Some Basic Concepts of Chemistry NEET PYQs – Chapter-wise MCQs with Answers

Some Basic Concepts of Chemistry NEET Previous Year Questions (PYQs)

Practice NEET previous year questions from Some Basic Concepts of Chemistry. These MCQs cover mole concept, stoichiometry, limiting reagent, concentration terms and calculations with answers and clear explanations.

👉 Read theory:
Some Basic Concepts of Chemistry Chapter Notes


Q1. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with concentrated H2SO4. The evolved gases are passed through KOH pellets. The mass of remaining gas at STP is:

(NEET 2018)

  • (A) 1.4 g
  • (B) 3.0 g
  • (C) 2.8 g
  • (D) 4.4 g
Show Answer

(C)

Show Explanation

Formic acid gives CO and oxalic acid gives CO + CO2. KOH absorbs CO2. Remaining gas is CO whose calculated mass is 2.8 g.


Q2. In which case is the number of water molecules maximum?

(NEET 2018)

  • (A) 18 mL of water
  • (B) 0.18 g of water
  • (C) 10−3 mol of water
  • (D) 0.00224 L of water vapour at STP
Show Answer

(A)

Show Explanation

18 mL of water ≈ 18 g = 1 mole, which has the maximum number of molecules.


Q3. Which of the following depends on temperature?

(NEET 2017)

  • (A) Molality
  • (B) Mole fraction
  • (C) Weight percentage
  • (D) Molarity
Show Answer

(D)

Show Explanation

Molarity depends on volume, which changes with temperature.


Q4. A 20.0 g sample of magnesium carbonate decomposes to give 8.0 g MgO. Percentage purity of MgCO3 is:

(NEET 2015)

  • (A) 75%
  • (B) 96%
  • (C) 60%
  • (D) 84%
Show Answer

(D)

Show Explanation

Theoretical MgCO3 needed to form 8 g MgO is 16.8 g. Purity = (16.8/20)×100 = 84%.


Q5. The number of water molecules is maximum in:

(NEET 2015)

  • (A) 18 molecules of water
  • (B) 1.8 g of water
  • (C) 18 g of water
  • (D) 18 moles of water
Show Answer

(D)

Show Explanation

18 moles contain 18 × 6.022 × 1023 molecules, the maximum.


Q6. Mole fraction of solute in a 1.00 m aqueous solution is:

(NEET 2015)

  • (A) 0.177
  • (B) 1.770
  • (C) 0.0354
  • (D) 0.0177
Show Answer

(D)

Show Explanation

1 m means 1 mole solute in 1000 g water (≈55.5 moles). Mole fraction = 1/(1+55.5).


Q7. If Avogadro number changes from 6.022×1023 to 6.022×1020, it would change:

(NEET 2015)

  • (A) Mass of one mole of carbon
  • (B) Definition of mass in grams
  • (C) Ratio in balanced equation
  • (D) Ratio of elements in a compound
Show Answer

(A)

Show Explanation

Mass of one mole depends directly on Avogadro number.


Q8. What mass of AgCl precipitate is formed when 50 mL of 16.9% AgNO3 is mixed with 50 mL of 5.8% NaCl?

(NEET 2015)

  • (A) 28 g
  • (B) 3.5 g
  • (C) 7 g
  • (D) 14 g
Show Answer

(C)

Show Explanation

AgNO3 is limiting. Moles of AgCl formed = moles of AgNO3 = 0.049. Mass ≈ 7 g.


Q9. Equal masses of H2, O2 and CH4 are taken at same conditions. Volume ratio H2 : O2 : CH4 is:

(NEET 2014)

  • (A) 8 : 16 : 1
  • (B) 16 : 8 : 1
  • (C) 16 : 1 : 2
  • (D) 8 : 1 : 2
Show Answer

(C)

Show Explanation

Volume ∝ moles = mass/molar mass. Ratio = (1/2):(1/32):(1/16) = 16:1:2.


Q10. When 22.4 L H2 reacts with 11.2 L Cl2 at STP, moles of HCl formed are:

(NEET 2014)

  • (A) 1 mol
  • (B) 2 mol
  • (C) 0.5 mol
  • (D) 1.5 mol
Show Answer

(A)

Show Explanation

Cl2 is limiting (0.5 mol). Hence 1 mol of HCl is formed.


Q11. 1.0 g Mg burns with 0.56 g O2. Excess reactant and amount left is:

(NEET 2014)

  • (A) Mg, 0.16 g
  • (B) O2, 0.16 g
  • (C) Mg, 0.44 g
  • (D) O2, 0.28 g
Show Answer

(A)

Show Explanation

Oxygen is limiting. Magnesium left = 0.16 g.


Q12. One mole of carbon weighs 12 g. Number of atoms present is:

(NEET 2020)

  • (A) 1.2×1023
  • (B) 6.022×1022
  • (C) 12×1022
  • (D) 6.022×1023
Show Answer

(D)

Show Explanation

One mole always contains Avogadro number of atoms.


Q13. An organic compound contains 78% carbon and rest hydrogen. Its empirical formula is:

(NEET 2023)

  • (A) CH
  • (B) CH2
  • (C) C2H3
  • (D) C3H8
Show Answer

(B)

Show Explanation

Dividing percentage by atomic mass gives ratio C:H ≈ 1:2.


Q14. What mass of 95% pure CaCO3 is required to neutralise 50 mL of 0.5 M HCl?

(NEET 2022)

  • (A) 1.25 g
  • (B) 1.32 g
  • (C) 3.65 g
  • (D) 9.50 g
Show Answer

(B)

Show Explanation

Moles of HCl = 0.025. Required CaCO3 = 0.0125 mol → 1.25 g. Adjusting purity gives 1.32 g.


👉 More Some Basic Concepts of Chemistry NEET PYQs will be added gradually.

Scroll to Top

🎯 NEET Chemistry PYQ Pack

Chapter-wise PYQs,
explanations & mock tests.

₹149 Only

View Pack