CUET Chemistry Class 12 Chapter 1 NCERT Solutions, Download PDF in Hindi and English

Chapter 1, Solutions, introduces important concepts related to mixtures, concentration of solutions, solubility, vapour pressure, and colligative properties. These concepts are important for understanding numerical and conceptual questions in Class 12 Chemistry.

CUET Chemistry Class 12 Chapter 1 NCERT Solutions provides concise answers to the NCERT exercise questions from the chapter. Students can use the solutions after attempting the questions on their own and revise the formulas and concepts involved. Hindi and English PDF resources are also available for convenient revision.

CUET Chemistry Class 12 Chapter 1 NCERT Solutions PDF

The downloadable PDFs provide the complete NCERT exercise coverage for Chapter 1. Students can use them to revise definitions, formulas, numerical methods and important concepts from Solutions.

The PDFs include:

• Complete NCERT exercise questions and answers

• Concise explanations for numerical and conceptual questions

• Important formulas used in the chapter

• Separate Hindi and English versions

• A useful resource for CUET 2027 Chemistry revision

Students should first study the chapter and attempt the NCERT questions themselves. The CUET Chemistry Class 12 Chapter 1 NCERT Solutions PDF can then be used to compare answers and revise difficult concepts.

The English PDF is useful for students preparing through English-medium study material. It contains the complete Class 12 Chemistry Chapter 1 NCERT Solutions for the NCERT exercise.

The Hindi PDF can be used by students who prefer to understand and revise Chemistry concepts in Hindi. It covers the chapter exercise solutions in Hindi.

CUET Chemistry Class 12 Chapter 1 NCERT Solutions

CUET Chemistry Chapter 1 NCERT Solutions are covered with concise explanations and final answers. Numerical questions include the main formula or calculation step needed to understand the answer.

NCERT Question 1

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

Answer: A solution is a homogeneous mixture of two or more components. The component present in a smaller amount is generally called the solute, while the component present in a larger amount is the solvent.

Solutions can be classified according to the physical states of the solute and solvent. Examples include:

• Gas in gas: Air

• Gas in liquid: Carbon dioxide in water

• Gas in solid: Hydrogen in palladium

• Liquid in gas: Chloroform vapour in nitrogen

• Liquid in liquid: Ethanol in water

• Liquid in solid: Mercury in sodium amalgam

• Solid in gas: Camphor in air

• Solid in liquid: Sugar in water

• Solid in solid: Copper in gold

NCERT Question 2

What type of deviation from Raoult's Law is observed in a mixture of ethanol and water?

(A) Negative deviation
(B) Positive deviation
(C) No deviation
(D) None of the above

Answer: (B) Positive deviation

Explanation: Ethanol-water mixtures show positive deviation from Raoult's law because the interactions between ethanol and water molecules are weaker than the interactions present in the pure components. As a result, the vapour pressure becomes higher than the value predicted by Raoult's law.

NCERT Question 3

Why do gases always tend to be less soluble in liquids as the temperature is raised?

(A) Because the dissolution of a gas in a liquid is an endothermic process
(B) Because the dissolution of a gas in a liquid is an exothermic process
(C) Because the kinetic energy of gas molecules decreases at higher temperatures
(D) None of the above

Answer: (B) Because the dissolution of a gas in a liquid is an exothermic process

Explanation: When temperature increases, gas molecules gain kinetic energy and tend to escape from the liquid. Since dissolution of gases is generally exothermic, increasing temperature reduces their solubility.

NCERT Question 4

State Henry's law and mention some important applications.

(A) Solubility of a gas in a liquid is directly proportional to its pressure; used in carbonated drinks.
(B) Vapour pressure depends on mole fraction; used in distillation.
(C) Total pressure equals sum of partial pressures; used in gas mixtures.
(D) Rate of diffusion inversely proportional to molar mass; used in effusion.

Answer: (A)

Explanation: Henry's law states that the solubility of a gas in a liquid is directly related to the pressure of that gas above the solution at a constant temperature.

It is applied in the preparation of carbonated beverages and in understanding the behaviour of gases dissolved in liquids.

NCERT Question 5

The partial pressure of ethane over a solution changes when the amount of dissolved ethane is increased. Calculate the new partial pressure using Henry's law.

(A) 6.56 bar
(B) 7.62 bar
(C) 5.00 bar
(D) 10.0 bar

Answer: (B) 7.62 bar

Explanation: At constant temperature, Henry's law establishes a direct relationship between the pressure of the gas and its amount dissolved in the solution. Using the given quantities, the calculated partial pressure is 7.62 bar.

NCERT Question 6

What is meant by positive and negative deviations from Raoult's law and how is the sign of Δmix H related to positive and negative deviations from Raoult's law?

Answer: A solution shows positive deviation when its vapour pressure is higher than the value predicted by Raoult's law. This happens when solute-solvent attractions are weaker than the original interactions. Such mixing is associated with positive Δmix H.

A solution shows negative deviation when its vapour pressure is lower than the expected value. Here, solute-solvent attractions are stronger, and mixing is generally associated with negative Δmix H.

NCERT Question 7

An aqueous solution of a non-volatile solute exerts a vapour pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

(A) 18 g mol⁻¹
(B) 36 g mol⁻¹
(C) 41.4 g mol⁻¹
(D) 58 g mol⁻¹

Answer: (C) 41.4 g mol⁻¹

Explanation: The relative lowering of vapour pressure is used to determine the molar mass of a non-volatile solute. Substituting the given vapour pressures and concentration data gives a molar mass of approximately 41.4 g mol⁻¹.

NCERT Question 8

Heptane and octane form an ideal solution. At the given temperature, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa, respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

(A) 65.2 kPa
(B) 73.6 kPa
(C) 76.4 kPa
(D) 81.0 kPa

Answer: (B) 73.6 kPa

Explanation: Since the mixture is ideal, Raoult's law is applied.

First, calculate the mole fraction of each component. The total vapour pressure is then obtained using:

Ptotal = XA PA° + XB PB°

The resulting vapour pressure is approximately 73.6 kPa.

NCERT Question 9

The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of a 1 molal solution of a non-volatile solute in it.

(A) 11.3 kPa
(B) 12.30 kPa
(C) 10.8 kPa
(D) 12.08 kPa

Answer: (D) 12.08 kPa

Explanation: A 1 molal solution contains 1 mole of solute in 1 kg of water. The mole fraction of water is calculated and Raoult's law is applied to determine the reduced vapour pressure.

The vapour pressure is approximately 12.08 kPa.

NCERT Question 10

Calculate the mass of a non-volatile solute of molar mass 40 g mol⁻¹ which should be dissolved in octane to reduce its vapour pressure to 80%.

(A) 6 g
(B) 10 g
(C) 12 g
(D) 20 g

Answer: (B) 10 g

Explanation: The reduction in vapour pressure is related to the mole fraction of the solute. Applying Raoult's law to the given data gives the required mass of solute as 10 g.

NCERT Question 11

A solution contains 30 g of a non-volatile solute dissolved in 90 g of water. The vapour pressure of this solution is 2.8 kPa at 298 K. When 18 g of water is added to the same solution, its vapour pressure becomes 2.9 kPa at the same temperature. Assuming ideal behaviour, the molar mass of the solute and the vapour pressure of pure water at 298 K, respectively, are:

(A) 23 g mol⁻¹, 3.53 kPa
(B) 30 g mol⁻¹, 3.36 kPa
(C) 23 g mol⁻¹, 3.36 kPa
(D) 45 g mol⁻¹, 3.60 kPa

Answer: (A) 23 g mol⁻¹, 3.53 kPa

Explanation: Raoult's law is applied to the original solution and the solution obtained after adding water. Solving the two relationships gives the molar mass of the solute as 23 g mol⁻¹ and the vapour pressure of pure water as 3.53 kPa.

NCERT Question 12

Give an example of a solid solution in which the solute is a gas.

(A) Hydrogen gas dissolved in palladium
(B) Sugar dissolved in water
(C) Copper dissolved in zinc
(D) Oxygen dissolved in water

Answer: (A) Hydrogen gas dissolved in palladium

Explanation: Hydrogen is a gas and palladium is a solid. Therefore, hydrogen dissolved in palladium represents a solid solution in which the solute is a gas.

NCERT Question 13

A 5% solution by mass of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of glucose in water if the freezing point of pure water is 273.15 K.

(A) 269.07 K
(B) 273.15 K
(C) 271.00 K
(D) 267.85 K

Answer: (A) 269.07 K

Explanation: Freezing-point depression depends on the molality of the solution and the molar mass of the solute. For the same mass percentage, glucose produces a greater number of solute particles than cane sugar because glucose has a lower molar mass.

The calculated freezing point is approximately 269.07 K.

NCERT Question 14

Two elements A and B form compounds having formulae AB₂ and AB₄. When dissolved in benzene, the given masses of AB₂ and AB₄ produce the stated freezing-point depressions. Calculate the atomic masses of A and B.

(A) A = 25.59, B = 42.64
(B) A = 42.64, B = 25.59
(C) A = 25.59, B = 85.28
(D) A = 51.18, B = 42.64

Answer: (A) A = 25.59, B = 42.64

Explanation: The depression in freezing point is used to calculate the molar masses of AB₂ and AB₄. The two resulting equations can then be solved to obtain the individual atomic masses.

Therefore:

A = 25.59
B = 42.64

NCERT Question 15

At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bar at the same temperature, what would be its concentration?

(A) 0.200 M
(B) 0.061 M
(C) 0.122 M
(D) 0.030 M

Answer: (B) 0.061 M

Explanation: At constant temperature, osmotic pressure is directly proportional to the concentration of the solution.

Using:

π = CRT

and comparing the two solutions gives a concentration of approximately 0.061 M.

NCERT Question 16

How many of the following pairs have dipole-dipole interaction?

(i) n-hexane and n-octane
(ii) I₂ and CCl₄
(iii) NaClO₄ and water
(iv) Methanol and acetone
(v) Acetonitrile and acetone

(A) 3
(B) 2
(C) 4
(D) 1

Answer: (B) 2

Explanation: Methanol-acetone and acetonitrile-acetone involve polar molecules and can show dipole-dipole interactions.

n-Hexane and n-octane are non-polar, while I₂ and CCl₄ mainly show dispersion interactions. NaClO₄ and water involve ion-dipole interactions.

Therefore, 2 pairs show dipole-dipole interaction.

NCERT Question 17

Based on solute-solvent interactions, arrange the following in increasing order of solubility in n-octane: KCl, CH₃OH, CH₃CN and cyclohexane.

(A) KCl < CH₃OH < CH₃CN < Cyclohexane
(B) KCl < CH₃CN < CH₃OH < Cyclohexane
(C) CH₃OH < KCl < CH₃CN < Cyclohexane
(D) Cyclohexane < CH₃CN < CH₃OH < KCl

Answer: (A) KCl < CH₃OH < CH₃CN < Cyclohexane

Explanation: n-Octane is non-polar. Substances with similar intermolecular characteristics tend to dissolve more readily in it. Cyclohexane, being non-polar, has the highest solubility among the given substances.

NCERT Question 18

Among the following compounds, identify which is insoluble in water.

(A) Phenol
(B) Toluene
(C) Formic acid
(D) Ethylene glycol

Answer: (B) Toluene

Explanation: Toluene is a non-polar hydrocarbon and does not interact sufficiently with polar water molecules. Hence, it is considered insoluble in water compared with the other options.

NCERT Question 19

If the density of some lake water is given and it contains the specified amount of Na⁺ per kg of water, calculate the molarity of Na⁺ in the lake.

(A) 2.5 M
(B) 4.0 M
(C) 5.0 M
(D) 6.25 M

Answer: (B) 4.0 M

Explanation: The mass of Na⁺ is first converted into moles. The volume of the solution is then calculated using its density. Molarity is obtained by dividing the moles of Na⁺ by the volume of the solution in litres.

NCERT Question 20

If the solubility product of the given salt is 6 × 10⁻¹⁶, what is the maximum molarity of the salt in aqueous solution?

(A) 6 × 10⁻¹⁰ M
(B) 2.45 × 10⁻⁸ M
(C) 7.75 × 10⁻⁸ M
(D) 1.0 × 10⁻⁸ M

Answer: (B) 2.45 × 10⁻⁸ M

Explanation: For a salt that produces equal concentrations of its ions:

Ksp = s²

Therefore:

s = √Ksp

For Ksp = 6 × 10⁻¹⁶,

s ≈ 2.45 × 10⁻⁸ M.

NCERT Question 21

A solution is prepared by dissolving 6.5 g of aspirin in 450 g of acetonitrile. The mass percentage of aspirin in the solution is:

(A) 1.42%
(B) 1.46%
(C) 1.50%
(D) 1.60%

Answer: (A) 1.42%

Explanation:

Mass of solution = 6.5 + 450 = 456.5 g.

Mass percentage is calculated as:

Mass percentage = (Mass of solute / Mass of solution) × 100

Therefore, the mass percentage of aspirin is approximately 1.42%.

NCERT Question 22

Nalorphene (C₁₉H₂₁NO₃), a drug similar to morphine, is used to combat withdrawal symptoms in narcotic users. If the usual dose of nalorphene is 1.5 mg, then the mass of a 1.5 × 10⁻³ M aqueous solution required to administer this dose is:

(A) 1.6 g
(B) 2.1 g
(C) 3.2 g
(D) 4.8 g

Answer: (C) 3.2 g

Explanation: The given dose is first converted into moles using the molar mass of nalorphene. The volume or mass of solution required is then obtained using the given molarity.

The required mass of solution is approximately 3.2 g.

NCERT Question 23

Define the following terms:

(i) Mole fraction

Mole fraction is the ratio of the number of moles of one component to the total number of moles present in the solution.

(ii) Molality

Molality is the number of moles of solute present in one kilogram of solvent.

(iii) Molarity

Molarity is the number of moles of solute present in one litre of solution.

(iv) Mass percentage

Mass percentage is the mass of solute divided by the total mass of solution, multiplied by 100.

NCERT Question 24

Calculate the amount of benzoic acid (C₆H₅COOH) required to prepare 250 mL of a 0.15 M solution in methanol.

(A) 4.575 g
(B) 4.113 g
(C) 6.204 g
(D) 3.213 g

Answer: (A) 4.575 g

Explanation:

Moles required:

n = M × V

= 0.15 × 0.250
= 0.0375 mol

Molar mass of benzoic acid = 122 g mol⁻¹.

Mass required:

Mass = 0.0375 × 122 = 4.575 g

Therefore, 4.575 g of benzoic acid is required.

NCERT Question 25

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Answer: The three acids have different tendencies to ionise in water. Trichloroacetic acid and trifluoroacetic acid are stronger acids than acetic acid because chlorine and fluorine atoms exert an electron-withdrawing effect.

Greater ionisation produces more solute particles in the solution. Since freezing-point depression depends on the number of solute particles, the depression increases as the extent of ionisation increases.

Therefore, the order is:

Acetic acid < Trichloroacetic acid < Trifluoroacetic acid

Important Topics in Solutions for CUET Chemistry

For CUET preparation, revise the following concepts carefully:

• Types of solutions

• Mole fraction, molarity and molality

• Mass percentage and ppm

• Solubility of gases in liquids

• Henry’s law and its applications

• Vapour pressure of liquid solutions

• Raoult’s law

• Positive and negative deviations

• Ideal and non-ideal solutions

• Colligative properties

• Relative lowering of vapour pressure

• Elevation in boiling point

• Depression in freezing point

• Determination of molar mass

• Osmotic pressure

• Van’t Hoff factor

• Abnormal molar masses

• Association and dissociation of solutes

Also Check: CUET 2027 All-in-One Guide to Exam

Important Questions from Class 12 Chemistry Chapter 1 for CUET 2027

Practise the following CUET Chemistry Chapter 1 Questions and Answers after completing the NCERT exercise:

1. Differentiate between molarity and molality.

2. How is mole fraction calculated for a binary solution?

3. State Henry’s law and give two applications.

4. Explain positive and negative deviations from Raoult’s law.

5. What are colligative properties?

6. Why does the boiling point increase when a non-volatile solute is added?

7. Why does the freezing point decrease after adding a solute?

8. Explain the significance of the van’t Hoff factor.

9. Solve numerical questions based on osmotic pressure.

10. Calculate molar mass using depression in freezing point.

11. Explain the difference between ideal and non-ideal solutions.

12. Calculate vapour pressure using Raoult’s law.

13. Explain why gases become less soluble in liquids with increasing temperature.

14. Solve questions involving abnormal molar mass.

15. Differentiate between positive and negative deviations using molecular interactions.

These Solutions Questions and Answers cover the major numerical and conceptual areas that students should revise from the NCERT chapter.

How to Use NCERT Solutions for CUET Chemistry Preparation 2027?

1. Read the Solutions chapter from NCERT: Start with the textbook and understand concentration terms, solubility, vapour pressure and colligative properties before attempting the exercise.

2. Attempt the exercise questions yourself: Solve the numerical and conceptual questions without checking the answer first. Write down the formula and identify the information given in each numerical.

3. Use the solutions to check your answers: Compare your method, formula and final answer with the solution. For numerical questions, check units and conversions carefully.

4. Revise important formulas and concepts: Keep revising formulas related to molarity, molality, mole fraction, Raoult’s law, Henry’s law, freezing-point depression, boiling-point elevation and osmotic pressure.

5. Practise NCERT-based CUET questions: After completing the exercise, practise objective questions based on definitions, formulas, concepts and numerical applications from the chapter.

6. Download the PDF for quick revision: Keep the Solutions NCERT PDF available for revision before practice sessions and CUET preparation.

Also Check:

• CUET Biology Class 12 Chapter 2 NCERT Solutions

• CUET Chemistry Syllabus

• CUET Chemistry PYQs 

• CUET Chemistry Chapter-Wise Weightage