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Some Basic Concepts of Chemistry: NEET previous year questions

Every Some Basic Concepts of Chemistry question from the official NEET papers on this site, with the answer and a worked solution.

The questions come from NEET 2026, Re-NEET 2026 and NEET 2020. Each shows its answer and solution when you open it, so you can attempt it first. For what the chapter covers and how NEET tests it, see the Some Basic Concepts of Chemistry guide.

Practise on the CBT screen5 questions, 5 minutes, marked +4 / −1Download PDF3 pages, 0.7 MB, with answers and solutions

NEET 2026 · the full paper

  1. NEET 2026 · Q61

    The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea = 60 g mol−160\ \text{g mol}^{-1}; NA=6.022×1023 particles mol−1N_A = 6.022 \times 10^{23}\ \text{particles mol}^{-1})

    1. (1)

      1.084×10231.084 \times 10^{23}

    2. (2)

      1.084×10221.084 \times 10^{22}

    3. (3)

      2.168×10222.168 \times 10^{22}

    4. (4)

      2.168×10232.168 \times 10^{23}

    Answer and solution

    Answer: (4)

    Moles of urea = 5.4/60 = 0.09 mol. Urea, NH2CONH2\mathrm{NH}_{2}\mathrm{CONH}_{2}, contains 4 hydrogen atoms per molecule. So number of H atoms = 0.09×4×6.022×1023≈2.168×10230.09 \times 4 \times 6.022 \times 10^{23} \approx 2.168 \times 10^{23}. The distractors arise from using the wrong number of H atoms per molecule or a wrong power of ten.

  2. NEET 2026 · Q84

    When 1 dm3^3 of CO2\mathrm{CO}_{2} gas is passed over hot coke, the volume of gaseous mixture after complete reaction at STP becomes 1.4 dm3^3. The composition of the gaseous mixture at STP is:

    1. (1)

      0.8 dm3^3 of CO, 0.8 dm3^3 of CO2\mathrm{CO}_{2}

    2. (2)

      0.8 dm3^3 of CO, 0.6 dm3^3 of CO2\mathrm{CO}_{2}

    3. (3)

      0.6 dm3^3 of CO, 0.8 dm3^3 of CO2\mathrm{CO}_{2}

    4. (4)

      0.6 dm3^3 of CO, 0.4 dm3^3 of CO2\mathrm{CO}_{2}

    Answer and solution

    Answer: (2)

    The reaction CO2+C(s)→2CO\mathrm{CO}_{2} + \mathrm{C}(\text{s}) \rightarrow 2\mathrm{CO} converts x dm3^3 of CO2\mathrm{CO}_{2} into 2x dm3^3 of CO. Total final volume =(1−x)+2x=1+x=1.4=(1-x)+2x = 1+x = 1.4, so x = 0.4 dm3^3. Therefore CO2\mathrm{CO}_{2} remaining =1−0.4=0.6=1-0.4=0.6 dm3^3 and CO formed =2×0.4=0.8=2\times0.4=0.8 dm3^3, which is option (2).

Re-NEET 2026 · the full paper

  1. Re-NEET 2026 · Q51

    The numbers 17.0145 and 21.0235 were rounded to three figures after the decimal point. The resulting numbers, respectively, are

    1. (1)

      17.014 and 21.023

    2. (2)

      17.015 and 21.023

    3. (3)

      17.014 and 21.024

    4. (4)

      17.015 and 21.024

    Answer and solution

    Answer: (3)

    Both numbers end in an exact 5 with nothing after it, which is the tie case, so the round-half-to-EVEN convention applies rather than always rounding up. In 17.0145 the digit to be retained is 4, already even, so it stays: 17.014. In 21.0235 the retained digit is 3, which is odd, so it is raised to the even value 4: 21.024. NCERT Class 11, Some Basic Concepts of Chemistry (significant figures).

  2. Re-NEET 2026 · Q63

    The amount of carbon dioxide evolved upon complete combustion of 116 g of n-butane is (Given: atomic mass in amu H = 1, C = 12 and O = 16)

    1. (1)

      352 g

    2. (2)

      322 g

    3. (3)

      176 g

    4. (4)

      362 g

    Answer and solution

    Answer: (1)

    n-Butane is C4H10\mathrm{C}_{4}\mathrm{H}_{10} with molar mass 58 g/mol, so 116 g is exactly 2 mol. The balanced combustion C4H10+13/2O2→4CO2+5H2O\mathrm{C}_{4}\mathrm{H}_{10} + 13/2 \mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 5\mathrm{H}_{2}\mathrm{O} yields 4 mol of CO2\mathrm{CO}_{2} per mole of butane, hence 8 mol here. Mass of CO2=8×44=352 g\mathrm{CO}_{2} = 8 \times 44 = 352\ \text{g}. NCERT Class 11, Some Basic Concepts of Chemistry (stoichiometry).

NEET 2020 · the full paper

  1. NEET 2020 · Q117

    Which one of the followings has maximum number of atoms ?

    1. (1)

      1 g of Mg(s) [Atomic mass of Mg = 24]

    2. (2)

      1 g of O2\mathrm{O}_{2}(g) [Atomic mass of O = 16]

    3. (3)

      1 g of Li(s) [Atomic mass of Li = 7]

    4. (4)

      1 g of Ag(s) [Atomic mass of Ag = 108]

    Answer and solution

    Answer: (3)

    To find the number of atoms, use n=mM×NAn = \frac{m}{M} \times N_A, where m is mass, M is molar mass, and NAN_A is Avogadro's number. For 1 g of each: - Ag: 1108×NA\frac{1}{108} \times N_A - Mg: 124×NA\frac{1}{24} \times N_A - O2\mathrm{O}_{2}: 132×2×NA=116×NA\frac{1}{32} \times 2 \times N_A = \frac{1}{16} \times N_A (since O2\mathrm{O}_{2} has a molar mass of 32 g/mol and two atoms per molecule) - Li: 17×NA\frac{1}{7} \times N_A Li has the highest number of atoms. Chapter: Some Basic Concepts of Chemistry.

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Questions

Are these the official NEET questions on Some Basic Concepts of Chemistry?

Yes. They are the questions NTA set in each paper, in its wording. The solutions are written by BrainIt.

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