Mole Concept — 50 Original Practice Questions in Competitive-Exam Style
The set below is something I wrote myself, question by question, to mirror the flavour of numericals you'll meet in JEE Main, JEE Advanced, NEET, and similar entrance papers on Mole Concept — the same style of limiting-reagent traps, empirical-formula-from-combustion problems, dilution and mixing calculations, POAC applications, and significant-figure catches that these exams lean on. I've simply built questions that test identical ideas in a comparable format, and tagged each one with the exam style and rough year-band it's modelled after, purely to give you a feel for how the difficulty has trended over time.
How to use this set: Attempt each question cold before reading the solution. If a pattern trips you up more than once, go back to the matching section in the main notes — every question here maps to a specific concept covered earlier (mole–mass conversions, POAC, limiting reagent, concentration terms, gravimetric analysis, significant figures, and so on).
Section A — Atomic Mass, Molecular Mass, and the Mole (Q1–Q10)
Q1.JEE Main pattern, ~2015–2017
An element X exists as two isotopes with masses 68.9 u and 70.9 u, present in the ratio 3:1 by abundance. What is the average atomic mass of X?
(a) 69.4 u (b) 69.9 u (c) 70.4 u (d) 68.9 u
Answer: (a). Average = (0.75 × 68.9) + (0.25 × 70.9) = 51.675 + 17.725 = 69.4 u.
Q2.NEET pattern, ~2016–2019
The number of moles of oxygen atoms present in 4.9 g of H₂SO₄ (molar mass 98) is:
(a) 0.05 (b) 0.1 (c) 0.2 (d) 0.4
Answer: (c). Moles of H₂SO₄ = 4.9/98 = 0.05. Each molecule has 4 O atoms, so moles of O = 0.05 × 4 = 0.2.
Q3.JEE Advanced pattern, ~2013–2016
Two isotopes of an element have masses 24.0 u and 26.0 u. If the average atomic mass of the element is 24.4 u, the percentage abundance of the lighter isotope is closest to:
(a) 20% (b) 40% (c) 60% (d) 80%
Answer: (d). Let x = fraction of lighter isotope: 24x + 26(1−x) = 24.4 ⟹ −2x = −1.6 ⟹ x = 0.8, i.e. 80%.
Q4.AIEEE-style, ~2005–2009
Which of the following samples contains the greatest number of atoms?
(a) 4 g of He (molar mass 4) (b) 12 g of Mg (molar mass 24) (c) 20 g of Ca (molar mass 40) (d) 8 g of O₂ (molar mass 32)
Answer: (a). He: 4/4 = 1 mol of atoms (monatomic). Mg: 12/24 = 0.5 mol atoms. Ca: 20/40 = 0.5 mol atoms. O₂: 8/32 = 0.25 mol molecules × 2 atoms/molecule = 0.5 mol atoms. He gives the most atoms at 1 mol, ahead of the other three at 0.5 mol each — a reminder to convert to actual atom count, not just moles of substance, before comparing.
Q5.JEE Main pattern, ~2018–2020
The mass of a single molecule of glucose (C₆H₁₂O₆, molar mass 180) is:
(a) 1.5 × 10⁻²² g (b) 2.99 × 10⁻²² g (c) 2.99 × 10⁻²³ g (d) 1.8 × 10⁻²² g
Answer: (c). Mass of one molecule = 180 / 6.023×10²³ = 2.99 × 10⁻²³ g.
Q6.NEET-style, ~2014–2017
The vapour density of a gaseous hydrocarbon is 21. Its molecular mass is:
(a) 21 (b) 10.5 (c) 42 (d) 84
Answer: (c). Molecular mass = 2 × vapour density = 2 × 21 = 42.
Q7.JEE Main pattern, ~2016–2019
Which of the following weighs the least?
(a) 1 mol of CO₂ (b) 1 mol of N₂ (c) 1 mol of O₂ (d) 1 mol of SO₂
Answer: (b). Molar masses: CO₂ = 44, N₂ = 28, O₂ = 32, SO₂ = 64. N₂ has the smallest molar mass.
Q8.AIEEE-style, ~2007–2011
How many moles of electrons weigh 1 kg? (Mass of an electron = 9.1 × 10⁻³¹ kg)
(a) 6.023 × 10²³ (b) 1/(9.1 × 6.023) × 10⁸ (c) 1.098 × 10⁵² (d) 0.155 × 10⁵⁴
Answer: (b). Number of electrons in 1 kg = 1/(9.1×10⁻³¹). Moles = that divided by 6.023×10²³, which simplifies to 1/(9.1 × 6.023) × 10⁸ mol.
Q9.JEE Advanced pattern, ~2014–2018
A sample of CaCO₃ and MgCO₃ weighing 2.21 g leaves a residue of 1.15 g on strong heating (both decompose to their oxides, releasing CO₂). Approximately what mass of CaCO₃ was present? (Molar masses: CaCO₃ = 100, CaO = 56, MgCO₃ = 84, MgO = 40)
(a) 0.5 g (b) 1.0 g (c) 1.5 g (d) 2.0 g
Answer: (b). Let x g = mass of CaCO₃, so (2.21 − x) g = mass of MgCO₃. Moles CaCO₃ = x/100 → gives x/100 mol CaO → mass (56x/100). Moles MgCO₃ = (2.21−x)/84 → mass MgO = 40(2.21−x)/84. Setting 0.56x + 0.476(2.21−x) = 1.15 and solving gives x ≈ 1.0 g.
Q10.NEET pattern, ~2019–2021
Number of moles present in 5.6 L of a gas at STP is:
(a) 0.10 (b) 0.25 (c) 0.50 (d) 1.00
Answer: (b). Moles = 5.6/22.4 = 0.25.
Section B — Empirical/Molecular Formula and Percentage Composition (Q11–Q18)
Q11.JEE Main pattern, ~2013–2016
An organic compound contains 40% C, 6.7% H, and the rest oxygen by mass. Its empirical formula is:
(a) CH₂O (b) C₂H₄O (c) CHO (d) C₂H₄O₂
Answer: (a). O% = 100 − 40 − 6.7 = 53.3%. Moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Ratio C:H:O = 1:2:1 → CH₂O.
Q12.JEE Advanced pattern, ~2011–2015
A compound with empirical formula CH₂O has a molecular mass of 180 g/mol. Its molecular formula is:
(a) CH₂O (b) C₂H₄O₂ (c) C₆H₁₂O₆ (d) C₃H₆O₃
Answer: (c). Empirical formula mass = 30. n = 180/30 = 6. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
Q13.NEET-style, ~2015–2018
The percentage of nitrogen in ammonium sulphate, (NH₄)₂SO₄ (molar mass 132), is closest to:
(a) 10.6% (b) 21.2% (c) 15.9% (d) 28%
Answer: (b). Mass of N = 2 × 14 = 28. % N = (28/132) × 100 ≈ 21.2%.
Q14.JEE Main pattern, ~2017–2020
Combustion of 4.5 g of an organic compound containing only C and H gives 13.2 g CO₂ and 8.1 g H₂O. The empirical formula is:
(a) CH₂ (b) CH₃ (c) C₂H₆ (d) C₂H₄
Answer: (b). Moles C = 13.2/44 = 0.3 → mass C = 3.6 g. Moles H = 2×(8.1/18) = 0.9 → mass H = 0.9 g. Total = 3.6 + 0.9 = 4.5 g, matching the sample mass exactly, confirming only C and H are present with no oxygen. Mole ratio C:H = 0.3:0.9 = 1:3, giving the empirical formula CH₃.
Q15.AIEEE-style, ~2006–2010
A hydrated salt MSO₄·xH₂O contains 45.32% water by mass. If the molar mass of anhydrous MSO₄ is 120 g/mol, the value of x is closest to:
(a) 5 (b) 6 (c) 7 (d) 4
Answer: (a). Let total molar mass = 120 + 18x. Water fraction: 18x/(120+18x) = 0.4532. Solving: 18x = 54.4 + 8.16x ⟹ 9.84x ≈ 54.4 ⟹ x ≈ 5.5, closest to 5 among the given options (rounding to the nearest whole-number hydrate).
Q16.JEE Main pattern, ~2019–2022
The percentage of water of crystallisation in CuSO₄·5H₂O (molar mass 250) is:
(a) 18% (b) 28% (c) 36% (d) 40%
Answer: (c). Mass of water = 5 × 18 = 90. % = (90/250) × 100 = 36%.
Q17.NEET pattern, ~2020–2023
30 cm³ of a gaseous hydrocarbon requires 105 cm³ of O₂ for complete combustion and produces 60 cm³ of CO₂ (volumes at same T, P). The molecular formula is:
(a) C₂H₄ (b) C₂H₆ (c) C₂H₂ (d) CH₄
Answer: (b). Using CₓHᵧ + (x + y/4)O₂ → xCO₂ + (y/2)H₂O, with volumes directly proportional to moles per Avogadro's law: x = 60/30 = 2. Also (x + y/4) = 105/30 = 3.5, so y/4 = 3.5 − 2 = 1.5 ⟹ y = 6. This gives C₂H₆.
Q18.JEE Advanced pattern, ~2012–2016
Two oxides of a metal contain 30% and 40% oxygen by mass respectively. This data illustrates:
(a) Law of conservation of mass (b) Law of definite proportions (c) Law of multiple proportions (d) Avogadro's law
Answer: (c). Same metal forming two different oxides with oxygen masses in a simple ratio (per fixed mass of metal) is the defining signature of the law of multiple proportions.
Section C — Stoichiometry, Limiting Reagent, and Yield (Q19–Q28)
Q19.JEE Main pattern, ~2015–2018
For N₂ + 3H₂ → 2NH₃, if 5 mol N₂ reacts with 9 mol H₂, the limiting reagent and the maximum moles of NH₃ formed are:
(a) N₂; 6 mol (b) H₂; 6 mol (c) H₂; 3 mol (d) N₂; 10 mol
Answer: (b). 9 mol H₂ needs 3 mol N₂ (ratio 3:1), leaving N₂ in excess; H₂ is limiting. Moles NH₃ = (2/3) × 9 = 6.
Q20.NEET-style, ~2016–2019
10 g of CaCO₃ is treated with excess dilute HCl. The volume of CO₂ released at STP (CaCO₃ molar mass 100) is:
(a) 1.12 L (b) 2.24 L (c) 4.48 L (d) 22.4 L
Answer: (b). Moles CaCO₃ = 10/100 = 0.1. By 1:1 stoichiometry, moles CO₂ = 0.1. Volume = 0.1 × 22.4 = 2.24 L.
Q21.JEE Main pattern, ~2014–2017
A reaction has a theoretical yield of 40 g. If the actual yield obtained is 34 g, the percentage yield is:
(a) 75% (b) 80% (c) 85% (d) 90%
Answer: (c). % yield = (34/40) × 100 = 85%.
Q22.AIEEE-style, ~2008–2012
5.6 g of Fe reacts with 4.0 g of S to form FeS (Fe = 56, S = 32, molar reaction 1:1). The mass of FeS formed and the excess reagent left are:
(a) 8.8 g FeS, 0 g excess (b) 8.8 g FeS, 0.8 g S left (c) 11.0 g FeS, 0.8 g Fe left (d) 8.8 g FeS, 0.8 g Fe left
Answer: (b). Moles Fe = 5.6/56 = 0.1. Moles S = 4.0/32 = 0.125. Fe is limiting (1:1 ratio). FeS formed = 0.1 mol × 88 g/mol = 8.8 g. S used = 0.1 mol × 32 = 3.2 g, so S left = 4.0 − 3.2 = 0.8 g.
Q23.JEE Advanced pattern, ~2013–2017
For the reaction 2Al + 3Cl₂ → 2AlCl₃, if 5.4 g Al (molar mass 27) reacts with 7.1 g Cl₂ (molar mass 71), the limiting reagent is:
(a) Al (b) Cl₂ (c) Both are exactly stoichiometric (d) Cannot be determined
Answer: (b). Moles Al = 5.4/27 = 0.2. Moles Cl₂ = 7.1/71 = 0.1. Required ratio is Al:Cl₂ = 2:3. For 0.2 mol Al, 0.3 mol Cl₂ is needed, but only 0.1 mol is available — Cl₂ is limiting.
Q24.NEET pattern, ~2018–2021
How many grams of water are produced when 4 g of H₂ reacts completely with excess O₂?
(a) 18 g (b) 36 g (c) 9 g (d) 72 g
Answer: (b). Moles H₂ = 4/2 = 2. By 2H₂ + O₂ → 2H₂O, moles H₂O = 2. Mass = 2 × 18 = 36 g.
Q25.JEE Main pattern, ~2016–2019
50 mL of 0.2 M AgNO₃ is mixed with 50 mL of 0.1 M NaCl. The mass of AgCl (molar mass 143.5) precipitated is:
(a) 0.7175 g (b) 1.435 g (c) 0.35 g (d) 1.0 g
Answer: (a). Moles AgNO₃ = 0.05 × 0.2 = 0.01. Moles NaCl = 0.05 × 0.1 = 0.005. NaCl is limiting (1:1 ratio). Mass AgCl = 0.005 × 143.5 = 0.7175 g.
Q26.AIEEE-style, ~2009–2012
The number of moles of KMnO₄ needed to oxidise 1 mole of FeSO₄ in acidic medium (5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O) is:
(a) 1 (b) 0.2 (c) 5 (d) 0.5
Answer: (b). Ratio Fe²⁺ : MnO₄⁻ = 5:1, so for 1 mol Fe²⁺, moles KMnO₄ needed = 1/5 = 0.2.
Q27.JEE Advanced pattern, ~2015–2019
A 2.0 g sample of an impure Na₂CO₃ sample is completely reacted with excess HCl, releasing 0.42 L of CO₂ at STP. The percentage purity of Na₂CO₃ (molar mass 106) is closest to:
(a) 90.4% (b) 95.5% (c) 99.4% (d) 85%
Answer: (c). Moles CO₂ = 0.42/22.4 = 0.01875. By 1:1 stoichiometry, moles Na₂CO₃ = 0.01875. Mass = 0.01875 × 106 = 1.9875 g. Purity = (1.9875/2.0) × 100 ≈ 99.4%.
Q28.NEET-style, ~2017–2020
The reaction 2KClO₃ → 2KCl + 3O₂ is carried out with 24.5 g of KClO₃ (molar mass 122.5). The volume of O₂ produced at STP is:
(a) 2.24 L (b) 4.48 L (c) 6.72 L (d) 11.2 L
Answer: (c). Moles KClO₃ = 24.5/122.5 = 0.2. Moles O₂ = (3/2) × 0.2 = 0.3. Volume = 0.3 × 22.4 = 6.72 L.
Section D — Principle of Atom Conservation, POAC (Q29–Q34)
Q29.JEE Main pattern, ~2017–2020
2 mol of Na₂CO₃ is converted, through an unspecified sequence of steps, entirely into NaCl. Applying POAC on sodium, the moles of NaCl formed are:
(a) 1 (b) 2 (c) 4 (d) 6
Answer: (c). Each mole of Na₂CO₃ has 2 mol Na. POAC: 2 × 2 = 1 × n(NaCl) ⟹ n = 4.
Q30.JEE Advanced pattern, ~2014–2018
1 mol of Al₂(SO₄)₃ is converted completely, via several steps, into BaSO₄. Applying POAC on sulphur, the moles of BaSO₄ formed are:
(a) 1 (b) 2 (c) 3 (d) 6
Answer: (c). Each mole of Al₂(SO₄)₃ has 3 mol S. POAC: 1 × 3 = 1 × n(BaSO₄) ⟹ n = 3.
Q31.NEET pattern, ~2019–2022
1 mol of C₃H₈ is completely combusted (unbalanced equation given): C₃H₈ + O₂ → CO₂ + H₂O. Using POAC on carbon and hydrogen, the moles of CO₂ and H₂O respectively are:
(a) 3, 4 (b) 3, 8 (c) 1, 4 (d) 3, 2
Answer: (a). POAC on C: 3×1 = 1×n(CO₂) ⟹ n = 3. POAC on H: 8×1 = 2×n(H₂O) ⟹ n = 4.
Q32.JEE Main pattern, ~2018–2021
Continuing from Q31, applying POAC on oxygen, the moles of O₂ consumed are:
(a) 3 (b) 4 (c) 5 (d) 6
Answer: (c). POAC on O: 2×n(O₂) = 2×n(CO₂) + 1×n(H₂O) = 2(3) + 4 = 10 ⟹ n(O₂) = 5.
Q33.AIEEE-style, ~2007–2010
Under what circumstance is POAC preferred over conventional stoichiometry?
(a) When the balanced equation is fully known (b) When the equation is unbalanced or the reaction pathway is unspecified (c) Only for acid-base titrations (d) Only when gases are involved
Answer: (b). POAC's main advantage is that it sidesteps the need to know or balance every intermediate step, as long as the element of interest converts completely.
Q34.JEE Advanced pattern, ~2016–2020
xNa₂S₂O₃ + yI₂ → Na₂S₄O₆ + zNaI. Using POAC on sodium, if 2 mol Na₂S₂O₃ reacts completely, the moles of NaI formed are:
(a) 1 (b) 2 (c) 3 (d) 4
Answer: (b). POAC on Na: 2×2 = 2×n(Na₂S₄O₆) + 1×n(NaI). With 1 mol Na₂S₄O₆ formed (standard iodometric stoichiometry, 2:1 ratio for thiosulfate to product), 4 = 2 + n(NaI) ⟹ n(NaI) = 2.
Section E — Concentration Terms: Molarity, Molality, Dilution, Mixing (Q35–Q44)
Q35.JEE Main pattern, ~2019–2022
What volume of 0.5 M H₂SO₄ must be diluted with water to prepare 250 mL of 0.1 M H₂SO₄?
(a) 25 mL (b) 50 mL (c) 100 mL (d) 125 mL
Answer: (b). M₁V₁ = M₂V₂ ⟹ 0.5 × V₁ = 0.1 × 250 ⟹ V₁ = 50 mL.
Q36.NEET-style, ~2015–2018
300 mL of 0.5 M NaOH is mixed with 200 mL of 1.0 M NaOH. The resulting molarity is:
(a) 0.60 M (b) 0.70 M (c) 0.75 M (d) 0.80 M
Answer: (b). Total moles = (0.3×0.5) + (0.2×1.0) = 0.15 + 0.2 = 0.35. Total volume = 0.5 L. M = 0.35/0.5 = 0.7 M.
Q37.JEE Advanced pattern, ~2012–2016
A 2 molal solution of a solute (molar mass 60) in water has a density of 1.12 g/mL. Its molarity is:
(a) 1.6 M (b) 1.8 M (c) 2.0 M (d) 2.2 M
Answer: (c). For 1 kg (1000 g) of solvent, moles of solute = 2, so mass of solute = 2 × 60 = 120 g. Total solution mass = 1000 + 120 = 1120 g. Volume of solution = mass/density = 1120/1.12 = 1000 mL = 1 L. Molarity = 2 mol / 1 L = 2.0 M.
Q38.AIEEE-style, ~2008–2011
The mole fraction of ethanol in a solution containing 46 g ethanol (molar mass 46) and 90 g water (molar mass 18) is:
(a) 0.091 (b) 0.167 (c) 0.200 (d) 0.500
Answer: (b). Moles ethanol = 46/46 = 1. Moles water = 90/18 = 5. Mole fraction ethanol = 1/(1+5) = 0.167.
Q39.JEE Main pattern, ~2014–2017
The normality of a 0.1 M H₂SO₄ solution is:
(a) 0.05 N (b) 0.1 N (c) 0.2 N (d) 0.4 N
Answer: (c). H₂SO₄ has n-factor 2 (diprotic). Normality = Molarity × n-factor = 0.1 × 2 = 0.2 N.
Q40.NEET pattern, ~2020–2023
How many grams of NaCl (molar mass 58.5) are needed to prepare 500 mL of a 0.9% (w/V) saline solution?
(a) 2.25 g (b) 3.5 g (c) 4.5 g (d) 9.0 g
Answer: (c). % w/V = (mass/volume in mL)×100 ⟹ mass = 0.9 × 500/100 = 4.5 g.
Q41.JEE Advanced pattern, ~2011–2015
A solution is 20% NaOH by mass with density 1.2 g/mL. Its molarity (NaOH molar mass 40) is:
(a) 4.0 M (b) 5.0 M (c) 6.0 M (d) 8.0 M
Answer: (c). In 100 g solution, NaOH = 20 g. Volume of solution = 100/1.2 = 83.33 mL. Moles NaOH = 20/40 = 0.5. Molarity = 0.5/(83.33/1000) = 6.0 M.
Q42.AIEEE-style, ~2009–2013
The molality of pure water (density 1 g/mL) is closest to:
(a) 18 m (b) 55.5 m (c) 100 m (d) Undefined, since water is the solvent
Answer: (d). Molality requires a distinguishable solute dissolved in a solvent; pure water alone has no solute, so molality in the conventional sense isn't defined here (this contrasts with molarity, which can still be computed as roughly 55.5 M treating water as "solute" in itself for demonstration purposes only).
Q43.JEE Main pattern, ~2016–2019
150 mL of 0.2 M HCl and 100 mL of 0.4 M HCl are mixed. The final molarity is:
(a) 0.26 M (b) 0.28 M (c) 0.30 M (d) 0.32 M
Answer: (b). Total moles = (0.15×0.2) + (0.1×0.4) = 0.03 + 0.04 = 0.07. Total volume = 0.25 L. Molarity = 0.07/0.25 = 0.28 M.
Q44.NEET-style, ~2017–2020
Which of the following concentration terms remains unaffected by a change in temperature?
(a) Molarity (b) Normality (c) Molality (d) % (V/V)
Answer: (c). Molality is based on mass of solvent, not volume, so it is unaffected by thermal expansion or contraction.
Section F — Significant Figures, Units, and Gravimetric Analysis (Q45–Q50+)
Q45.JEE Main pattern, ~2015–2018
The number of significant figures in 6.023 × 10²³ is:
(a) 3 (b) 4 (c) 23 (d) 27
Answer: (b). Only the coefficient's digits count: 6, 0, 2, 3 — four significant figures.
Q46.AIEEE-style, ~2006–2010
2.005 × 3.1 (both measured quantities) should be reported to how many significant figures?
(a) 2 (b) 3 (c) 4 (d) 5
Answer: (a). In multiplication, the answer is limited to the fewest significant figures among the factors — 3.1 has only 2, so the product must be reported to 2 significant figures.
Q47.JEE Advanced pattern, ~2013–2017
12.11 g + 18.0 g + 1.013 g, all measured quantities, should be reported to how many decimal places?
(a) 1 (b) 2 (c) 3 (d) 0
Answer: (a). In addition, the answer's decimal places are limited by the term with the fewest — 18.0 has just one decimal place.
Q48.NEET pattern, ~2018–2021
A pure sample of MCO₃ (2.0 g) is heated completely to MO + CO₂, leaving a residue of 1.12 g. If the atomic mass of M is x, which equation correctly relates x to the given data?
(a) (x+16)/(x+60) = 1.12/2.0 (b) (x+60)/(x+16) = 2.0/1.12 (c) x/(x+60) = 1.12/2.0 (d) (x+16)/60 = 1.12/2.0
Answer: (a). Molar mass of MCO₃ = x + 60; molar mass of MO (the residue) = x + 16. Since moles are conserved between MCO₃ and MO, mass ratio equals molar mass ratio: (x+16)/(x+60) = 1.12/2.0.
Q49.JEE Main pattern, ~2020–2023
In gravimetric estimation of sulphate as BaSO₄, which property of BaSO₄ makes it suitable as the weighed precipitate?
(a) High solubility in water (b) Low solubility, chemical stability, and known fixed composition (c) Volatility on heating (d) Strong colour for visual detection
Answer: (b). A good gravimetric precipitate must be nearly insoluble (so nothing is lost to the filtrate), chemically stable on drying, and of exactly known stoichiometric composition so its weighed mass converts cleanly to moles of analyte.
Q50.AIEEE-style, ~2007–2011
10 mL of 1 cm diameter spherical marbles are close-packed (centres within the boundary) along one edge of a 5 cm square tray. Approximately how many marbles fit per unit area, following the same reasoning used for close-packing problems in the mole concept chapter?
(a) 5 (b) 10 (c) 25 (d) 50
Answer: (c). Along a 5 cm edge, marbles of 1 cm diameter fit 5 to a row (centres spaced 1 cm apart within the boundary), giving a 5×5 grid — 25 marbles per unit area, following the same square-packing logic used for surface-area/molecule-size estimation problems.
Q51.JEE Advanced pattern, ~2016–2020
0.01 mol of an acid (0.05 M, 200 mL) is adsorbed by activated charcoal such that its concentration drops to 0.04 M in the same volume. If the charcoal's surface area is 2.0 × 10² m², the area occupied by one adsorbed molecule is closest to:
(a) 1.66 × 10⁻¹⁹ m² (b) 3.32 × 10⁻¹⁹ m² (c) 1.66 × 10⁻²⁰ m² (d) 8.3 × 10⁻²⁰ m²
Answer: (a). Moles adsorbed = 0.2 L × (0.05 − 0.04) M = 0.002 mol. Molecules adsorbed = 0.002 × 6.023×10²³ = 1.2×10²¹. Area per molecule = (2.0×10²)/(1.2×10²¹) ≈ 1.66 × 10⁻¹⁹ m².
Q52.NEET-style, ~2019–2022
Which SI prefix corresponds to a factor of 10⁻⁹?
(a) micro (b) nano (c) pico (d) femto
Answer: (b). nano = 10⁻⁹.
Q53.JEE Main pattern, ~2013–2016
Convert 72 km/h to m/s.
(a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 25 m/s
Answer: (c). 72 km/h × (1000 m/1 km) × (1 h/3600 s) = 20 m/s.
Q54.AIEEE-style, ~2010–2013
The density of a metal is 8.9 g/cm³. Its density in SI units (kg/m³) is:
(a) 89 (b) 890 (c) 8900 (d) 89000
Answer: (c). 1 g/cm³ = 1000 kg/m³, so 8.9 g/cm³ = 8900 kg/m³.
