Important Questions Class 12 Chemistry Chapter 2

Important Questions for CBSE Class 12 Chemistry Chapter 2 – Solutions

Important Questions Class 12 Chemistry Chapter 2 helps students prepare well for exams. These notes consist of marks distribution of Important Questions, Henry’s law, Raoult’s Law, Ideal Solutions and other concepts. Chapter 2 Class 12 Chemistry Important Questions prepared by Extramarks subject matter experts will help students to practise questions and better understand the above-mentioned topics.

These questions are created in accordance with the CBSE Syllabus. Class 12 Chemistry Chapter 2 Important Questions are presented with step-by-step solutions. Students can refer to them from the Extramarks website. 

CBSE Class 12 Chemistry Chapter-2 Important Questions

Study Important Questions for Class 12 Chemistry Chapter 2 – Solutions

Below are a few examples of 1 mark, 2 marks, 3 marks and 5 marks questions. Access the link provided to review the full set of questions. 

Very Short Answer Questions
1 Mark

1) 10 g glucose is dissolved in 400 g of solution. Calculate the percentage concentration of the solution.

Ans. The percentage is the ratio of a component to the total solution multiplied by 100. 

Formula to calculate Mass of a component in percentage = Mass of the component in the solutionTotal mass of the solution100

Thus, the percentage concentration of the solution is given as

Percentage of concentration of the solution =10400100=2.5%

2) Calculate the mole fraction of  water in C₂H₅OH a solution containing an equal number of moles of water and C₂H₅OH?

Ans. The proportion of particular moles of a component to the total moles of the solution is  the mole fraction of that component. If we take x moles of water then we will have equal i.e., x moles of  C₂H₅OH. Therefore,

Mole fraction = xx+x=12=0.5

3) AgNO₃ on reaction with NaCl in aqueous solution gives white precipitate. If the two solutions are separated by a semipermeable membrane, will there be the appearance of a white ppt. in the side ‘X’ due to osmosis? 

Ans. There will be no appearance of precipitate when the solutions of AgNO₃ and NaCl are split up by a semipermeable membrane as the motion of only solvent particles takes place  through the membrane and not that of solute particles. 

4) State the unit of ebullioscopic constant (molal boiling point elevation constant.)

Ans:  K kg mol⁻¹ is the unit of the ebullioscopic constant.

Short Answer questions
2 Marks

1) What are azeotropes? Give an example.

Ans: Azeotropes are the binary mixtures of solutions created after some liquids are mixed. These binary mixtures contain similar compositions in liquid and vapour phases at a steady boiling temperature. It is not feasible to segregate the components by fractional distillation. 

  • Minimum Boiling Azeotropes – Solutions that present a huge positive deviation from Raoult’s law at a specific composition create minimum boiling azeotrope. For example, a water and ethanol mixture.
  • Maximum Boiling Azeotropes –  Solutions that present a huge negative deviation from Raoult’s law at a specific composition create maximum boiling azeotrope. For example, water and HCl mixture.

2) Show that the relative lowering of vapour pressure of a solvent is a colligative property.

Ans: Colligative properties are connected to the number of solute particles in the solution. The proportion of the number of solute particles to the number of solvent particles in the solution is a colligative property..

Relative lowering of vapour pressure is presented as;

PP0=P0– PP0

As Raoult’s law states,

P0– PP0=x

In the above formula, x is the mole fraction of any component.

Therefore, the relative lowering of vapour pressure of a solvent can be considered a colligative  property.

Short Answer Questions
3 Marks

1) For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2℃. Assuming the concentration of solute is much lower than the concentration of solvent, determine the vapour pressure (mm of Hg) of the solution. 

(Given: Kbfor water =0.76Kg Mol⁻¹)

Ans: From the above-given information,




P0=1 atm=760mmHg

Kb=0.76Kg Mol⁻¹

M1=18g mol⁻¹

Now, we have elevation in boiling point as:



M2= 9.5g mol⁻¹

Now, for dilute solutions:



Vapour pressure of the solution, 


2) Glycerin, ethylene glycol, and methanol are sold at the same price per kg. Which would be priced the least among all the three solutes mentioned above for making an antifreeze solution for the radiator of an automobile?

Ans: The lowering of the freezing point of water is possible with the help of the antifreeze solution which is needed for the proper functioning of the radiator in an automobile.

According to the formula;


where, m = molality and is stated as; 


Molar masses are given as;

Glycerin = 92 g/mol

Ethylene glycol = 62 g/mol

Methanol = 32 g/mol,

Therefore, methanol will be comparatively cheaper than glycerin and ethylene glycol for making an antifreeze solution for the radiator of an automobile.

Long Answer Type Questions
5 Marks

1) (a) Explain a method of measuring the molar mass of a non-volatile solute through vapour pressure lowering.

Ans: The method of measuring the molar mass of a non–volatile solute from vapour pressure is stated below:

By Raoult’s law;

P⁰-PᣵP⁰= x

P⁰-PᣵP⁰= n₂n₁+n₂

If n₂ << n₁ then,




Now we can evaluate the M₁.

(b) State the amount of urea (mol. mass 60 g/mol) that must be dissolved in 50g of water so as the vapour pressure at the room temperature is reduced by 25%? Evaluate the molality of the solution obtained as well.

Ans. Given that,


   M₁=18g mol⁻¹ 

  M₂=60g mol⁻¹

The relative lowering in vapour pressure is given as

P⁰-PᣵP⁰= W₂M₁M₂W₁

This is reduced to 25% which can be given as;


∴ W₂=41.67g

Molality is given as;



m= 41.6710006050=14m 

Numerical Problems

1) The solubility of oxygen in water is 1.3510⁻³molL⁻¹ at 20℃ and 1 atm pressure. Measure the concentration of oxygen at  20℃  and  0.2 atm pressure.

Ans. Given that,

C=1.35 10⁻³molL⁻¹

P = 1 atm

We know Henry’s law;



KH=CP= 1.3510⁻³M atm⁻¹

Now, at 0.2 atm;


C= 2.710⁻³ M

2) The vapour pressure of pure liquids A and B are 450 and 750 mm Hg respectively, at 350K. Evaluate the composition of the liquid mixture if the total vapour pressure is 600 mm Hg. Compute the composition of the vapour phase as well. 

Ans. Given that,




Now, from Raoult’s law;






PT= PA0xA+PB01-xA


From this, 



This gives us;




Composition in the vapour phase is given as;



3) The osmotic pressure of a 0.0103 molar solution of an electrolyte was found to be 0.75 atm at 27℃. Compute Van’t Hoff factor.

Ans. Given that, 



R=0.0820Latm Mol⁻¹K⁻¹


By the formula of osmotic pressure, we get;


∴ i=𝛑CRT=0.750.01030.082300=2.95

Class 12 Chemistry Solutions Important Questions 

Several important questions, terminologies and concepts are included in Class 12 Chemistry Chapter 2 Important Questions. These questions cover the important concepts of the chapter along with necessary equations and numericals. It also includes other important topics such as the application of Henry’s law, Raoult’s Law,  the concept of molarity, etc. Below are some of the topics covered in this set. 

A solute and a solvent combined together make a solution. Class 12 Chemistry Chapter 2 includes the concept of solutions and its three types:

  • Solid solutions
  • Gaseous solutions
  • Liquid solutions

The molarity, molality, mole fraction and percentages can be depicted by the concentration of a solution. This chapter includes some important concepts like the following:

  • Henry’s law
  • Raoult’s Law
  • Ideal solutions

Henry’s law

Henry’s law states that in a solvent, there is a quantitative connection between pressure and solubility of a gas. According to Henry’s law, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas at a steady temperature usually above the surface of liquid or solution. It is represented in the equation:

                              p =KHx


KH is the Henry’s law constant,

p is the partial pressure of the gas in the vapour phase, and

x is the mole fraction of the gas in the solution.

Raoult’s Law

The solvent’s lowering of the vapour pressure was described in Raoult’s law. 

Raoult’s law states that the relative lowering of the vapour pressure of the solvent above a solution and the mole fraction of a non-volatile solute existing in the solution are equivalent. Mathematically it can be equated as follows:


Ideal Solutions

The solutions that follow Raoult’s law out of the whole range of concentrations are considered to be Ideal solutions. There are two types of deviations under Raoult’s law:

  • Positive deviations
  • Negative deviations

This set of Important Questions contain solutions for each question. Therefore, students can improve the quality of their answers by practising and revising the answers provided. 

Benefits of Class 12 Chemistry Solutions Important Questions

Here are some benefits of utilising Class 12 Chemistry Important Questions during exam preparations:- 

  • These questions follow a step-by-step approach while presenting answers for important concepts and numericals for easier understanding. 
  • The important questions are arranged as per the topic weightage, so students can focus on writing precise answers and manage time efficiently. 
  • Students can practise these important questions and score excellent grades in their Class 12 board exams.  
  • The solutions include key terms and formulas for each question. Students can fetch more marks in the exam by mentioning them. 

CBSE Class 12 Chemistry Important Questions

Class 12 Chemistry Chapter 2 Important Questions are compiled by subject matter experts and are available on the Extramarks website. These questions are made with due diligence according to the CBSE Syllabus, while ensuring in-depth coverage of important concepts.

FAQs (Frequently Asked Questions)

1. What are solvents and solutes?

Solutions are made up of homogeneous mixtures of two or more components. Homogeneous mixture means that its composition and properties are steady throughout the mixture. A solvent is the component that is given in the largest quantity, whereas solutes are one or more components existing in the solution other than the solvent.

2. Why should Class 12 students refer to important questions for Class 12 Chemistry Chapter 2 Solutions?

Class 12 is an important academic year for every student. Therefore, students should keep practising the papers of previous years as well as Important Questions for their board exams. Class 12 Chemistry Chapter 2 Important Questions will be beneficial for CBSE Class 12 students to score excellent marks in their board exams. These questions consist of important topics from Chemistry Chapter 2 which will help students revise the chapter’s key concepts and clear their doubts with the solutions given. Class 12 Chemistry Chapter 2 Important Questions are made in accordance with the CBSE syllabus.