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 Organization | NCERT |
| Name of Class | 12th Class |
| Name of Subject | Chemistry |
| Name of Chapter | Chapter 2 |
| Name of Content | Solutions |
| Name of Category | NCERT Solutions |
| Official site | http://ncert.nic.in/ |
Chapter 2 Solutions covers numerous Questions and answers from all topics and sub-topics which are given below
| Section Name | Topic Name |
| 2 | Solutions |
| 2.1 | Types of Solutions |
| 2.2 | Expressing Concentration of Solutions |
| 2.3 | Solubility |
| 2.4 | Vapour Pressure of Liquid Solutions |
| 2.5 | Ideal and Non-ideal Solutions |
| 2.6 | Colligative Properties and Determination of Molar Mass |
| 2.7 | Abnormal Molar Masses |
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.
