NCERT Solutions for class 12 Chemistry Chapter 2

NCERT Solutions Class 12 Chemistry Chapter 2

Adda247 provides NCERT Solutions for Class 12 Chemistry Chapter 2. These NCERT Solutions for class 12 Chemistry Chapter 2 will not only help students to boost their board exams and score brilliant marks but for competitive exams. The Solutions are according to the NCERT guidelines.

Complete 16 chapters wise solutions were provided for the benefit of students.12^th class sets the base for higher education for every student. This makes it the most crucial class for any student who is aiming for his/her dream of quality education. Scoring good marks in the 12^th class is equivalent to quality high education. So, it becomes extremely important for students to give a boost to their class 12 Chemistry preparation with Adda247 NCERT solutions.

Chemistry NCERT Solutions class 12 offers students an in-depth knowledge of different aspects of chemistry.

The students can access the solutions anywhere while browsing the web easily. The solutions are very precise and accurate.

NCERT Solution for Class 12 Chemistry Chapter 2: Solutions PDF

NCERT Solutions for Class 12 Chemistry Chapter 2 is a homogeneous mixture of one or more solutes dissolved in a solvent.

• Solvent: the substance in which a solute dissolves to produce a homogeneous mixture.
• Solute: the substance that dissolves in a solvent to produce a homogeneous mixture.

Many different kinds of solutions exist. For example, a solute can be a gas, a liquid or a solid. Solvents can also be gases, liquids or solids. Chapter 2 of NCERT Chemistry is based on the basic idea of Solutions and its types. With the help of NCERT Solutions for class 12 Chemistry Chapter 2, students will be able to understand types of solutions, expressing the concentration of the solutions, solubility, ideal and non-ideal solutions, colligative properties and abnormal molar mass.

NCERT solution chapter 2 provide greater insight into the topics covered in the chapter.

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NCERT Solutions for Class 12 Chemistry Chapter 2: Key features

• The NCERT solution provides clear and precise answers.
• The columns are used wherever necessary.
• Give step-by-step solutions to questions and problems.
• It gives the students a scope to optimize their marks in the board.

 

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Class 12 NCERT solutions for Chemistry Chapter 2: Important Questions

 

Question:1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Ans.

Solutions are homogeneous mixtures of two or more than two components.  A homogenous mixture we means that its composition and properties are uniform throughout the mixture.

There are nine types of solutions:

Gaseous solutions:

• Gas in gas : mixture of oxygen and nitrogen gases

 

• Liquids in gases : Chloroform mixed with nitrogen gas

 

• Solid in gas : Camphor in nitrogen gas

 

 Liquid solutions:

• Gas in liquid : oxygen dissolved in water

 

• Liquid in liquid : Ethanol dissolved in water
• Solid in liquid : Glucose dissolved in water

 

Solid solutions:

• Gas in solid : solutions of hydrogen in plladium

 

• Liquid in solid : Amalgam of mercury with sodium

 

• Solid in solid : Copper dissolved in gold

 

 

Question:2 Suppose a solid solution is formed between two substances, one whose   particles are very large and the other whose particles are very small. What kind of

solid solution is this likely to be?

Ans –

Solution of hydrogen in palladium and dissolved gases in minerals.

 

Question:3 Define the following terms:

i.)Mole fraction 

ii.)Molality 

iii.)Molarity 

iv.) Mass percentage

Ans –Mole fraction: It is defined as: Mole fraction of a component = Number of moles of the component /Total number of moles of all the components

For example, in a binary mixture, if the number of moles of A and B are nA and nB respectively, the mole fraction of A will be

Xa = nA/nA+nB

For a solution containing I number of components, we have

X =ni/n1+n2+——ni

It can be shown that in a given solution sum of all the mole fractions is unity, 1.e.

x1 + x2 + ——-+ xi= 1

 

Molality:  Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:

Molality (m): Moles of solute/ Mass of solvent in kg

 

Molarity: Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.

Molarity= Moles of solute/Volume of solution in litre

 

Mass percentage (w/w): The mass percentage of a component of a solution is defined as:

Mass % of a component = Mass of the component in the solution × 100 /Total mass of the solution.

Question:4 Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^-1?

Solution:

68 % nitric acid by mass means that 68g mass of nitric acid is dissolved in 100g mass the solution. Molar mass of HNO₃ = 6 mol^-1

:. 68g of HNO₃ = 68/63 = 1.079mole

Density of solution =1.504 g mL^-1 given

: Volume of solution = mass/density =100/1.504=66.5 mL

: Molarity of solution:

Moles of slolute x 1000/Volume of solution in mL

• x 1000/ 65 =16.23M

 

Question:5 A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of ton is 1.2 g mol^-1, then what shall be the molarity of the solution?

Solution 5.

10 percent w/w solution of glucose in water means 10g glucose and 90g of     water.

: 10g of glucose=10/180- 0.0555 moles

And 90g of H20-90/18-5 moles

: Molality of solution

= Moles of solutex1000/Mass of solution in grams

0.0555/90 x 1000=0.617 m

Moles fraction of glucose

= X g = no of moles of glucose/No.of moles of glucose + No.of moles of water

=0.0555/5+0.0555=-0.01

Mole fraction of water:

=X g = No. of males of water/No of modes of glucose + No of moles of water

=5/5+0.0555 = 0.99

Volume of 100g solution

Mass of solution/density= 100/1.2= 83.33mL

: Molarity of solution =0.0555/83.33 x 1000

= 0.67M

 

Question:6 How many mL of 0.1 M HCI are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Solution:

Calculation of no. of moles of components in the mixture.

Let x g of Na₂CO₃ is present in the mixture.

: (1-x)g of NaHCO₃ is present in the mixture.

Molar mass of Na₂CO₃

=2 x 23 + 12 +3 x 16= 84g mol_-1

And molar mass of NaHCO₃

= 23 x 1+1+12+3 x 16 = 84g mol_-1

No. of moles of Na₂CO₃ in x g = x/106

No. of moles of NaHCO₃ in (1-x) g-(1-x)/84

As given that the mixture contains equimolar amounts of Na₂CO₃ and NaHCO₃, therefore

x/106 = 1-x /84

106-106x = 84x

106-190 x

: x = 106/190= 0.558g

No of moles of Na₂CO₃ present

0.558/106 = 0.00526

And no. Of moles of NaHCO₃ present

= 1 – 0.558/84 = 0.00526

Calculation of no. of moles of HCL required

Na₂CO₃ + 2HCI  ->   2NaCI+ H₂O + CO₂

NaHCO₃ + HCI  ->  NaCl + H₂O + CO₂

As can be seen, each mole of Na₂CO₃ needs 2 moles of HCl,

: 0.00526 mole of NaCO₃ needs = 0.00526 x 2 = 0.01052mole

Each mole of NaHCO₃ needs 1 mole of HCI

:  0.00526 mole of NaCHO₃ needs =1× 0.00526 = 0.00526 mole

Total amount of HCI needed will be

= 0.01052 + 0.00526 = 0.01578 mole.

0.1 mole of 0.1 M HCI are present in 1000 mL of HCI

:   0.01578 mole of 0.1 M HCI will be present in

1000/0.1× 0.01578 = 157.8 mL

 

Question:7 A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

Solution:

300 g of 25% solution will contain = 25 × 300 / 100 = 75 g of solute

400g of 40% solution will contain = 40 × 400 / 100 = 160g of solute

:  Total mass of solute=160+ 75= 235g

Total mass of solution =300 + 400 = 700g

Now, the percentage of solute in solution= 235/700 x 100 = 33.5 %

And the percentage of water in solution =100-33.5 = 66.5%

 

Question:8 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C₂H₆O₂) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, then what shall be the molarity of the solution?

Solution:

Mass of solute = 222.6g

Molar mass of solute C₂H₆O₂ = 12 x 2 + 4 + 2 (12+1) = 62 g mol^-1

Moles of solute = 222.6/62= 3.59

Mass of solvent = 200 g

Molality= 3.59/200 x 1000= 17.95 mol kg^-1

Total mass of solution = 422.6g

Volume of solution = 422.6/1.072 = 394.21 mL

Molarity= 3.59 / 394.2 x1000 = 9.1 mol L^-1

 

Question:9 A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass) :

i.) Express this in percent by mass

ii.) Determine the modality of chloroform in the water sample.

Solution:

15 ppm means 15 parts in million (10*6) by mass in the solution

: percentage by mass = 15/10⁶ × 100 = 15×10^-4 %

As only 15g of chloroform is present in 10⁶g of the solution, mass of the solvent = 106g

Molar mass of CHCI₃ -12+1+3 ×35.5 = 119.5 mol^-1

Moles of CHCl₃ = 15 /119.5

: Molality = 15/119.5 × 1000 / 10⁶ = 1.25 × 10^-4 m

 

Question:10 What role does the molecular interaction play in a solution of alcohol and water?

Solution:

Alcohol and water both have strong tendency to form intermolecular hydrogen bonding. On mixing the two, a solution is formed as a result of formation of H-bonds between alcohol and H₂O molecules but these interactions are weaker and less extensive than those in pure H2p. Thus they show a positive deviation from ideal behaviour. As a result of this, the solution of alcohol and water will have higher vapour pressure and lower boiling point than that of water and alcohol.

Related Post:

Class 12 Chemistry Chapter 1: The Solid State

Class 12 Chemistry Chapter 2: Solutions

Class 12 Chemistry Chapter 3: Electrochemistry

Class 12 Chemistry Chapter 4: Chemical Kinetics

Class 12 Chemistry Chapter 5: Surface Chemistry

Class 12 Chemistry Chapter 6: General Principles and Processes of Isolation of Elements

Class 12 Chemistry Chapter 7: The p-Block Elements

Class 12 Chemistry Chapter 8: The d & f Block Elements

Class 12 Chemistry Chapter 9: Coordination Compounds

Class 12 Chemistry Chapter 10: Haloalkanes and Haloarenes

Class 12 Chemistry Chapter 11: Alcohols, Phenols, and Ethers

Class 12 Chemistry Chapter 12: Aldehydes, Ketones, and Carboxylic Acids

Class 12 Chemistry Chapter 13: Amines

Class 12 Chemistry Chapter 14: Biomolecules

Class 12 Chemistry Chapter 15: Polymers

Class 12 Chemistry Chapter 16: Chemistry in Everyday Life

Class 12 Chemistry Chapter 2 Ncert Solutions: FAQs

 

1. What are the topics covered in chapter 2 of NCERT class 12 Chemistry?

Ans. Topics covered in the NCERT solution of class 12 chapter 2 of chemistry are as follows:

1. Solution
2. Types of solutions.
3. Solubility
4. Ideal and Non-ideal solutions.
5. Concentration of solutions.
6. Colligative properties and determination.
7. Vapour pressure of liquid solutions.
8. Abnormal Molar Masses.

Other than these topics class 12 chemistry Chapter 2 also includes the complicated Henry’s laws. Chapter 2 of solutions occupy 5 marks of the board paper. Hence, it is must to study the chapter thoroughly.

The NCERT solution provide question in detail. Hence it is essential to follow the solution in the time of revision.

 

2. Where do I get the PDF of NCERT solutions class 12 Chapter 2?

Ans. NCERT solutions class 12 chapter 2 is available online on many sites. These solutions are available online for easy download. A student can download PDF file of NCERT solutions and save it for future use also.

NCERT solution PDF file (downloaded) has many advantages:

• It can be accessed from anywhere
• Any time
• Even in offline mode

It is very helpful for the students who are doing self-studies, or revision work. It also saves a lot of money. Therefore, the pdf of class 12 Chemistry Solution is commonly in demand.

 

3. How are NCERT solutions chemistry class 12 helpful in exams?

Ans. The NCERT Solutions for class 12 chemistry chapter 2 is very helpful for the students. The expert faculty at Adda247 make use of a simple language, which makes it easy for the students while learning. These solutions provide students with a strong foundation of fundamental concepts which are important from the exam point of view. By using these solutions, students can save a lot of time in searching the correct answer as per CBSE syllabus and guidelines.

 

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