Some Basic Concepts of Chemistry NEET PYQReal NEET & AIPMT 2007–2020 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Some Basic Concepts of Chemistry question from the NEET & AIPMT papers on this site (2007–2020), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Some Basic Concepts of Chemistry previous year questions
20 questions. Tap an option to check your answer.
Q1NEET 2020 · Q123
Which one of the followings has maximum number of atoms ?
Not quite — the answer is D.
Moles of atoms: Ag 1/108, Mg 1/24, O₂ gives 2 × 1/32 = 1/16, and Li 1/7. The largest is 1/7 mol, so 1 g of lithium has the most atoms. For O₂ one must count two atoms per molecule, but even then it is less than lithium.
Q2NEET 2019 · Q68
The number of moles of hydrogen molecules required to produce 20 moles of ammonia through Haber's process is:
Not quite — the answer is C.
N₂ + 3H₂ → 2NH₃: 3 mol H₂ give 2 mol NH₃, so 20 mol NH₃ need 20 × 3/2 = 30 mol H₂. 10 mol is the N₂ requirement.
Q3NEET 2018 · Q57
A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be
Not quite — the answer is A.
HCOOH (2.3/46 = 0.05 mol) → CO + H₂O; H₂C₂O₄ (4.5/90 = 0.05 mol) → CO + CO₂ + H₂O. KOH absorbs CO₂ (and water), leaving 0.1 mol CO = 0.1 × 28 = 2.8 g. 4.4 g would be the CO₂ mass if it were wrongly counted.
Q4NEET 2018 · Q78
In which case is the number of molecules of water maximum?
Not quite — the answer is C.
18 mL of liquid water (density 1 g/mL) is 18 g = 1 mol. The others are 10⁻⁴ mol (0.00224 L of vapour at STP), 0.01 mol (0.18 g) and 10⁻³ mol, so 18 mL has the most molecules.
Q5NEET 2017 (Gujarati paper) · Q15
A hydrocarbon contains 85.7% C. If 42 mg of the compound contains 3.01 × 10²⁰ molecules, the molecular formula of the compound will be:
Not quite — the answer is B.
Moles = 3.01 × 10²⁰ / 6.02 × 10²³ = 5 × 10⁻⁴ mol, so molar mass = 0.042 g / 5 × 10⁻⁴ mol = 84 g mol⁻¹. 85.7% C corresponds to the empirical formula CH₂ (14 g mol⁻¹), and 84/14 = 6 gives C₆H₁₂. All options have 85.7% C, so only the molar mass separates them.
Q6NEET 2016 Phase 2 · Q59
Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y are
Not quite — the answer is A.
Molar masses: XY₂ = 10/0.1 = 100 and X₃Y₂ = 9/0.05 = 180. So X + 2Y = 100 and 3X + 2Y = 180; subtracting gives 2X = 80, X = 40 and Y = 30. The pair 30, 20 fails the first equation (30 + 40 = 70).
Q7AIPMT 2015
A mixture of gases contains H₂ and O₂ gases in the ratio of 1 : 4 (w/w). What is the molar ratio of the two gases in the mixture?
Not quite — the answer is C.
Take 1 g H₂ and 4 g O₂: moles = 1/2 = 0.5 and 4/32 = 0.125. Mole ratio 0.5 : 0.125 = 4 : 1. 1 : 4 is just the mass ratio.
Q8AIPMT 2014
1.0 g of magnesium is burnt with 0.56 g O₂ in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16)
Not quite — the answer is D.
2Mg + O₂ → 2MgO. 0.56 g O₂ = 0.0175 mol, which needs 0.035 mol Mg = 0.84 g, while 1.0 g Mg is available, so O₂ is limiting. Mg left over = 1.0 − 0.84 = 0.16 g.
Q9AIPMT 2014
When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g), each at S.T.P., the moles of HCl(g) formed is equal to:
Not quite — the answer is D.
H₂ + Cl₂ → 2HCl. 22.4 L H₂ = 1 mol and 11.2 L Cl₂ = 0.5 mol, so Cl₂ is limiting and gives 2 × 0.5 = 1 mol HCl, not 2 mol.
Q10NEET 2013
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO₃? The concentrated acid is 70% HNO₃.
Not quite — the answer is A.
Moles of HNO₃ = 2.0 × 0.250 = 0.50 mol = 0.50 × 63 = 31.5 g. The 70% solution needed = 31.5/0.70 = 45.0 g. 90 g comes from doubling the moles.
Q11AIPMT 2011 Mains
Which has the maximum number of molecules among the following?
Not quite — the answer is A.
Moles: 8 g H₂ = 8/2 = 4 mol; 64 g SO₂, 44 g CO₂ and 48 g O₃ are each 1 mol. The largest number of moles means the most molecules, so 8 g of H₂ has the most, even though it has the smallest mass.
Q12AIPMT 2010 Prelims · Q52
25.3 g of sodium carbonate, Na₂CO₃ is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na⁺ and carbonate ions, CO₃²⁻ are respectively (Molar mass of Na₂CO₃ = 106 g mol⁻¹)
Not quite — the answer is C.
Moles = 25.3/106 = 0.2387 mol in 0.250 L, so [Na₂CO₃] = 0.955 M. Each formula unit gives 2 Na⁺ and 1 CO₃²⁻, so [Na⁺] = 1.910 M and [CO₃²⁻] = 0.955 M. The reversed pair swaps the ions.
Q13AIPMT 2010 Prelims · Q73
The number of atoms in 0.1 mol of a triatomic gas is (NA = 6.02 × 10²³ mol⁻¹)
Not quite — the answer is C.
0.1 mol of molecules × 3 atoms × 6.02 × 10²³ = 1.806 × 10²³ atoms. 6.02 × 10²² is the number of molecules, not atoms.
Q14AIPMT 2009 Prelims · Q192
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be:
Not quite — the answer is B.
2H₂ + O₂ → 2H₂O. 10 g H₂ = 5 mol and 64 g O₂ = 2 mol; 2 mol O₂ needs only 4 mol H₂, so O₂ is limiting and gives 4 mol of water. Taking H₂ as limiting would wrongly give 5 mol.
Q15AIPMT 2008 Prelims · Q59
What volume of oxygen gas (O₂) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane gas (C₃H₈) measured under the same conditions?
Not quite — the answer is D.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. At the same temperature and pressure, volumes are in the same ratio as moles, so 1 L of propane needs 5 L of O₂. 7 L comes from adding the volumes of the products, which is not what the question asks.
Q16AIPMT 2008 Prelims · Q72
How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?
Not quite — the answer is A.
PbO + 2HCl → PbCl₂ + H₂O. Moles of PbO = 6.5/223 ≈ 0.029, and of HCl = 3.2/36.5 ≈ 0.088, more than the 0.058 needed. PbO is the limiting reagent, so 0.029 mol PbCl₂ forms. 0.044 would wrongly treat HCl as limiting (0.088/2).
Q17AIPMT 2008 Prelims · Q74
Volume occupied by one molecule of water (density = 1 g cm⁻³) is
Not quite — the answer is D.
One mole of water (18 g) occupies 18 cm³ at density 1 g cm⁻³. Volume per molecule = 18/(6.022 × 10²³) ≈ 3.0 × 10⁻²³ cm³. 6.023 × 10⁻²³ cm³ just reuses the digits of Avogadro's number.
Q18AIPMT 2008 Prelims · Q95
An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be
Not quite — the answer is B.
O = 100 − 38.71 − 9.67 = 51.62%. Moles: C = 38.71/12 = 3.23, H = 9.67/1 = 9.67, O = 51.62/16 = 3.23, a ratio of 1 : 3 : 1, so the empirical formula is CH₃O. CH₄O (methanol) would need about 12.5% H.
Q19AIPMT 2007 Prelims · Q65
Concentrated aqueous sulphuric acid is 98% H₂SO₄ by mass and has a density of 1.80 g mL⁻¹. Volume of acid required to make one litre of 0.1 M H₂SO₄ solution is
Not quite — the answer is C.
Molarity of the concentrated acid = (98/98) mol per 100 g × 1.80 g/mL × 1000/100 = 18.0 M. V = (0.1 × 1000)/18.0 = 5.55 mL. 11.10 mL would result from treating the acid as 9 M.
Q20AIPMT 2007 Prelims · Q95
An element, X has the following isotopic composition: ²⁰⁰X : 90%, ¹⁹⁹X : 8.0%, ²⁰²X : 2.0% The weighted average atomic mass of the naturally occurring element X is closest to
Not quite — the answer is D.
Average mass = 200 × 0.90 + 199 × 0.08 + 202 × 0.02 = 180 + 15.92 + 4.04 = 199.96 ≈ 200 amu. 201 amu would be a simple average of the three masses.
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