NCERT Solutions for class 12 Chemistry Chapter 2: Solutions

Education Journalist | Study Abroad Strategy Lead

NCERT Solutions for class 12 chemistry chapter 2: Solutions are provided in this article which includes important formulas, chemical equations and numericals. Some of the important topics of Class 12 Chemistry Chapter 2 Solutions include:

Expected No. of questions: 2 to 4 questions of 1 to 5 marks

Download PDF: NCERT Solutions for Class 12 Chemistry Chapter 2 pdf

NCERT Solutions for Class 12 Chemistry Chapter 2

The NCERT solutions for class 12 chemistry chapter 2: Solutions is given below in pdf as well as image form.

NCERT Solutions for Class 12 Chemistry Chapter 2: Solutions – Important Topics

A Solution is a homogeneous mixture of two or more chemically non-reacting substances whose composition can be varied within certain limits. A solution is regarded as a simple phase having more than one component.

Types of Solutions: Based on the states of matter, solutions can be divided into solid solutions, liquid solutions and gaseous solutions.

Solid Solutions: The solvent is in solid state

Liquid Solutions: The solvent is in liquid state

Gaseous Solutions: The solvent is in gaseous state

Solubility: The amount of solute that can dissolve in a solvent to form a solution is called solubility. Solubility depends on factors like temperature, pressure, nature of solvent and solute.

Unsaturated Solution: A solution in which more solute can be added at a given temperature.

Saturated Solution: A solution in which no more solute can be added at a given temperature.

Supersaturated solution: A solution in which more solute than the limited amount is present at a given temperature.

Concentration of Solution: The ratio of the solute to the solvent is known as concentration of a solution

Henry’s Law: Henry’s Law states that the partial pressure (p) of the gas in the vapour phase is proportional to the mole fraction (x) of the gas in the solution. Mathematically,

p = K_H. x

Raoult’s Law: Raoult’s Law states that the Partial vapour pressure of every component in a solution is directly proportional to the mole fraction of that component. Mathematically,

P = P₀X

Also Read:

Related Articles
Colligative properties	Azeotropic Distillation	Solvent Examples
Mass Percent Formula	Melting and Boiling Point	Mole Fraction
Factors affecting Solubility	Freezing Point Depression	Percent by Weight Formula

Check-Out:

PCMB Study Guides
NCERT Solutions for Class 12 Physics	Class 12 Physics Notes	Formulas in Physics
Topics for Comparison in Physics	Choice-based questions in physics	Topics in relation to physics
Physics Study Notes	Class 12 Physics Book PDF	Class 12 Physics Practicals
Important Derivations in Physics	SI units in Physics	Important Physics Constants and Units
Class 12 Physics Syllabus	Mendeleev's Periodic Table	NCERT Solutions for Class 12 Biology
NCERT Solutions for Class 12 English	NCERT Solutions for Class 11 Chemistry	Class 12 Chemistry Practicals
Class 12 Chemistry Notes	Important Chemical Reactions	Class 12 Maths Notes
Class 12 PCMB Notes	NCERT Class 12 Biology Book	NCERT Class 12 Maths Book
NCERT Class 12 Textbooks	NCERT Solutions for Class 11 Maths	NCERT Solutions for Class 11 Physics
NCERT Solutions for Class 12 Maths	NCERT Solutions for Class 11 Biology	NCERT Solutions for Class 11 English
NCERT Class 11 Chemistry Book	NCERT Class 11 Physics Book	NCERT Class 11 Biology Book
Class 11 Notes	Important Maths Formulas	Important Chemistry Formulas
Class 11 PCMB Syllabus	Chemistry Study Notes	Biology Study Notes
Periodic Table in Chemistry	Important Chemical Reactions	Comparison topics in Chemistry
Comparison Topics in Maths	Biology MCQs	Important Named Reactions
Maths MCQs	Comparison Topics in Biology	Chemistry MCQs

CBSE CLASS XII Related Questions

1.
Which of the following is most basic?

$ Mn_2O_7 $
$ MnO_2 $
$ MnO $
$ Mn_2O_3 $

View Solution

2.
Assertion (A) : Actinoids show irregularities in their electronic configurations.
Reason (R) : In actinoids 5f, 6d and 7s orbitals are of comparable energies.

View Solution

3.
Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 $ g mol^{-1} $) in 5 g of $ CS_2 $ in which it dimerises to 88%. $ T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} $.

View Solution

4.
Which of the following reagents are used to prepare primary amines by Hofmann bromamide degradation reaction?

(i), (ii) and (iv)
(i) and (iii)
(i), (ii) and (iii)
(i), (iii) and (iv)

View Solution

5.
State Henry’s law. Calculate the mole fraction of CO$_2$ in water at 298 K under 700 mm Hg pressure. (Given: Henry’s constant for CO$_2$ in water at 298 K = $ 1.25 \times 10^6 $ mm Hg)

View Solution

6.
Consider the following reaction: \[ \text{Zn}_{(s)} + \text{Ag}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2\text{Ag}_{(s)} + 2\text{OH}^-_{(aq)} \] Given: \[ E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \text{ V} \] \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \] \[ 1F = 96500 \text{ C mol}^{-1} \] $\Delta G^\circ$ for above reaction is:

$-301.080 \text{ kJ mol}^{-1}$
$+310.080 \text{ kJ mol}^{-1}$
$-326.070 \text{ kJ mol}^{-1}$
$375.060 \text{ kJ mol}^{-1}$

View Solution

NCERT Solutions for class 12 Chemistry Chapter 2: Solutions

Download PDF: NCERT Solutions for Class 12 Chemistry Chapter 2 pdf

NCERT Solutions for Class 12 Chemistry Chapter 2

NCERT Solutions for Class 12 Chemistry Chapter 2: Solutions – Important Topics

CBSE CLASS XII Related Questions

1.
Which of the following is most basic?

2.
Assertion (A) : Actinoids show irregularities in their electronic configurations.
Reason (R) : In actinoids 5f, 6d and 7s orbitals are of comparable energies.

3.
Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 \( g mol^{-1} \)) in 5 g of \( CS_2 \) in which it dimerises to 88%. \( T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} \).

4.
Which of the following reagents are used to prepare primary amines by Hofmann bromamide degradation reaction?

5.
State Henry’s law. Calculate the mole fraction of CO\(_2\) in water at 298 K under 700 mm Hg pressure. (Given: Henry’s constant for CO\(_2\) in water at 298 K = \( 1.25 \times 10^6 \) mm Hg)

Similar Chemistry Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

NCERT Solutions for class 12 Chemistry Chapter 2: Solutions

Download PDF: NCERT Solutions for Class 12 Chemistry Chapter 2 pdf

NCERT Solutions for Class 12 Chemistry Chapter 2

NCERT Solutions for Class 12 Chemistry Chapter 2: Solutions – Important Topics

CBSE CLASS XII Related Questions

1.Which of the following is most basic?

2.Assertion (A) : Actinoids show irregularities in their electronic configurations. Reason (R) : In actinoids 5f, 6d and 7s orbitals are of comparable energies.

3.Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 \( g mol^{-1} \)) in 5 g of \( CS_2 \) in which it dimerises to 88%. \( T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} \).

4.Which of the following reagents are used to prepare primary amines by Hofmann bromamide degradation reaction?

5.State Henry’s law. Calculate the mole fraction of CO\(_2\) in water at 298 K under 700 mm Hg pressure. (Given: Henry’s constant for CO\(_2\) in water at 298 K = \( 1.25 \times 10^6 \) mm Hg)

Similar Chemistry Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Which of the following is most basic?

2.
Assertion (A) : Actinoids show irregularities in their electronic configurations.
Reason (R) : In actinoids 5f, 6d and 7s orbitals are of comparable energies.

3.
Calculate the boiling point of a solution containing 0.61 g of benzoic acid (122 \( g mol^{-1} \)) in 5 g of \( CS_2 \) in which it dimerises to 88%. \( T_b^\circ(CS_2) = 46.2^\circ C, K_b = 2.3 K kg mol^{-1} \).

4.
Which of the following reagents are used to prepare primary amines by Hofmann bromamide degradation reaction?

5.
State Henry’s law. Calculate the mole fraction of CO\(_2\) in water at 298 K under 700 mm Hg pressure. (Given: Henry’s constant for CO\(_2\) in water at 298 K = \( 1.25 \times 10^6 \) mm Hg)