Class 12 Chemistry Chapter 1 Solutions – NCERT Answers & Key Concepts

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Students can find the solutions for the Chemistry Class 12 Solution Chapter 2 exercises below. The solutions are prepared by the renowned teachers. Solving these questions, students can prepare themselves for the CBSE Class 12 board exam. In case of any confusion, seek the help of the class teacher to understand complex problems easily.

Exercise Q 2.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example

A 2.1 A solution is a homogeneous mixture of two or more than two substances on molecular level whose composition can vary within certain limits. The part or component of the mixture present in a lesser amount is called the SOLUTE and the one present in larger amount is called the SOLVENT. For eg- small amount of salt [solute] dissolved in water [solvent]. There are nine types of solutions formed. They are:       Sr.No.     State of solute     State of solvent     Examples     1     GAS     GAS     Air     2     GAS     LIQUID     Oxygen in water, carbonated water     3     SOLID     GAS     Smoke particles in air, dust particles in air     4     LIQUID     GAS     Mist     5     LIQUID     LIQUID     Alcohol in water       6     LIQUID     SOLID     Mercury in silver     7     GAS     SOLID     Adsorption of hydrogen over palladium or platinum     8     SOLID     LIQUID     Sugar in water     9     SOLID     SOLID     Carbon in Iron(steel), Alloy Out of these nine types solution, solid in liquid, liquid in liquid & gas in liquid are very common. When the components of the solution are mixed, the resulting solution may exist in any of the three possible states of matter that is solid, liquid or gaseous. They are: (1) Gaseous solution: In such solutions solvent is Since the solvent is gas,the solute can be solid, liquid or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution. (2) Liquid solution: In such type of solutions liquid acts as the solvent. The solute in these solutions may be gas, liquid, or solid. (3) Solid solutions: As the name suggests, in such solutions solid acts as the solvent. The solute in these solutions may be a gas, liquid or solid. For example, a solution of copper in gold is a solid solution.

Q 2.2 Give an example of a solid solution in which the solute is a gas.

A 2.2 As the name signifies, a solid solution is one in which solvent is solid.So considering this aspect absorption of hydrogen over platinum or palladium is an example of such solution. Platinum or palladium is used as a catalyst in hydrogenation processes.

Q 2.3 Define the following terms: (i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage

A 2.3 (1) Mole fraction - The mole fraction of a particular component in a solution is the ratio of the number of moles of that component to the total number of moles of all the components present in the solution. Mathematically, Mole Fraction of component = Number of moles of given component / Total number of moles in solution Mole Fraction is independent of temperature. (2) Molality - Molality of a solution is defined as the number of moles of solute dissolved per 1000g [1kg] of the It is represented by m. Molality actually represents the concentration of solution in mol / kg. Mathematically, Molality = Number of moles of solute/ Mass of solvent in kg It is represented by m. (3) Molarity- The number of moles of solute dissolved per litre of the solution at a particular temperature is called the molarity of the solution at that Molarity actually represents the concentration of a solution in mol / L. Mathematically, Molarity = Number of moles of solute/ Volume of solution in litres (4) Mass percentage - Mass percentage is defined as the mass of the solute in grams dissolved per 100g of the It is also referred to as weight percentage [w/w]. For example, 10% [by mass] urea solution means that 10 g of urea are present in 100 g of solution, the solvent being only 100-10 = 90 g. Mathematically, the mass percentage of a solute in a solution is given by: Mass Percentage of Solute = Mass of solute / Mass of solute + Mass of solvent X 100 Or Mass Percentage of Solute = Mass of solute / Mass of solution X 100

Q 2.4 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?

A 2.4 Given: Concentration of Nitric Acid, HNO3 = 68% Density of solution, d = 1.504 g/ml To find: Molarity, Mo Formula: Density, d = Mass (M) / volume (V) Molarity, Mo  = Number of moles of solute/ Volume of solution in litres Solution: 68% of Nitric acid by mass in aqueous solution means that 68g [[68 × 100]/100] of Nitric acid present in 100g of solution. ⇒ Molecular mass of Nitric Acid, HNO3 = [1 × 1] + [1 × 14] + [16 × 3] = 63g ⇒ Number of moles of Nitric Acid = [68/63] = 1.079 moles ⇒ Given Density, d = 1.504 g/ml ⇒ Volume, v = [100/1.504] = 66.489 ml ⇒ Molarity, Mo = [1.079/66.489] × 1000 = 16.23 M Therefore the molarity of the sample is 16.24 M.

 

Your one-stop solution for all the requirements to clear the Class 12 Chemistry exam with excellent marks is here. The given table consists of all that you need, including assignment, study notes, NCERT Solutions, NCERT Notes, quick revision notes and important formulas for Chapter 1 Solutions.

Particular	Link
Class 12 Chemistry Solutions Notes	Download Here
Chapter 1 Solutions Quick Revision Notes	Download Here
Chemistry Solutions Important Formulas	Download Here
Solutions Important Questions for CBSE, JEE, and NEET	Download Here
Class 12 Chapter 1 NCERT Exemplar Solutions	Download Here

This chapter covers fundamentaly important topics to understand liquid solutions and formations, colligative properties, vapour pressure and osmosis. Check the list of key topics below.

Key Topics in Class 12 Chemistry Chapter 1 Solutions

• Solutions and Their Types
• Concentration of Solutions
• Solubility
  • Solubility of a Solid in a Liquid
  • Solubility of a Gas in a Liquid
  • Henry’s Law
• Vapour Pressure of Pressure of Pressure of Liquid Solutions
  • Vapour Pressure of Liquid - Liquid Solutions
  • Vapour Pressure of Solid - Liquid Solutions 
  • Raoult’s Law
• Ideal and Non-ideal Solutions
• Colligative Properties
  • Relative Lowering of Vapour Pressure
  • Elevation of Boiling Point
  • Depression of Freezing Point
  • Determination of Molar Mass
• Osmosis and Osmotic Pressure
• Reverse Osmosis and Water Purification

Important Formulae of Solutions for CBSE and Competitive Exams

• Concentration Terms

Important Concentration Terms Formulae
Mass Percentage (%)	$\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
Mole Fraction ( $X$)	$X_A = \frac{n_A}{n_A + n_B}$
Molarity (M)	$M = \frac{\text{moles of solute}}{\text{litres of solution}}$
Molality (m)	$m = \frac{\text{moles of solute}}{\text{kg of solvent}}$
Normality (N)	$N = \frac{\text{equivalents of solute}}{\text{litres of solution}}$

• Colligative Properties
  • Relative Lowering of Vapour Pressure

$$\frac{P^o_A - P_A}{P^o_A} = X_B$$

 

• • Elevation of Boiling Point ( $\Delta T_b$)

$$\Delta T_b = K_b \cdot m$$

$K_b$ = Molal elevation constant

• • Depression of Freezing Point 

$$\Delta T_f = K_f \cdot m$$

$K_f$ = Molal depression constant

• • Osmotic Pressure ( $\pi$)

$$\pi = MRT$$

• • Where, 

$M$ = Molarity, $R$ = Gas constant (0.0821 L atm/mol K), $T$ = Temperature

• Abnormal Molar Mass & Van’t Hoff Factor ( $i$)
  • If the observed molar mass differs from the expected value.

$$i = \frac{\text{Observed Colligative Property}}{\text{Calculated Colligative Property}}$$

• Corrected Colligative Property Formulae

  • Boiling Point Elevation:

    $$\Delta T_b = i K_b m$$

  • Freezing Point Depression:

    $$\Delta T_f = i K_f m$$

  • Osmotic Pressure:

    $$\pi = i M R T$$

• • Solubility and Henry’s Law

$$C=k_{H}P$$

• • Raoult’s Law (Vapour Pressure of Solutions)

$$P_A = X_A P_A^o$$ $$P_B = X_B P_B^o$$

(Total vapour pressure: $P_{\text{total}}=P_A+P_B$

The Class 12 Chemistry Chapter 1 Solutions is important for understanding concentration terms, colligative properties, Raoult’s Law, and ideal/non-ideal solutions. The Chemistry Chapter 1 Class 12 is crucial for Competitive Exams. Check the topics covered in this chapter:

Chemistry Chapter 1 Solutions Weightage in NEET, JEE Main Exam

In both these competitive exams, this chapter holds moderate weightage. Refer to the table below to see the weightage:

Exam	Number of Questions
CBSE	4 - 7 Marks
NEET	1-2 questions
JEE Main	1-2 questions

Chapter 1 Solutions Important Questions