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Some Basic Concepts of Chemistry — JEE Main Chemistry MCQs with Solutions
Free JEE Main Chemistry Some Basic Concepts of Chemistry MCQs with step-by-step solutions covering Development & Importance of Chemistry, Nature of Matter, Properties of Matter & Measurement, Uncertainty in Measurement, Laws of Chemical Combination, Dalton's Atomic Theory. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — Development & Importance of Chemistry · hard · theory
Acharya Kanad (born 600 BCE) proposed the concept of 'Paramanu'. Relative to John Dalton (1766–1844), Kanad conceptualised this atomic theory approximately:
A. 250 years before Dalton
B. 2500 years before Dalton ✓ Correct
C. 500 years after Dalton
D. at the same time as Dalton
Solution: The chapter states that Kanad conceptualised the theory of Paramanu around 2500 years before John Dalton. Having been born in 600 BCE and predating Dalton (1766–1844), the gap is roughly 2500 years, not a few hundred years.
Q2 — Development & Importance of Chemistry · hard · theory
Chemistry developed mainly in the forms of Alchemy and Iatrochemistry during the period:
A. 1300–1600 CE ✓ Correct
B. 600–200 BCE
C. 1600–1800 CE
D. 1000–1300 CE
Solution: The chapter states that chemistry developed mainly in the form of Alchemy and Iatrochemistry during 1300–1600 CE, with modern chemistry taking shape in 18th-century Europe. The other date ranges do not match the text.
Q3 — Development & Importance of Chemistry · hard · numerical
Saltpetre (potassium nitrate, KNO₃) was used in ancient Indian fireworks. Using K = 39, N = 14, O = 16, the mass of oxygen present in 202 g of KNO₃ is:
A. 96 g ✓ Correct
B. 32 g
C. 48 g
D. 101 g
Solution: Molar mass of KNO₃ = 39 + 14 + (3 × 16) = 101 g mol⁻¹. Moles = 202 ÷ 101 = 2 mol. Each mole contains 3 oxygen atoms = 48 g of oxygen, so oxygen mass = 2 × 48 = 96 g. Using only 1 mol gives 48 g; counting a single O atom gives 32 g; 101 g is merely the molar mass.
Q4 — Nature of Matter · hard · theory
Which of the following statements is INCORRECT?
A. The constituents of a compound can be separated by simple physical methods ✓ Correct
B. Air is a mixture, not a compound
C. A compound has a fixed composition, whereas a mixture has a variable composition
D. The properties of a compound are usually different from those of its constituent elements
Solution: The incorrect statement is that the constituents of a compound can be separated by simple physical methods — in fact they can be separated only by chemical methods. Compounds do have a fixed composition, their properties differ from those of their elements (e.g. water vs hydrogen and oxygen gases), and air is indeed a mixture.
Q5 — Nature of Matter · hard · theory
Consider the statements: (i) a mixture has its components in any ratio, (ii) a compound has its elements in a fixed definite ratio, (iii) a mixture can be separated by physical methods, (iv) a compound can be separated into its elements by physical methods. How many of these statements are correct?
A. Four
B. One
C. Two
D. Three ✓ Correct
Solution: Statements (i), (ii) and (iii) are correct — mixtures have a variable composition and can be separated physically, while a compound has a fixed ratio of elements. Statement (iv) is wrong: a compound can be separated into its elements only by chemical methods, not physical ones. Hence three statements are correct.
Q6 — Properties of Matter & Measurement · hard · numerical
The density of aluminium is 2.7 g cm⁻³. Expressed in SI units (kg m⁻³), this density is:
A. 2700 kg m⁻³ ✓ Correct
B. 27 kg m⁻³
C. 270000 kg m⁻³
D. 0.0027 kg m⁻³
Solution: Since 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³, we have 1 g cm⁻³ = 10⁻³ ÷ 10⁻⁶ = 10³ kg m⁻³. Therefore 2.7 g cm⁻³ = 2.7 × 1000 = 2700 kg m⁻³.
Q7 — Properties of Matter & Measurement · hard · numerical
At what Celsius temperature does the Fahrenheit reading become exactly twice the Celsius reading? (Use °F = (9/5)°C + 32.)
A. 160 °C ✓ Correct
B. 80 °C
C. 320 °C
D. 40 °C
Solution: Set °F = 2 × °C: 2C = (9/5)C + 32 ⇒ 2C − 1.8C = 32 ⇒ 0.2C = 32 ⇒ C = 160 °C. Check: °F = 1.8 × 160 + 32 = 320 = 2 × 160. (320 is the Fahrenheit reading, not the Celsius answer.)
Q8 — Uncertainty in Measurement · hard · numerical
Evaluate (6.4 × 10⁻³) ÷ (8.0 × 10⁵) and express the answer in scientific notation.
A. 8.0 × 10⁻⁷
B. 8.0 × 10⁻⁹ ✓ Correct
C. 1.25 × 10⁻⁸
D. 8.0 × 10⁻⁸
Solution: Divide the digit terms and subtract the exponents: (6.4 ÷ 8.0) × 10^(−3 − 5) = 0.8 × 10⁻⁸. To make the digit term lie between 1 and 10, write 0.8 as 8.0 × 10⁻¹, which lowers the exponent by one: 0.8 × 10⁻⁸ = 8.0 × 10⁻⁹. Dividing 8.0 ÷ 6.4 the wrong way gives the 1.25 × 10⁻⁸ trap.
Q9 — Uncertainty in Measurement · hard · numerical
Evaluate (4.5 × 10⁻³) + (5.0 × 10⁻⁴) and express the answer in scientific notation.
A. 5.0 × 10⁻⁴
B. 5.0 × 10⁻⁷
C. 9.5 × 10⁻³
D. 5.0 × 10⁻³ ✓ Correct
Solution: For addition, first make the exponents equal: 5.0 × 10⁻⁴ = 0.50 × 10⁻³. Then add the digit terms: (4.5 + 0.50) × 10⁻³ = 5.0 × 10⁻³. Adding the digit terms without aligning exponents (4.5 + 5.0) wrongly gives 9.5 × 10⁻³.
Q10 — Uncertainty in Measurement · hard · numerical
A car moves at 72 km h⁻¹. Using 1 km = 1000 m and 1 h = 3600 s, its speed in m s⁻¹ is:
A. 200 m s⁻¹
B. 2.0 m s⁻¹
C. 20 m s⁻¹ ✓ Correct
D. 259.2 m s⁻¹
Solution: By the unit-factor method: 72 (km/h) × (1000 m ÷ 1 km) × (1 h ÷ 3600 s) = 72 × 1000 ÷ 3600 = 72000 ÷ 3600 = 20 m s⁻¹. Multiplying by 3.6 instead of dividing gives the wrong 259.2 m s⁻¹.
Q11 — Laws of Chemical Combination · hard · numerical
Two oxides of nitrogen are analysed. Oxide A contains 63.6% nitrogen and oxide B contains 46.7% nitrogen by mass. For a fixed mass of nitrogen, the ratio of the mass of oxygen in A to that in B is:
A. 1 : 2 ✓ Correct
B. 1 : 1
C. 3 : 4
D. 2 : 1
Solution: In A, per 63.6 g N there is 36.4 g O, so O per gram of N = 36.4 ÷ 63.6 = 0.572. In B, per 46.7 g N there is 53.3 g O, so O per gram of N = 53.3 ÷ 46.7 = 1.143. Ratio = 0.572 : 1.143 = 1 : 2 (A is N₂O, B is NO), confirming the law of multiple proportions.
Q12 — Laws of Chemical Combination · hard · numerical
10 mL of methane is burnt in excess oxygen at 120 °C: CH₄ + 2O₂ → CO₂ + 2H₂O. With all gases measured at the same temperature and pressure, the volume of oxygen consumed and the total volume of gaseous products are respectively:
A. 20 mL and 30 mL ✓ Correct
B. 20 mL and 10 mL
C. 40 mL and 30 mL
D. 10 mL and 20 mL
Solution: By Gay-Lussac’s law volumes follow the coefficients. For 10 mL CH₄: oxygen = 2 × 10 = 20 mL. Products = CO₂ (10 mL) + H₂O vapour (2 × 10 = 20 mL) = 30 mL. Forgetting the water vapour (counting only 10 mL CO₂) gives the wrong 10 mL for products.
Q13 — Dalton's Atomic Theory · hard · theory
The species ⁴⁰Ar, ⁴⁰K and ⁴⁰Ca have practically the same mass but belong to different elements. This observation is inconsistent with Dalton’s assumption that:
A. compounds are formed in a fixed ratio
B. atoms of different elements differ in mass ✓ Correct
C. atoms of the same element are identical
D. atoms are indivisible
Solution: Atoms of different elements having the same mass are called isobars (here all have mass number 40). Their existence contradicts Dalton’s postulate that atoms of different elements must differ in mass. The other postulates are not challenged by isobars.
Q14 — Dalton's Atomic Theory · hard · theory
The discovery of electrons, protons and neutrons directly overturned which postulate of Dalton’s atomic theory?
A. Atoms are indivisible ✓ Correct
B. Atoms are rearranged in chemical reactions
C. Atoms combine in a fixed ratio to form compounds
D. Atoms of different elements differ in mass
Solution: Dalton’s first postulate held that atoms are the ultimate, indivisible units of matter. The discovery of subatomic particles (electrons, protons, neutrons) showed that atoms are in fact divisible, overturning this postulate while leaving the atom’s role as the unit of chemical combination intact.
Q15 — Dalton's Atomic Theory · hard · theory
Which statement correctly distinguishes Dalton’s view of gaseous elements from Avogadro’s and explains why Dalton could not account for combining gas volumes?
A. Dalton believed atoms of the same kind could not combine, so molecules such as H₂ and O₂ did not exist, whereas Avogadro allowed gases to be polyatomic ✓ Correct
B. Avogadro denied the existence of atoms, unlike Dalton
C. Dalton proposed diatomic molecules while Avogadro insisted every gas is monatomic
D. Both held that gaseous elements are always monatomic
Solution: Dalton (and others of his time) believed atoms of the same kind could not combine, so diatomic molecules like H₂ and O₂ did not exist. Avogadro instead proposed that gaseous elements could be polyatomic; treating H₂ and O₂ as diatomic makes the 2:1:2 volume relationship of hydrogen, oxygen and water vapour understandable — something Dalton’s picture could not do.
Q16 — Dalton's Atomic Theory · hard · numerical
Two oxides of nitrogen contain 63.6% and 46.7% nitrogen by mass respectively. For a fixed mass of nitrogen, the ratio of the mass of oxygen in the first oxide to that in the second (illustrating the law of multiple proportions) is:
A. 3:4
B. 2:1
C. 1:2 ✓ Correct
D. 2:3
Solution: Oxygen combined per unit nitrogen = (%O)/(%N). First oxide: 36.4 ÷ 63.6 = 0.572; second oxide: 53.3 ÷ 46.7 = 1.14. The ratio 0.572:1.14 = 1:2 (the oxides are N₂O and NO). The nitrogen mass must be fixed first — comparing the oxygen percentages directly (36.4:53.3 ≈ 2:3) is the common mistake.
Q17 — Atomic & Molecular Masses · hard · numerical
An element has two isotopes of atomic masses 63 u and 65 u, and its average atomic mass is 63.5 u. The percentage abundance of the lighter (63 u) isotope is:
A. 25%
B. 50%
C. 75% ✓ Correct
D. 65%
Solution: Let the fraction of the 63 u isotope be x. Then 63x + 65(1 − x) = 63.5 ⇒ 65 − 2x = 63.5 ⇒ 2x = 1.5 ⇒ x = 0.75, i.e. 75% (the element is copper, Cu = 63.5 u). The value 25% is the abundance of the heavier isotope, and 50% would give a plain average of 64 u, not 63.5 u.
Q18 — Atomic & Molecular Masses · hard · numerical
The molecular mass of ammonium sulphate, (NH₄)₂SO₄, using N = 14, H = 1, S = 32 and O = 16 (all in u), is:
A. 96 u
B. 124 u
C. 132 u ✓ Correct
D. 114 u
Solution: The subscript 2 outside the bracket doubles both N and H: (NH₄)₂SO₄ has 2 N, 8 H, 1 S and 4 O. Mass = 2(14) + 8(1) + 32 + 4(16) = 28 + 8 + 32 + 64 = 132 u. Using only one NH₄ group (14 + 4 + 32 + 64) gives 114 u; forgetting the 8 hydrogens (28 + 32 + 64) gives 124 u; counting only the sulphate part (32 + 64) gives 96 u.
Q19 — Atomic & Molecular Masses · hard · numerical
Neon consists of three isotopes: ²⁰Ne (90.9%), ²¹Ne (0.3%) and ²²Ne (8.8%), with isotopic masses 20, 21 and 22 u respectively. The average atomic mass of neon is closest to:
A. 21.00 u
B. 20.00 u
C. 20.18 u ✓ Correct
D. 20.90 u
Solution: Average atomic mass = (0.909 × 20) + (0.003 × 21) + (0.088 × 22) = 18.18 + 0.063 + 1.936 = 20.18 u. A plain average of 20, 21 and 22 would give 21.00 u, and simply taking the most abundant isotope would give 20.00 u — both ignore the weighting by abundance.
Q20 — Atomic & Molecular Masses · hard · numerical
A binary oxide has the formula X₂O₃ and a molecular mass of 102 u (O = 16 u). The atomic mass of element X is:
A. 54 u
B. 51 u
C. 27 u ✓ Correct
D. 43 u
Solution: X₂O₃ mass = 2X + (3 × 16) = 2X + 48 = 102 ⇒ 2X = 54 ⇒ X = 27 u (the oxide is Al₂O₃). Forgetting to divide by 2 gives 54 u; using only one oxygen (2X + 16 = 102) gives 43 u; ignoring the oxygen entirely (2X = 102) gives 51 u.
Q21 — Mole Concept & Molar Masses · hard · numerical
The total number of atoms present in 3.6 g of glucose (C₆H₁₂O₆, molar mass = 180 g mol⁻¹) is: (Nₐ = 6.022 × 10²³)
A. 2.891 × 10²³ ✓ Correct
B. 2.891 × 10²²
C. 1.2044 × 10²²
D. 2.891 × 10²⁴
Solution: n = 3.6 ÷ 180 = 0.02 mol, giving 0.02 × 6.022 × 10²³ = 1.2044 × 10²² molecules. Each glucose molecule contains 6 + 12 + 6 = 24 atoms, so total atoms = 24 × 1.2044 × 10²² = 2.891 × 10²³. Forgetting to multiply by 24 atoms per molecule leaves 1.2044 × 10²² (the molecule count); the others are power-of-ten slips.
Q22 — Mole Concept & Molar Masses · hard · numerical
What mass of magnesium (Mg = 24) contains the same number of atoms as are present in 12 g of carbon (C = 12)?
A. 48 g
B. 6 g
C. 24 g ✓ Correct
D. 12 g
Solution: 12 g of carbon = 12 ÷ 12 = 1 mol = 6.022 × 10²³ atoms. To hold the same number of atoms we need 1 mol of Mg = 1 × 24 = 24 g. Assuming equal masses give equal atom counts (12 g) ignores the different atomic masses; 48 g and 6 g come from doubling or halving in error.
Q23 — Mole Concept & Molar Masses · hard · numerical
At STP (molar volume = 22.7 L mol⁻¹), the number of molecules present in 5.675 L of nitrogen gas (N₂) is: (Nₐ = 6.022 × 10²³)
A. 1.5055 × 10²²
B. 1.5055 × 10²³ ✓ Correct
C. 6.022 × 10²³
D. 3.011 × 10²³
Solution: n = volume ÷ molar volume = 5.675 ÷ 22.7 = 0.25 mol. Molecules = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³. Treating the sample as 1 mol gives 6.022 × 10²³; using 0.5 mol gives 3.011 × 10²³; a power-of-ten slip gives 1.5055 × 10²².
Q24 — Mole Concept & Molar Masses · hard · numerical
3.4 g of a gas occupies 2.27 L at STP (molar volume = 22.7 L mol⁻¹). The molar mass of the gas is:
A. 68 g mol⁻¹
B. 34 g mol⁻¹ ✓ Correct
C. 17 g mol⁻¹
D. 3.4 g mol⁻¹
Solution: n = 2.27 ÷ 22.7 = 0.1 mol, so molar mass = mass ÷ moles = 3.4 ÷ 0.1 = 34 g mol⁻¹ (the gas is H₂S). Dividing by 0.2 instead gives 17 g mol⁻¹; taking the raw mass as the molar mass gives 3.4 g mol⁻¹; doubling gives 68 g mol⁻¹.
Q25 — Mole Concept & Molar Masses · hard · numerical
The number of oxygen atoms present in 9.8 g of sulphuric acid (H₂SO₄, molar mass = 98 g mol⁻¹) is: (Nₐ = 6.022 × 10²³)
A. 6.022 × 10²²
B. 2.4088 × 10²²
C. 6.022 × 10²³
D. 2.4088 × 10²³ ✓ Correct
Solution: n = 9.8 ÷ 98 = 0.1 mol, giving 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules. Each H₂SO₄ molecule has 4 oxygen atoms, so oxygen atoms = 4 × 6.022 × 10²² = 2.4088 × 10²³. Forgetting the factor of 4 leaves 6.022 × 10²² (the molecule count); the others are scaling slips.
Q26 — Mole Concept & Molar Masses · hard · numerical
A container holds 1.7 g of ammonia (NH₃, molar mass = 17 g mol⁻¹). The total number of hydrogen atoms in the sample is: (Nₐ = 6.022 × 10²³)
A. 1.8066 × 10²²
B. 6.022 × 10²³
C. 6.022 × 10²²
D. 1.8066 × 10²³ ✓ Correct
Solution: n = 1.7 ÷ 17 = 0.1 mol, giving 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules. Each NH₃ molecule has 3 hydrogen atoms, so H atoms = 3 × 6.022 × 10²² = 1.8066 × 10²³. Forgetting to multiply by 3 leaves 6.022 × 10²² (the molecule count).
Q27 — Percentage Composition & Formulae · hard · numerical
A compound contains 26.7% carbon, 2.2% hydrogen and 71.1% oxygen, and its molar mass is 90 g mol⁻¹. Its molecular formula is (C = 12, H = 1, O = 16):
A. C₃H₃O₆
B. CHO₂
C. C₂H₂O₄ ✓ Correct
D. C₄H₄O₈
Solution: In 100 g: moles C = 26.7 ÷ 12 = 2.22, H = 2.2 ÷ 1 = 2.2, O = 71.1 ÷ 16 = 4.44. Divide by smallest (2.2): C = 1, H = 1, O = 2 → empirical formula CHO₂ (mass = 12 + 1 + 32 = 45). n = 90 ÷ 45 = 2, so molecular formula = C₂H₂O₄ (oxalic acid). Choosing CHO₂ forgets to multiply by n.
Q28 — Percentage Composition & Formulae · hard · numerical
A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass and has a molar mass of 56 g mol⁻¹. Its molecular formula is (C = 12, H = 1):
A. C₄H₈ ✓ Correct
B. CH₂
C. C₃H₆
D. C₂H₄
Solution: Moles in 100 g: C = 85.7 ÷ 12 = 7.14, H = 14.3 ÷ 1 = 14.3. Divide by 7.14: C = 1, H = 2 → empirical formula CH₂ (mass = 14). n = 56 ÷ 14 = 4, so molecular formula = C₄H₈. Using n = 3 gives C₃H₆ (molar mass 42) and n = 2 gives C₂H₄ (molar mass 28) — neither matches 56.
Q29 — Percentage Composition & Formulae · hard · numerical
What is the mass percentage of water of crystallisation in copper sulphate pentahydrate, CuSO₄·5H₂O? (Cu = 63.5, S = 32, O = 16, H = 1)
A. 56.4%
B. 25.5%
C. 63.9%
D. 36.1% ✓ Correct
Solution: Molar mass = 63.5 + 32 + 4(16) + 5(18) = 159.5 + 90 = 249.5. Mass of the 5 water molecules = 90. Mass % water = (90 ÷ 249.5) × 100 = 36.1%. Dividing the water mass by the anhydrous salt mass (90 ÷ 159.5) wrongly gives 56.4%; 25.5% is the mass % of copper (63.5 ÷ 249.5).
Q30 — Percentage Composition & Formulae · hard · numerical
A compound has the empirical formula CH₂O and a vapour density of 30. Its molecular formula is (C = 12, H = 1, O = 16):
A. C₂H₄O₂ ✓ Correct
B. CH₂O
C. C₃H₆O₃
D. C₄H₈O₄
Solution: Molar mass = 2 × vapour density = 2 × 30 = 60 g mol⁻¹. Empirical formula mass of CH₂O = 30, so n = 60 ÷ 30 = 2. Molecular formula = C₂H₄O₂ (acetic acid). The trap CH₂O comes from using the vapour density (30) itself as the molar mass, which wrongly gives n = 1.