Chemistry Class 11 Chapter 1 Notes PDF Download
FAQs on Some Basic Concepts of Chemistry Class 11 Notes: CBSE Chemistry Chapter 1
1. How can I quickly revise the key topics in 'Some Basic Concepts of Chemistry'?
For a quick revision of this chapter, focus on the core logical flow. Start with the fundamental laws of chemical combination, which provide the rules for reactions. Next, understand Dalton's Atomic Theory as the basis for matter. Then, master the mole concept and Avogadro's number, as this is central to all calculations. Finally, apply this knowledge to solve problems related to stoichiometry and determine solution concentrations using Molarity and Molality.
2. What is the best way to summarise the mole concept for quick revision?
Think of a mole as a chemist's counting unit, similar to how a 'dozen' means 12. For revision, remember these three key points:
- A mole of any substance contains exactly 6.022 x 10²³ particles (atoms or molecules). This is Avogadro's number.
- The mass of one mole of a substance in grams is numerically equal to its atomic or molecular mass in amu.
- For any gas at Standard Temperature and Pressure (STP), one mole occupies a volume of 22.4 litres.
3. Why is understanding the mole concept so crucial for the rest of chemistry?
The mole concept is fundamental because it provides a bridge between the macroscopic world (what we can measure, like grams) and the microscopic world of atoms and molecules. It allows chemists to 'count' particles by weighing them, which is essential for stoichiometry. Without it, we could not predict the amounts of reactants needed or products formed in a chemical reaction, making quantitative chemistry impossible.
4. What is the main difference between Molarity and Molality in solution concentration?
The key difference lies in what the solute is dissolved in. Molarity (M) is the number of moles of solute per litre of the solution. Molality (m) is the number of moles of solute per kilogram of the solvent. A crucial point for revision is that Molarity is affected by temperature changes (as volume can expand or contract), while Molality is not, because mass remains constant.
5. How does stoichiometry differ from simply balancing a chemical equation?
Balancing an equation ensures the Law of Conservation of Mass is followed—making sure the number of atoms of each element is the same on both sides. Stoichiometry uses the mole ratios from that balanced equation to perform quantitative calculations. In short, balancing sets the rules, while stoichiometry uses those rules to calculate the specific amounts (like mass, moles, or volume) of reactants and products involved in the reaction.
6. What is a limiting reagent, and how does it affect a chemical reaction?
The limiting reagent (or limiting reactant) is the substance that is completely consumed first in a chemical reaction. It is important because it dictates the maximum amount of product that can be formed. Even if you have a large excess of all other reactants, the reaction will stop once the limiting reagent runs out. The amount of product formed is therefore always determined by the initial amount of the limiting reagent.
7. How did Dalton's Atomic Theory help explain the Law of Definite Proportions?
The Law of Definite Proportions states that a chemical compound always contains its component elements in a fixed ratio by mass. Dalton's Atomic Theory explained this by postulating that compounds are formed when atoms of different elements combine in simple, fixed, whole-number ratios. Since atoms of a specific element have a characteristic mass, their fixed combination ratio naturally results in the compound having a fixed composition by mass.
8. What is the practical difference between an empirical formula and a molecular formula?
The empirical formula represents the simplest whole-number ratio of atoms present in a compound. For example, the empirical formula for both benzene (C₆H₆) and acetylene (C₂H₂) is just CH. The molecular formula, however, shows the actual number of atoms of each element in one molecule of the compound. It provides the true composition, which is essential for understanding the molecule's actual mass and structure.











