Course Content
CHAPTER 10: HALOALKANES AND HALOARENES
Topic Name 10 Haloalkanes and Haloarenes 10.1 Classification 10.2 Nomenclature 10.3 Nature of C–X Bond 10.4 Methods of Preparation of Haloalkanes 10.5 Preparation of Haloarenes 10.6 Physical Properties 10.7 Chemical Reactions
0/4
CHAPTER 11: ALCOHOLS, PHENOLS AND ETHERS
Topic Name 11 Alcohols, Phenols and Ethers 11.1 Classification 11.2 Nomenclature 11.3 Structures of Functional Groups 11.4 Alcohols and Phenols 11.5 Some Commercially Important Alcohols 11.6 Ethers
0/3
CHAPTER 13: AMINES
Topic Name 13 Amines 13.1 Structure of Amines 13.2 Classification 13.3 Nomenclature 13.4 Preparation of Amines 13.5 Physical Properties 13.6 Chemical Reactions 13.7 Method of Preparation of Diazonium Salts 13.8 Physical Properties 13.9 Chemical Reactions 13.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds
0/3
CHAPTER 14: BIOMOLECULES
Topic Name 14 Biomolecules 14.1 Carbohydrates 14.2 Proteins 14.3 Enzymes 14.4 Vitamins 14.5 Nucleic Acids 14.6 Hormones
0/3
CHAPTER 15: POLYMERS
Topic Name 15 Polymers 15.1 Classification of Polymers 15.2 Types of Polymerisation Reactions 15.3 Molecular Mass of Polymers 15.4 Biodegradable Polymers 15.5 Polymers of Commercial Importance
0/3
CHAPTER 16: CHEMISTRY IN EVERYDAY LIFE
Topic Name 16 Chemistry in Everyday Life 16.1 Drugs and their Classification 16.2 Drug-Target Interaction 16.3 Therapeutic Action of Different Classes of Drugs 16.4 Chemicals in Food 16.5 Cleansing Agents
0/3
Class 12th Chemistry Online Class: Excelling in CBSE Board Exams
About Lesson

Q. What is main advantage of molality over molarity? [Delhi 2014C/2011/2010]

Ans. Molality does not change with in temperature while, molarity decrease with rise in temperature. This is because volume depends on temperature ( V proportional T ) but mass does not.

NCERT BOOK Question

2.4. Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH2CONH2) = 60 g mol-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

2.3 Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Sol: (i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute

Some Extra Question For Practice:-

Wisdom TechSavvy Academy