Free NEET Chemistry practice
Some Basic Concepts of Chemistry: MCQs with Solutions
Convert the information in a question into moles before using a balanced chemical equation. These original questions separate mass, number of particles, and concentration so that similar-looking quantities do not get used interchangeably.
5 original practice questions — not previous-year paper reproductions. No account required.
By NEET Ninja · Updated
Before you answer
- Amount in moles is mass divided by molar mass. Multiply moles by the Avogadro constant to obtain the number of specified particles.
- Coefficients in a balanced equation give mole ratios, not mass ratios. Find the limiting reactant by comparing the available moles with the required ratio.
- Molarity is moles of solute divided by the final solution volume in litres. Solvent volume alone is not the definition.
Questions and worked answers
Choose an option before reading the answer. For a wrong answer, identify the step or concept that changed the result.
QUESTION 1
How many moles are present in 9.0 g of water? Use M(H₂O) = 18.0 g mol⁻¹.
- 0.25 mol
- 0.50 mol
- 1.0 mol
- 2.0 mol
Answer: B — 0.50 mol
Amount of substance is n = m/M = 9.0 g / 18.0 g mol⁻¹ = 0.50 mol. The gram units cancel, leaving moles. The number 18 is the molar mass of water, not the mass of a single water molecule in grams.
QUESTION 2
Using Nₐ = 6.022 × 10²³ mol⁻¹, how many molecules are in 0.25 mol of CO₂?
- 1.5055 × 10²³
- 6.022 × 10²³
- 2.4088 × 10²⁴
- 0.25 × 10⁻²³
Answer: A — 1.5055 × 10²³
The number of molecules is nNₐ = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³. This counts CO₂ molecules. Counting all atoms would require multiplying this result by three because each CO₂ molecule contains one carbon atom and two oxygen atoms.
QUESTION 3
For 2H₂ + O₂ → 2H₂O, what is the maximum amount of water formed from 3 mol H₂ and 2 mol O₂?
- 1 mol
- 2 mol
- 3 mol
- 4 mol
Answer: C — 3 mol
Three moles of H₂ require 1.5 mol of O₂, which is less than the 2 mol available. Hydrogen is therefore limiting. The H₂:H₂O mole ratio is 2:2, so 3 mol of H₂ form 3 mol of water. Half a mole of oxygen remains unreacted.
QUESTION 4
A solution contains 0.20 mol NaCl in a final volume of 500 mL. What is its molarity?
- 0.10 mol L⁻¹
- 0.20 mol L⁻¹
- 0.40 mol L⁻¹
- 4.0 mol L⁻¹
Answer: C — 0.40 mol L⁻¹
First convert the final solution volume: 500 mL = 0.500 L. Molarity is n/V = 0.20/0.500 = 0.40 mol L⁻¹. Using the value 500 directly would leave the volume in millilitres and introduce a factor-of-1000 error.
QUESTION 5
What is the mass percentage of carbon in CO₂? Use atomic masses C = 12 and O = 16.
- 12.0%
- 27.3%
- 50.0%
- 72.7%
Answer: B — 27.3%
The molar mass of CO₂ is 12 + 2 × 16 = 44 g mol⁻¹. Carbon contributes 12 g per mole of CO₂, so its mass percentage is (12/44) × 100 ≈ 27.3%. The remaining approximately 72.7% is oxygen; equal atom counts would not imply equal mass percentages.
Study source and corrections
Use the matching chapter in NCERT Class XI Chemistry, Part I — Some Basic Concepts of Chemistry for the underlying concepts. These questions are written for NEET Ninja, not endorsed or supplied by NCERT or NTA.
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