Some Basic Concepts of Chemistry NEET PYQs (2005–2026) with Solutions

10 solved Some Basic Concepts of Chemistry PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyRounding off numbers, significant figures

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

   1. AOption A: $17.014$ and $21.023$
   2. BOption B: $17.015$ and $21.023$
   3. COption C: $17.014$ and $21.024$
   4. DOption D: $17.015$ and $21.024$

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   Correct answer: (C) $17.014$ and $21.024$

   Explanation

   NCERT rule: if the digit to be removed is 5, the preceding digit is left unchanged if it is even and increased by one if it is odd. $17.0145$: the digit removed is 5 and the preceding digit is 4 (even), so it stays the same: $17.014$. $21.0235$: the digit removed is 5 and the preceding digit is 3 (odd), so it goes up by one: $21.024$. So the answer is $17.014$ and $21.024$, option (C).

2. Question 2 (NEET 2026)

   NEET 2026EasyMole Concept & Molar Mass

   The number of hydrogen atoms present in 5.4 g of urea is: (Given: Molar mass of urea = 60 g mol⁻¹; Nₐ = 6.022 × 10²³ particles mol⁻¹)

   1. AOption A: 1.084 × 10²³
   2. BOption B: 1.084 × 10²²
   3. COption C: 2.168 × 10²²
   4. DOption D: 2.168 × 10²³

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   Correct answer: (D) 2.168 × 10²³

   Explanation

   Moles of urea = 5.4/60=0.09 mol. Each urea molecule contains 4 H atoms. Total H atoms =0.09×4×6.022×10²³≈2.168×10²³. Hence option (D) is correct.

3. Question 3 (NEET 2026)

   NEET 2026EasyStoichiometry of combustion reactions

   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. AOption A: $352\ \mathrm{g}$
   2. BOption B: $322\ \mathrm{g}$
   3. COption C: $176\ \mathrm{g}$
   4. DOption D: $362\ \mathrm{g}$

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   Correct answer: (A) $352\ \mathrm{g}$

   Explanation

   Balanced equation: $$\mathrm{2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O}$$ Molar mass of n-butane $= 4(12) + 10(1) = 58\ \mathrm{g\ mol^{-1}}$, so$116\ \mathrm{g} = 2\ \mathrm{mol}$. 2 mol of butane gives 8 mol of$\mathrm{CO_2}$. Mass of$\mathrm{CO_2} = 8 \times 44 = 352\ \mathrm{g}$, option (A).

4. Question 4 (NEET 2025)

   NEET 2025EasyLaws of Chemical Combination

   Dalton’s Atomic theory could not explain which of the following?

   1. AOption A: Law of conservation of mass
   2. BOption B: Law of constant proportion
   3. COption C: Law of multiple proportion
   4. DOption D: Law of gaseous volume

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   Correct answer: (D) Law of gaseous volume

   Explanation

   Dalton’s atomic theory successfully explained the laws of conservation of mass, constant proportions, and multiple proportions. However, it failed to explain Gay-Lussac’s law of gaseous volumes.

5. Question 5 (NEET 2025)

   NEET 2025EasyMole Concept & Molar Mass

   Among the following, choose the ones with equal number of atoms. A. 212 g of Na₂CO₃(s) [molar mass = 106 g] B. 248 g of Na₂O(s) [molar mass = 62 g] C. 240 g of NaOH(s) [molar mass = 40 g] D. 12 g of H₂(g) [molar mass = 2 g] E. 220 g of CO₂(g) [molar mass = 44 g] Choose the correct answer from the options given below :

   1. AOption A: A, B, and C only
   2. BOption B: A, B, and D only
   3. COption C: B, C, and D only
   4. DOption D: B, D, and E only

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   Correct answer: (B) A, B, and D only

   Explanation

   Calculate the number of moles and total atoms: A: 212 g Na₂CO₃ = 2 mol, each molecule has 6 atoms → 12 mol atoms B: 248 g Na₂O = 4 mol, each molecule has 3 atoms → 12 mol atoms C: 240 g NaOH = 6 mol, each molecule has 3 atoms → 18 mol atoms D: 12 g H₂ = 6 mol, each molecule has 2 atoms → 12 mol atoms E: 220 g CO₂ = 5 mol, each molecule has 3 atoms → 15 mol atoms Hence A, B, and D have equal number of atoms.

6. Question 6 (NEET 2024)

   NEET 2024EasyMole Concept & Molar Mass

   The highest number of helium atoms is in:

   1. AOption A: 4 g of helium
   2. BOption B: 2.271098 L of helium at STP
   3. COption C: 4 mol of helium
   4. DOption D: 4 u of helium

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   Correct answer: (C) 4 mol of helium

   Explanation

   Number of atoms depends on the number of moles. (1) 4 g He: $$\text{moles} = \frac{4}{4} = 1\,mol$$ (2) 2.271098 L at STP: $$\text{moles} = \frac{2.271098}{22.7} \approx 0.1\,mol$$ (3) 4 mol He: Contains $$4N_A$$ atoms. (4) 4 u He: Represents approximately one atom. Thus, the maximum number of helium atoms is present in 4 mol of helium. Hence, option (C) is correct.

7. Question 7 (NEET 2024)

   NEET 2024MediumPercentage Composition & Stoichiometry

   1 gram of sodium hydroxide was treated with 25 mL of 0.75 M HCl solution, the mass of sodium hydroxide left unreacted is equal to:

   1. AOption A: Zero mg
   2. BOption B: 200 mg
   3. COption C: 750 mg
   4. DOption D: 250 mg

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   Correct answer: (D) 250 mg

   Explanation

   Reaction: $$NaOH + HCl \rightarrow NaCl + H_2O$$ Moles of NaOH: $$\frac{1}{40} = 0.025\,mol$$ Moles of HCl: $$0.75 \times 0.025 = 0.01875\,mol$$ Since reaction ratio is 1:1, moles of NaOH reacted: $$0.01875\,mol$$ Remaining NaOH: $$0.025 - 0.01875 = 0.00625\,mol$$ Mass left: $$0.00625 \times 40 = 0.25\,g = 250\,mg$$ Hence, option (D) is correct.

8. Question 8 (NEET 2022)

   NEET 2022MediumDalton's Law of Partial Pressure

   Which one is not the correct mathematical equation for Dalton's Law of partial pressure? Here p = total pressure of gaseous mixture.

   1. AOption A: Option 1
   2. BOption B: Option 2
   3. COption C: Option 3
   4. DOption D: Option 4

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   Correct answer: (A) Option 1

   Explanation

   Dalton's law states: pᵢ = χᵢp where p is total pressure. Expression pᵢ = χᵢpᵢ° corresponds to Raoult's law and is incorrect here. Hence option (A) is correct.

9. Question 9 (NEET 2022)

   NEET 2022MediumIdeal Gas Equation

   A 10.0 L flask contains 64 g of oxygen at 27°C. (Assume O₂ gas is behaving ideally). The pressure inside the flask in bar is: [Given R = 0.0831 L bar K⁻¹ mol⁻¹]

   1. AOption A: 4.9
   2. BOption B: 2.5
   3. COption C: 498.6
   4. DOption D: 49.8

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   Correct answer: (A) 4.9

   Explanation

   Moles of O₂: n = 64/32 = 2 mol Using PV = nRT: P = (2 × 0.0831 × 300)/10 ≈ 4.99 bar ≈ 4.9 bar Hence option (A) is correct.

10. Question 10 (NEET 2021)

    NEET 2021EasyBoyle's Law

    Choose the correct option for graphical representation of Boyle's law, which shows a graph of pressure vs. volume of a gas at different temperatures:

    1. AOption A: Option (1)
    2. BOption B: Option (2)
    3. COption C: Option (3)
    4. DOption D: Option (4)

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    Correct answer: (A) Option (1)

    Explanation

    According to Boyle's law, at constant temperature: P ∝ 1/V Therefore, PV = k where k = nRT. Hence, the graph between pressure and volume is a rectangular hyperbola. Since k = nRT increases with temperature, higher temperature curves lie above lower temperature curves. Therefore, option (A) is correct.