NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions

NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions provided here is extremely helpful in revising complete syllabus and getting a strong base on it. NCERT 12th Class Chemistry Solutions all questions are solved with detailed explanation available for students. In this article we had given of NCERT solutions for Solutions Class 12 Chemistry step by step solution for each and every question of the chapter. These solutions will also help you with your homework. Best teachers across the India created NCERT solutions for Class 12 Chemistry Chapter 2 Solutions according to curriculum and pattern of syllabus as per guidelines of NCERT (CBSE) Books.

12th Class NCERT Chemistry Chapter 2 Solutions

Name of OrganizationNCERT
Name of Class12th Class
Name of SubjectChemistry
Name of ChapterChapter 2
Name of ContentSolutions
Name of CategoryNCERT Solutions
Official sitehttp://ncert.nic.in/

Chapter 2 Solutions covers numerous Questions and answers from all topics and sub-topics which are given below

Section NameTopic Name
2Solutions
2.1Types of Solutions
2.2Expressing Concentration of Solutions
2.3Solubility
2.4Vapour Pressure of Liquid Solutions
2.5Ideal and Non-ideal Solutions
2.6Colligative Properties and Determination of Molar Mass
2.7Abnormal Molar Masses

NCERT Solutions for Class 12th Chemistry
NCERT Solutions for Class 12th Chemistry
NCERT Solutions for Class 12th Chemistry
Question 2.23:
Suggest the most important type of intermolecular attractive interaction in the following pairs.
(i) n-hexane and n-octane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).
Answer
(i) Van der Wall’s forces of attraction.
(ii) Van der Wall’s forces of attraction.
(iii) Ion-diople interaction.
(iv) Dipole-dipole interaction.
(v) Dipole-dipole interaction.

Question 2.24:
Based on solute-solvent interactions, arrange the following in order of increasing
solubility in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer
n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than that of a polar solute in the n-octane.
The order of increasing polarity is:
Cyclohexane < CH3CN < CH3OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH3OH < CH3CN < Cyclohexane

Question 2.25:
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer
(i) Phenol (C6H5OH) has the polar group −OH and non-polar group −C6H5. Thus, phenol is partially soluble in water.
(ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group −OH and can form H-bond with water.
Thus, formic acid is highly soluble in water.
(iv) Ethylene glycol has polar −OH group and can form H−bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky nonpolar −C5H11 group. Thus, pentanol is partially soluble in water.
NCERT Solutions for Class 12th Chemistry
NCERT Solutions for Class 12th Chemistry
NCERT Solutions for Class 12th Chemistry

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