Class 12 Chemistry Chapter 2 Solutions Detailed Step By Step NCERT Examples, Intext Questions, Exercises Answers

Chemistry / Unit 2 / SOLUTIONS / Detailed Answers

Through this post, we have covered every NCERT question including all examples, intext questions, exercise questions for Chapter Solutions from Class 12 Chemistry 2021 NCERT Textbook in a very simplified, precise, accurate and concise manner which students can easily understand and remember their approach to the question. The post aims to save students time spent on finding reliable and trustworthy answers to NCERT questions. 

Topics To Be Learned in the Unit ~ Chapter : 

Serial Number	Topic	Weightage
1	What are solutions	Low
2	Solvent & Solute	High
3	Binary Solutions	Low
4	Types of Solutions	Low
5	Concentration of Solutions and Concentration Terms	Very High
6	Solubility	High
7	Factors on which Solubility depends	High
8	Solubility of Solids in Liquids	Low
9	Solubility of Gases in Liquids	High
10	Effect of Pressure and Temperature on Solubility	High
11	Henry's Law	Very High
12	Vapor Pressure	Very High
13	Raoult's Law	Very High
14	Dalton's Law of Partial Pressures	High
15	Finding Total Pressure with graph plot	High
16	Raoult's Law: Special Case of Henry's Law	Low
17	Ideal and Non-Ideal Solutions	High
18	Positive and Negative Deviations	High
19	Azeotropes	High
20	Colligative Properties	Very High
21	Isotonic, Hypotonic, Hypertonic Solutions	Low
22	Abnormal Molar Mass	Low
23	Vant Hoff Factor	High

Points to Ponder & Formula List

To get all the formulae list sheets and detailed notes of this chapter click on this    and reach to the webpage containing the required information.

NCERT Examples step by step solutions 

Example Question 1 : Calculate the mole fraction of ethylene glycol (C2H6O2) in a solution containing 20% of C2H6O2 by mass.

Answer :   

◉ Assume we have a 100 g solution in total.

                 => 20 g = Ethylene glycol

                 => 80 g = Water

◉ Molar mass of C2H6O2 = 2(12) + 6(1) + 2(16) = 62 g/mol

◉ Molar mass of H2O = 2(1) + 1(16) = 18 g/mol 

◉ Number of moles of C2H6O2 = ' n_c2h6o2 ' 

                => n_c2h6o2 = ( [ 20g ] / [ 62 g/mol ] ) = 0.322 mol 

◉ Number of moles of H2O = ' n_H2O ' 

               => n_H2O = ( [ 80g ] / [ 18 g/mol ] ) = 4.444 mol 

◉ Total Moles = ( n_c2h6o2 ) + ( n_H2O ) = n_Total 

                    = 4.444 + 0.322 = 4.76 moles 

◉ Mole fraction of Ethylene Glycol = X_C2H6O2 = ( [ n_c2h6o2 ] / [ n_Total ] ) = ( [ 0.322 moles ] / [ 4.76 moles ] )  = 0.068 

◉ Mole Fraction of Water = X_H2O 

∵ It is a binary solution , So X_H2O + X_C2H6O2 = 1 

Hence, X_H2O = ( [ 1 - X_C2H6O2 ] ) = ( 1 - 0.068 ) = 0.932 

➤ So the Mole Fraction of Ethylene Glycol is 0.068 and the Mole Fraction of Water is 0.932 .