Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
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CHAPTER 7: EQUILIBRIUM
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
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CHAPTER 10: S-BLOCK ELEMENTS
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
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Class 11th Chemistry Online Class For 100% Result
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NCERT TEXTBOOK QUESTIONS SOLVED

Question 1. Calculate the molecular mass of the following:
(i) H20(ii) C02(iii) CH4

Answer:  (i) Molecular mass of H2O = 2(1.008 amu) + 16.00 amu=18.016 amu
(ii) Molecular mass of CO2= 12.01 amu + 2 x 16.00 amu = 44.01 amu
(iii) Molecular mass of CH4= 12.01 amu + 4 (1.008 amu) = 16.042 amu

(ii)

Hence, O2 is the limiting reagent.
∵ 32 g O2 reacts with C to produce 44 g of CO2
∵ 16 g O2 reacts with C to produce \frac { 44 }{ 32 } \times 16=22g\quad of\quad { CO }_{ 2 }

(iii)

∵ 64 g O2 reacts with C to produce 88 g of CO2
∵ 16 g O2 reacts with C to produce \frac { 88 }{ 64 } \times 16 22g\quad of\quad { CO }_{ 2 }

Question 7. How much copper can be obtained from 100 g of copper sulphate (CuSO4 )? (Atomic mass of Cu= 63.5 amu)
Answer:  1 mole of CuS0contains 1 mole (1 g atom) of Cu

Molar mass of CuS04= 63.5 + 32 + 4 x 16 = 159.5 g mol-1

Thus, Cu that can be obtained from 159.5 g of CuS0= 63.5 g

Question 23. In the reaction, A + B2——> AB2, identify the limiting reagent, if any, in the following mixtures
(i) 300 atoms of A + 200 molecules ofB
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules ofB
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Answer:  (i) According to the given reaction, 1 atom of A reacts with 1 molecule of B
.•. 200 molecules of B will react with 200 atoms of A and 100 atoms of A will be
left unreacted. Hence, B is the limiting reagent while A is the excess reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B
.•. 2 mol of A will react with 2 mol of B. Hence, A is the limiting reactant.
(iii) No limiting reagent.
(iv) 2.5 mol of B will react with 2.5 mol of A. Hence, B is the limiting reagent.
(v) 2.5 mol of A will react with 2.5 mol of B. Hence, A is the limiting reagent.

Question 25. How are 0.50 mol Na2C0and 0.50 M Na2C0different?
Answer: Molar mass of Na2C03= 2 x 23 + 12 + 3 x 16 = 106g mol-1 0.50 mol Na2C0means 0.50 x 105 g = 53 g 0. 50 M Na2C03 means 0.50 mol, i.e., 53 g Na2C0are present in 1 litre of the solution.

Question 26. If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour could be produced?
Answer: Hand 02 react according to the equation

2H2(g) + 02 (g) ——>2H2O (g) 

Thus, 2 volumes of H2 react with 1 volume of 02 to produce 2 volumes of water vapour. Hence, 10 volumes of H2 will react completely with 5 volumes of 02 to produce 10 volumes of water vapour.


MORE QUESTIONS SOLVED

I.Very Short Answer Type Questions
Question 1. What is the SI unit of molarity?
Answer: SI unit of molarity = mol dm-3

Question 2. What do you understand by stoichiometric coefficients in a chemical equation?
Answer: The coefficients of reactant and product involved in a chemical equation represented by the balanced form, are known as stoichiometric coefficients.
For example, N2(g) + 3H2(g) ———–> 2 NH3(g)
The stoichiometric coefficients are 1, 3 and 2 respectively.

Question 3. Give an example of a molecule in which the ratio of the molecular formula is six times the empirical formula.
Answer: The compound is glucose. Its molecular formula is C6H12O6, while empirical formula is CH2O.

Question 4. What is an atom according to Dalton’s atomic theory?
Answer: According to Dalton’s atomic theory, an atom is the ultimate particle of matter which cannot be further divided.

Question 5. Why air is not always regarded as homogeneous mixture?
Answer: This is due to the presence of dust particles.

Question 6. Define the term ‘unit’ of measurement.
Answer: It is defined as the standard of reference chosen to measure a physical quantity.

Question 7. Define law of conservation of mass.
Answer: It states that matter can neither be created nor destroyed.

Question 8. How is empirical formula of a compound related to its molecular formula?
Answer: Molecular formula = (Empirical formula)n where n is positive integer.

Question 9. How many oxygen atoms are there in 18 g of water? 
Answer: Molar mass of water is 18 g/mol.
Number of oxygen atoms is 18 g of water = 6.02 x 1023

Question 10. Name two factors that introduce uncertainty into measured figures.
Answer: (i) Reliability of measuring instrument.
(ii) Skill of the person making the measurement.

Question 11. State Avogadro’s law.
Answer: Equal volumes of all gases under the conditions of same temperature and pressure contain the same number of molecules.

Question 12. How are 0.5 ml of NaOH differents from 0.5 M of NaOH?
Answer: 0.5 ml of NaOH means 0.5 mole (20.0 g) of NaOH, 0.5M of NaOH means that 0.5 mole (20.0g) of NaOH are dissolved in 1L of its solution.

Question 13. What is one a.m.u. or one ‘u’?
Answer: 1 a.m.u. or 1 u = 1/12 th mass of an atom of carbon 12.

Question 14. What is the number of significant figures in 1.050 x 104?
Answer: Four.

 

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