Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry – Extra Questions with Answers

Extra practice questions for Class 11 Chemistry Chapter 1 (Some Basic Concepts of Chemistry), beyond the textbook. These Class 11 Chemistry Chapter 1 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. What is Avogadro’s number?
Ans: 6.022×10²³.

Q2. Define molarity.
Ans: Number of moles of solute dissolved per litre of solution.

Q3. What is the SI unit of amount of substance?
Ans: Mole (mol).

Q4. State the law of multiple proportions.
Ans: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other bear a simple whole-number ratio.

Q5. What is empirical formula?
Ans: The simplest whole-number ratio of atoms of each element in a compound.

Short Answer Questions (2–3 marks)

Q6. Calculate the number of moles in 22 g of CO₂ (molar mass = 44 g/mol).
Ans: n=mass/molar mass=22/44=0.5 mol.

Q7. Find the molarity of a solution containing 4 g NaOH (molar mass=40 g/mol) dissolved in 500 mL of solution.
Ans: Moles=4/40=0.1 mol. Volume=0.5 L. M=0.1/0.5=0.2 mol/L.

Q8. Calculate the mass percentage of water in a solution containing 20 g of salt in 180 g of water.
Ans: Total mass=200 g. Mass % of water=(180/200)×100=90%.

Higher-Order Thinking / Application Questions

Q9. In the reaction N₂+3H₂→2NH₃, if 2 mol N₂ reacts with 3 mol H₂, identify the limiting reagent and calculate the moles of NH₃ formed, explaining your reasoning.
Ans: From the balanced equation, 1 mol N₂ requires 3 mol H₂. For 2 mol N₂, we would need 6 mol H₂, but only 3 mol H₂ is available. So H₂ is the limiting reagent (it runs out first). Using H₂ as the basis: 3 mol H₂ produces (2/3)×3=2 mol NH₃ (since 3 mol H₂ gives 2 mol NH₃ per the equation’s ratio). So only 2 mol NH₃ is formed, and 2−1=1 mol of N₂ remains unreacted (since only 1 mol N₂ was actually consumed, using 3 mol H₂).

Q10. A compound contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula (atomic masses: C=12, H=1, O=16), showing your working.
Ans: Assume 100 g sample: C=40 g, H=6.7 g, O=53.3 g. Moles: C=40/12=3.33; H=6.7/1=6.7; O=53.3/16=3.33. Divide by smallest (3.33): C=1, H=2.01≈2, O=1. Empirical formula = CH₂O. This matches formaldehyde/its polymers, illustrating how mass percentage data is converted to mole ratios and then simplified to the smallest whole-number ratio to determine empirical formula.

Written by Satish

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