 12.25 u.
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Question 8

Match each physical quantity with its SI unit. Quantities: (a) Speed (b) Amount of substance (c) Electric current (d) Density. Units: (i) mol (ii) kg m⁻³ (iii) A (iv) m s⁻¹
  1. a–iv, b–iii, c–i, d–ii
  2. a–iv, b–i, c–iii, d–ii
  3. a–ii, b–i, c–iii, d–iv
  4. a–i, b–iv, c–ii, d–iii
Show answer
Answer: (B) a–iv, b–i, c–iii, d–ii
Explanation: Speed is distance per time, m s⁻¹, and density is mass per volume, kg m⁻³. Amount of substance and electric current are base quantities with units mole and ampere. So a–iv, b–i, c–iii and d–ii.

Question 9

A length of 40 pm, expressed in nanometres, is:
  1. 4 × 10⁻³ nm
  2. 4 × 10⁻¹ nm
  3. 4 × 10⁻² nm
  4. 4 × 10⁴ nm
Show answer
Answer: (C) 4 × 10⁻² nm
Explanation: 1 pm = 10⁻¹² m and 1 nm = 10⁻⁹ m, so 1 pm = 10⁻³ nm. Hence 40 pm = 40 × 10⁻³ nm = 4 × 10⁻² nm.

Question 10

Match each observation in List-I with the law it illustrates in List-II. List-I: (A) In SO₂ and SO₃, the masses of oxygen combining with a fixed mass of sulphur are in the ratio 2 : 3. (B) Every pure sample of magnesium oxide, whatever its origin, contains 60% magnesium by mass. (C) 6 g of carbon burns completely in 16 g of oxygen to give 22 g of carbon dioxide. (D) At the same temperature and pressure, 2 volumes of hydrogen combine with 1 volume of oxygen to give 2 volumes of steam. List-II: (i) Law of constant composition (ii) Law of multiple proportions (iii) Gay-Lussac's law of gaseous volumes (iv) Law of conservation of mass
  1. A-i, B-ii, C-iv, D-iii
  2. A-ii, B-iv, C-i, D-iii
  3. A-ii, B-i, C-iv, D-iii
  4. A-iii, B-i, C-iv, D-ii
Show answer
Answer: (C) A-ii, B-i, C-iv, D-iii
Explanation: Two compounds of the same pair of elements with oxygen masses in a simple whole-number ratio show multiple proportions. A fixed percentage by mass in every pure sample shows constant composition. Since 6 g + 16 g = 22 g, no mass is lost or gained, which is conservation of mass. Gases reacting in simple whole-number volume ratios at the same temperature and pressure follow Gay-Lussac's law.

Question 11

At the same temperature and pressure, 200 mL of hydrogen reacts with 100 mL of oxygen to give 200 mL of water vapour. Which statement about this observation is correct?
  1. The observation shows that gases always react in a 1 : 1 ratio by volume
  2. Since 300 mL of reactants gives only 200 mL of product, the law of conservation of mass is violated
  3. The observation shows that equal volumes of different gases at the same temperature and pressure have equal masses
  4. The volumes are in the simple ratio 2 : 1 : 2, as expected from Gay-Lussac's law of gaseous volumes
Show answer
Answer: (D) The volumes are in the simple ratio 2 : 1 : 2, as expected from Gay-Lussac's law of gaseous volumes
Explanation: Gases react and form gaseous products in simple whole-number ratios by volume at the same temperature and pressure. Mass is conserved, but total volume need not be, because the number of molecules changes. Equal volumes hold equal numbers of molecules, not equal masses.

Question 12

Carbon dioxide made only from ¹²C and ¹⁶O has a C : O mass ratio of 3 : 8. If it is instead made only from ¹⁴C and ¹⁶O, the C : O mass ratio and the correct conclusion are
  1. 7 : 16; this sample of carbon dioxide has a different ratio of C atoms to O atoms
  2. 3 : 8; the mass ratio of elements in a compound never changes
  3. 7 : 8; each molecule now contains one oxygen atom fewer
  4. 7 : 16; the mass composition of a compound depends on the isotopes it contains
Show answer
Answer: (D) 7 : 16; the mass composition of a compound depends on the isotopes it contains
Explanation: Each molecule still has one carbon atom and two oxygen atoms, so the atom ratio stays 1 : 2. The mass ratio becomes 14 : 32, which is 7 : 16. The law of definite proportions therefore holds strictly only when the isotopic composition is fixed.

Question 13

Four samples of black or red copper oxide were analysed, with these masses of copper and oxygen. Sample P: 1.60 g Cu, 0.40 g O. Sample Q: 3.20 g Cu, 0.80 g O. Sample R: 4.00 g Cu, 0.50 g O. Sample S: 2.40 g Cu, 0.60 g O. Which sample cannot be the same compound as the other three?
  1. Sample Q
  2. Sample S
  3. Sample R
  4. Sample P
Show answer
Answer: (C) Sample R
Explanation: A given compound has the same mass ratio of its elements whatever its source, by the law of definite proportions. Samples P, Q and S all have Cu : O = 4 : 1, while sample R has Cu : O = 8 : 1, so R is a different oxide. For a fixed mass of copper, the oxygen in R is half that in the others, which is a simple ratio of 1 : 2 as the law of multiple proportions expects.

Question 14

At a fixed temperature and pressure, 200 mL of hydrogen combines with 100 mL of oxygen to give 200 mL of water vapour. Statement I: These data fit the law of gaseous volumes, which is in effect a law of definite proportions by volume. Statement II: The hydrogen, oxygen and water vapour taking part are also in the simple ratio 2 : 1 : 2 by mass.
  1. Statement I is true but Statement II is false
  2. Both Statement I and Statement II are true
  3. Both Statement I and Statement II are false
  4. Statement I is false but Statement II is true
Show answer
Answer: (A) Statement I is true but Statement II is false
Explanation: The volumes are in the simple ratio 2 : 1 : 2 at the same temperature and pressure, which is what Gay-Lussac's law describes. The ratio applies to volumes of gases only. By mass, 2 mol H₂ (4 g), 1 mol O₂ (32 g) and 2 mol H₂O (36 g) are in the ratio 1 : 8 : 9.

Question 15

At the same temperature and pressure, one volume of hydrogen combines with one volume of chlorine to give two volumes of hydrogen chloride. What does Avogadro's law allow us to conclude from this result?
  1. Each molecule of hydrogen and of chlorine must contain at least two atoms
  2. The mass of hydrogen used equals the mass of chlorine used
  3. Hydrogen and chlorine both exist as free single atoms
  4. The product contains more molecules than the reactants because atoms are created
Show answer
Answer: (A) Each molecule of hydrogen and of chlorine must contain at least two atoms
Explanation: Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, so n molecules of hydrogen and n of chlorine give 2n molecules of hydrogen chloride. Each product molecule needs at least one atom of each element, so n molecules of hydrogen must supply 2n hydrogen atoms. Hence each reactant molecule must contain at least two atoms and must split during the reaction. This is how Avogadro separated the idea of a molecule from that of an atom.

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