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Some Basic Concepts of Chemistry

Punjab Board · Class 11 · Chemistry

Practice quiz for Some Basic Concepts of Chemistry — Punjab Board Class 11 Chemistry. MCQs and questions with answers to test your preparation.

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Quick Quiz: Some Basic Concepts of Chemistry

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1

50.0 kg of N₂ and 10.0 kg of H₂ are mixed to produce NH₃. Which of the following correctly identifies the limiting reagent and the mass of NH₃ produced?

2

A compound contains 4.07% H, 24.27% C, and 71.65% Cl. Its molar mass is 98.96 g/mol. What is the molecular formula of this compound?

3

The density of a 3 M solution of NaCl is 1.25 g/mL. What is the molality of this solution?

4

Which of the following correctly represents the number of significant figures in the measurement 0.00302 kg?

44 Questions·
multiple choice

Sample Questions

1multiple choice
1 marks

According to Avogadro's Law, 2 volumes of H₂ react with 1 volume of O₂ to give 2 volumes of water vapour. Which statement BEST explains why Dalton's original atomic theory FAILED to predict this result?

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Dalton believed atoms of the same element could not combine with each other, so he rejected diatomic molecules like H₂ and O₂

Step 1 - Recall Dalton's atomic theory: atoms are indivisible and atoms of the same element are identical; compounds form when atoms of DIFFERENT elements combine. Step 2 - The key flaw: Dalton believed same-type atoms CANNOT combine. This means he rejected the existence of H₂ and O₂ as diatomic molecules. Step 3 - Without diatomic molecules, the volume ratios cannot be explained. For example, if O is a single atom, you cannot get 2 water molecules from 1 oxygen. Step 4 - Avogadro later correctly proposed diatomic molecules (H₂, O₂) and his law of equal volumes containing equal molecules expla

2multiple choice
1 marks

A student dissolves 4 g of NaOH in enough water to prepare 250 mL of solution. What is the molarity of this NaOH solution? (Molar mass of NaOH = 40 g/mol)

Show answer

0.4 M

Step 1 - Write the formula: Molarity = moles of solute / volume of solution in litres. Step 2 - Calculate moles of NaOH: moles = mass / molar mass = 4 g / 40 g/mol = 0.1 mol. Step 3 - Convert volume to litres: 250 mL = 250/1000 = 0.25 L. Step 4 - Calculate molarity: M = 0.1 mol / 0.25 L = 0.4 mol/L = 0.4 M. Step 5 - Check: 0.1 M is wrong because it forgets to divide by 0.25 (not 1 L). 1.6 M multiplies instead of divides. 0.04 M divides moles by 2.5 instead of 0.25, a decimal point error.

3multiple choice
1 marks

Which of the following statements about the Law of Multiple Proportions is CORRECTLY illustrated?

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Carbon dioxide (CO₂) and carbon monoxide (CO): masses of oxygen combining with 12g of carbon are 32g and 16g — ratio 2:1

Step 1 - Recall Law of Multiple Proportions (Dalton, 1803): When two elements form more than one compound, masses of one element combining with a fixed mass of the other are in a simple whole number ratio. Step 2 - In CO: 12g C combines with 16g O. In CO₂: 12g C combines with 32g O. Step 3 - Ratio of oxygen masses = 16:32 = 1:2, a simple whole number ratio. This perfectly illustrates the law. Step 4 - Option B describes the Law of Definite Proportions (Proust), not multiple proportions. Step 5 - Option C describes Avogadro's Law. Option D describes Law of Conservation of Mass (Lavoisier). Each

4multiple choice
1 marks

The average atomic mass of chlorine is 35.5 u. Chlorine has two isotopes: ³⁵Cl (mass = 34.97 u) and ³⁷Cl (mass = 36.97 u). What is the approximate percentage abundance of ³⁵Cl?

Show answer

75%

Step 1 - Let the fractional abundance of ³⁵Cl = x. Then abundance of ³⁷Cl = (1 - x). Step 2 - Set up the weighted average equation: 34.97x + 36.97(1 - x) = 35.5. Step 3 - Expand: 34.97x + 36.97 - 36.97x = 35.5. Simplify: -2x = 35.5 - 36.97 = -1.47. Step 4 - Solve: x = 1.47/2 = 0.735 ≈ 0.75. Step 5 - % abundance of ³⁵Cl = 75%. This makes physical sense because the average (35.5) is much closer to 35 than to 37, indicating ³⁵Cl is more abundant. 50% would give average = 35.97, not 35.5. 25% is the abundance of ³⁷Cl.

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Frequently Asked Questions

What are the important topics in Some Basic Concepts of Chemistry for Punjab Board Class 11 Chemistry?
Key topics in Some Basic Concepts of Chemistry include Classification of Matter and Basic Chemistry Concepts, Chapter Overview: Some Basic Concepts of Chemistry, Classification of Matter — Complete Overview. These are the concepts Punjab Board Class 11 examiners draw on most — study them first, then practise related questions.
How to score full marks in Some Basic Concepts of Chemistry — Punjab Board Class 11 Chemistry?
Understand the core concepts first, then work through the 44 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.

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