NCERT Class 12 Chemistry Chapter 2 Notes - Download PDF

NCERT Class 12 Chemistry Chapter 2 Notes - Download PDF

Edited By Sumit Saini | Updated on Mar 31, 2024 11:31 PM IST

Solutions is the second chapter of Class 12 NCERT Chemistry. Students will get to know about Concepts like types of solutions, various concentration terms, colligative properties etc . This chapter holds significant weightage in Board exams and competitive exams as well. The Solutions class 12 notes given below provides short description of all the important topics covered in the NCERT Book. Solutions class 12 notes will be useful for quick revision before exam. Class 12 Chemistry chapter 2 notes can be of great help for students to boost their marks in the final examination.

Also see, NCERT Solutions for Class 12

Definition of Solutions:

The Heterogeneous Mixtures of two or more than two compounds are called Solutions.
Ex: Water + Salt , Alloys .

TYPES OF SOLUTION
Binary Solutions :

The Solutions which has only 2 components is known as binary solution. i.e. solute + solvent.

CONCENTRATION OF SOLUTION

Concentration of solution can be defined in the following ways:
1. Molarity (M)
2. Molality (m)
3. Mole fraction
4. Normality (N)
5. Formality
6. W/W = Mass Percentage
7. V/V = Volume Percentage
8. W/V = Gram
9. ppm = Parts per million
10. ppb = Parts per billion

MOLARITY (M)

It is number of moles of solute present in 1 L of solution. Its unit is mol/L.

MOLALITY (m)

It is number of moles of solute present in 1 Kg of solvent. Its unit is mol/Kg.

MOLE
The SI unit of the amount of the substance is mol . It has fixed Number (Avagadro Number).

No. of moles = Mass in gram /Molar mass

MOLE FRACTION
Mole Fraction of Solute = Moles of Solute / Total Moles
Mole Fraction of Solvent = Moles of Solvent / Total moles

SOLUBILITY
It is the maximum amount of a solute that can be dissolve in specific amount of solvent at specific temperature and pressure.
Factors
1. Nature of the Solute
2. Nature of the Solvent
3. Temperature
4 . Pressure

Various Terms Related To Solutions

DISSOLUTION

Dissolution refers to the process in which solute is added to solvent to increase its concentration in solution.

CRYSTALLISATION

Crystallisation is the process in which some solute particles in the solution collide with the solid solute particles and get separated out .

SATURATED SOLUTION

The solution in which no more solute can be dissolve at the same condition of temperature and pressure is known as saturated solution.

UNSATURATED SOLUTION

The solution in which more solute can be dissolve at the same condition of temperature and pressure is known as unsaturated solution.

Solubility of Gas in a Liquid

Solubility of a gas in a liquid is given by henry's Law.

HENRY’S LAW : The solubility of gas in liquid is directly proportional to the partial pressure of gas, present above the surface of the liquid or solution.
P ∝ n
P = KH .x
where ,
P = Partial pressure of gas
KH = Henry’s constant
x = Mole fraction of the solute { gas }

Dependence of KH

KH : KH T
KH ∝ 1/ Solubility
Hence , Solubility ∝ 1/T

  • Hence it can be concluded that O2 gas has more solubility in water in winter season, where temperature is less.
  • The value to KH for different gases at a particular temperature is different which signifies that KH is a function of nature of the gas.
  • KH for the same gas at the different Temperature is different which confirms that KH ∝ T.

APPLICATION OF HENRY’S LAW :

  • Scuba divers under high pressure in the deep sea have high concentration of dissoloved gases while breathing.
  • ANOXIA: Medical condition in which a climber become weak and unable to think clearly is known as Anoxia. At high altitude the partial pressure of O2 is less than the Partial pressure at ground level. This low concentration of O2 in blood and tissues of people living at high altitudes create anoxia.
  • In Soft Drinks : In order to increase the solubility of CO2 gas in soft drinks, the bottles are sealed under high pressure.


IDEAL SOLUTIONS

  • The solutions which obeys the Raoult’s Law over the entire range of concentration are known as Ideal Solution.

Properties of Ideal Solutions :
(i) Δmix H = 0
i.e. The enthalpy of mixing of the pure component to form solution is 0. It is neither exothermic nor endothermic.
(ii) Δmix V = 0
i.e. There is no change in volume when 2 components are mixed to form solution.

NON-IDEAL SOLUTIONS

  • The solutions which do not obey the Raoult’s Law over the entire range of concentration are Non-Ideal Solutions.

Types Of Non-Ideal Solutions :

a) Solutions Showing POSITIVE DEVIATION

  • The Vapour Pressure of such solutions are higher than predicted by Raoult’s Law.
  • Δmix H ≠ 0
  • Δmix V ≠ 0
  • Δmix H is +ve , therefore, it endothermic
  • Example: Mixture of Ethanol and Acetone.

b) Solutions Showing NEGATIVE DEVIATION

  • The Vapour Pressure of such solutions are lower than predicted by Raoult’s Law.
  • Δmix H ≠ 0
  • Δmix V ≠ 0
  • Δmix H is -ve ,therefore, it is exothermic
  • Example: When Phenol and Aniline are mixed

AZEOTROPES

  • Azeotropes are the binary Mixtures having same composition in Liquid & Vapour phrase.
  • They Boil at constant Temperature.
  • They can’t be separated by Fractional distillation.

COLLIGATIVE PROPERTY

  • In a solution, the properties which depend upon the number of solute particles are Known as Colligative Property.
  • Four Colligative Property are there in our sylabus :
    (i) Relative Lowering of Vapour Pressure of Solvent
    (ii) Elevation of Boiling Point of Solvent
    (iii) Depression in the Freezing Point of Solvent
    (iv) Osmosis & Osmotic Pressure of Solution

(i) Relative Lowering of Vapour Pressure

According to the Raoult’s Law , the vapour pressure of the solvent in the solution is less than the pure solvent.
The Lowering of vapour pressure depend upon only on the concentration solute particle & is independent of their of nature.

P1 = P1ox1 {Raoult’s law} ………………… (i)
P2 = P2ox2
The Lowering of vapour pressure,
ΔP = P10 – P1
putting value of P1 from eqn (i)
ΔP = P1o – P1ox1
ΔP = P1o (1-x1)
ΔP = P1ox2
x2 = ΔP /P1o
x2 = P10 – P1 /P1o {Relative Lowering of Vapour Pressure }
n2/(n1 + n2) = P10 – P1 /P1oFor very dilute solution ,n2 is very less than n1 and n2 can be ignored.
.’. P1oP1 / P1o = W2M2/W1M2
where,
W = Mass in gram
M = Molar mass

(ii) Elevation of Boiling Point
The Boiling point of solution is higher than the Boiling point of the pure solvent.
Mathematically ,
ΔTb = Tbo – Tb
where ,
ΔTb = Elevation of Boiling Point
Tbo = Boiling Point of pure solvent
Tb = Boiling Point of solution
also,
ΔTb ∝ m {molality}
ΔTb = Kb m
On simplifying,
ΔTb = Kb n2/W1
ΔTb = Kb W2/M2W1
where,
W2 = mass of solute
M2 = molar mass of solute
W1 = mass of solvent

(iii) Depression in the Freezing Point of Solvent

When a non-volatile solid is added to the solvent, its vapour pressure decreases . As a result, it would become equal to the vapour pressure of the solid solvent at low temperature and this is known as freezing point of the solvent.
Mathematically ,
ΔTf = Tfo – Tf
Tfo= freezing point of the pure solvent.
Tf = freezing point of the pure solvent
For dilute solution ,
ΔTf ∝ m
ΔTf = Kf m
ΔTf = Kf n2/W1
ΔTf = Kf W2 / M2 W1
here, Kf = molal depression constant or cryoscopic constant

(iv) Osmosis & Osmotic Pressure of Solution
The phenomenon in which the solvent molecules flows through the semi permeable membrane from pure solvent to solution is known as Osmosis.
Semi Permeable Membrane :

  • These membranes contains a large network of sub microscopic holes through which small solvent molecules like water can pass but passage of bigger molecules like solute is hindered.

HYPERTONIC SOLUTION

  • When cell is placed in the solution having more than 0.9% sodium Chloride solution then Exo-osmosis take place & the cell shrinks .Such a solution is called Hypertonic Solution.

HYPOTONIC SOLUTION

  • When a cell is placed in the solution less than 0.9% sodium Chloride solution, Endo-osmosis take place & the cell swells. Such a solution is called Hypotonic Solution.

OSMOTIC PRESSURE

The pressure that just stop the flow of the solvent is called Osmotic Pressure .It depends upon the concentration of the solution.

Mathematically,
Π = MRT
= (n2/V) RT
= (W2 / M2) RT
where,
M = Molality
R = Universal gas constant = .083 bar L/mol K or 8.134 J/mol K

Significance of NCERT Class 12 Chemistry Chapter 2 Notes

Solutions notes class 12 will help to revise the student and score good marks in the coming 12 board examination. cbse class 12 chemistry ch 2 notes covers all the important topics of solutions. ch 2 chemistry class 12 notes are also helpful for competitive exams like VITEEE, BITSAT, JEE Main, NEET, etc. chemistry class 12 chapter 2 notes pdf download can be used to prepare in offline mode.

NCERT Class 12 Notes Chapterwise

Also, Check the NCERT Solutions of Questions Given in the Book-

Also, see NCERT Exemplar Solutions -

Subject Wise NCERT Solutions

Frequently Asked Questions (FAQs)

1. What are the important topics of this chapter?
  • Henry’s law  
  • Solubility of gases in liquids 
  • Colligative properties 
  • Raoult's law  
  • Relative lowering of vapour pressure 
  • Elevation of boiling point 
  • Osmotic pressure 
  • Abnormal molecular mass 
  • Van't Hoff factor
2. Where can I find complete solutions of NCERT Class 12 Chemistry?
3. What is the weightage of NCERT class 12 Chemistry chapter 2 in JEE Mains?

4 marks

4. What is the weightage of NCERT Class 12 Chemistry chapter 2 in CBSE board exam ?

This chapter holds weightage of 5 marks

5. What is the weightage of NCERT Class 12 Chemistry chapter 2 in NEET?

The weightage of solutions is 5%

Articles

Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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