Solved Exercise of Class 12 Chemistry Chapter 2 Solutions

Maharashtra Board Class 12 Solutions for Chemistry Chapter 2 Solutions

Welcome to our comprehensive article on Maharashtra State Board 12th Chemistry Solved Exercise of Chapter 2: Solutions! In this article, we provide step-by-step solutions and explanations for the exercises in Chapter 2, allowing students to master the concepts and excel in their chemistry studies.
By meticulously solving each exercise problem, we aim to provide a comprehensive understanding of the topics covered in this chapter. Whether you're struggling with concentration units, solubility calculations, or colligative properties, our detailed solutions will guide you through the process, ensuring clarity and accuracy.
With Maharashtra State Board 12th Chemistry Solved Exercise of Chapter 2 Solutions, you can confidently tackle the exercise problems, reinforce your knowledge, and sharpen your problem-solving skills. Let's dive into the solutions and unlock your potential for success in chemistry!

Question 1. Choose the most correct answer.

i. The vapour pressure of a solution containing 2 moles of a solute in 2 moles of water (vapour pressure of pure water = 24 mm Hg) is

(a) 24 mm Hg

(b) 32 mm Hg

(c) 48 mm Hg

(d) 12 mm Hg

Answer: (d) 12 mm Hg

ii. The colligative property of a solution is

(a) vapour pressure

(b) boiling point

(c) osmotic pressure

(d) freezing point

Answer: (c) osmotic pressure

iii. In calculating osmotic pressure the concentration of solute is expressed in

(a) molarity

(b) molality

(c) mole fraction

(d) mass per cent

Answer: (a) molarity

iv. Ebullioscopic constant is the boiling point elevation when the concentration of solution is

(a) 1 m

(b) 1 M

(c) 1 mass%

(d) 1 mole fraction of solute

Answer: (a) 1 m

v. Cryoscopic constant depends on

(a) nature of solvent

(b) nature of solute

(c) nature of solution

(d) number of solvent molecules

Answer: (a) nature of solvent

vi. Identify the correct statement

(a) vapour pressure of solution is higher than that of pure solvent.

(b) boiling point of solvent is lower than that of solution

(c) osmotic pressure of solution is lower than that of solvent

(d) osmosis is a colligative property.

Answer: (b) boiling point of solvent is lower than that of solution

vii. A living cell contains a solution which is isotonic with 0.3 M sugar solution. What osmotic pressure develops when the cell is placed in 0.1 M KCl solution at body temperature ?

(a) 5.08 atm

(b) 2.54 atm

(c) 4.92 atm

(d) 2.46 atm

Answer: (c) 4.92 atm

viii. The osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose isotonic with blood has the percentage (by volume)

(a) 5.41%

(b) 3.54%

(c) 4.53%

(d) 53.4%

Answer: (a) 5.41%

ix. Vapour pressure of a solution is

(a) directly proportional to the mole fraction of the solute

(b) inversely proportional to the mole fraction of the solute

(c) inversely proportional to the mole fraction of the solvent

(d) directly proportional to the mole fraction of the solvent

Answer: (d) directly proportional to the mole fraction of the solvent

x. Pressure cooker reduces cooking time for food because

(a) boiling point of water involved in cooking is increased

(b) heat is more evenly distributed in the cooking space

(c) the higher pressure inside the cooker crushes the food material

(d) cooking involves chemical changes helped by a rise in temperature

Answer: (a) boiling point of water involved in cooking is increased

xi. Henry’s law constant for a gas CH3Br is 0.159 mol dm-3 atm at 250°C. What is the solubility of CH3Br in water at 25 °C and a partial pressure of 0.164 atm?

(a) 0.0159 mol L-1

(b) 0.164 mol L-1

(c) 0.026 M

(d) 0.042 M

Answer: (c) 0.026 M

xii. Which of the following statement is NOT correct for 0.1 M urea solution and 0.05 M sucrose solution ?

(a) osmotic pressure exhibited by urea solution is higher than that exhibited by sucrose solution

(b) urea solution is hypertonic to sucrose solution

(c) they are isotonic solutions

(d) sucrose solution is hypotonic to urea solution

Answer: (c) they are isotonic solutions

Question 2. Answer the following in one or two sentences

i. What is osmotic pressure ?

Answer: The hydrostatic pressure that stops osmosis is an osmotic pressure of the solution 

ii. A solution concentration is expressed in molarity and not in molality while considering osmotic pressure. Why ?

Answer: While calculating osmotic pressure the concentration is expressed in molarity but not in molality.

This is because the measurements of osmotic pressure are made at a specific constant temperature.

Molarity depends upon temperature but molality is independent of temperature.

Hence in osmotic pressure measurements, concentration is expressed in molarity.

iii. Write the equation relating boiling point elevation to the concentration of solution.

Answer: The elevation in the boiling point of a solution is directly proportional to the molal concentration (expressed in mol kg-1) of the solution.

Hence, if ΔTb is the elevation in the boiling point of a solution of molal concentration m then,

ΔTb ∝ m

∴ ΔTb = Kb m

where Kb is a proportionality constant.

If m = 1 molal,

ΔTb = Kb

Kb is called the ebullioscopic constant or molal elevation constant.

iv. A 0.1 m solution of K2SO4 in water has freezing point of -0.43 °C. What is the value of van’t Hoff factor if Kf for water is 1.86 K kg mol-1?

Answer: Given : m = 0.1 m, ΔTf = 0 – (-0.43) = 0.43 °C

Kf = 1.86 K kg mol-1, i = ?

ΔTf = i × Kf × m

∴ i = ΔTf × kf × m 

     = 0.431.86×0.1 = 2.312

van’t Hoff factor = i = 2.312

v. What is van’t Hoff factor?

Answer: It is defined as a ratio of the observed colligative property of electrolyte solution divided by the colligative property of nonelectrolyte solution of the same concentration.

vi. How is van’t Hoff factor related to degree of ionization?

Answer: Consider an electrolyte AxBy that dissociates to give x number of Ay+ ions and y number of Bx- ions. 

Let α is the degree of dissociation of electrolyte.

At equilibrium,

AxBy ⇌ xAy+ + yBx- .......(1)

initially: 1 mol   0   0

at equilibrium:  (1 – α) mol  (x α) mol  (y α) mol

If α is the degree of dissociation of electrolyte, then the moles of cations are αx and those of anions are αy at equilibrium.

α mol of electrolyte dissociates and (1 - α) mol remains undissociated at equilibrium.

Total moles after dissociation 

= (1 – α) + (xα) + (yα)

= 1 + α (x + y – 1)

= 1 + α (n– 1) .........(2)

where, n = (x + y) = moles of ions obtained from dissociation of 1 mole of electrolyte.

The van’t Hoff factor i given by, 

i= actual moles of particle after dissociation / moles of formula units

 i = 1 + α(n – 1)/1

Hence i = 1 + α(n – 1) or 

α = i−1/n−1 ……(3)

This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.

vii. Which of the following solutions will have higher freezing point depression and why ?

a. 0.1 m NaCl b. 0.05 m Al2(SO4)3

Answer: 1. Freezing point depression is a colligative property, hence depends on the number of solute particles in the solution.

2. More the number of solute particles in the solution, higher is the depression in freezing point.

3. The number of particles (ions) are more in 0.05 m Al2(SO4)3 than in 0.1 m NaCl.

4. Therefore Al2(SO4)3 solution will have higher freezing point depression.

viii. State Raoult’s law for a solution containing a nonvolatile solute.

Answer: The law states that the vapour pressure of a solvent over the solution of a nonvolatile solute is equal to the vapour pressure of the pure solvent multiplied by its mole fraction in the solution.

ix. What is the effect on the boiling point of water if 1 mole of methyl alcohol is added to 1 dm3 of water? Why?

Answer: The boiling point of water (or any liquid) depends on its vapour pressure.

Higher the vapour pressure, lower is the boiling point.

When 1 mole of volatile methyl alcohol is added to 1 dm3 of water, its vapour pressure is increased which decreases the boiling point of water.

x. Which of the four colligative properties is most often used for molecular mass determination? Why?

Answer: Osmotic pressure has much larger values and therefore more precisely measurable property than other colligative properties.

The osmotic pressure can be measured at a suitable constant temperature.

The molecular masses can be measured more accurately using osmotic pressure.

Therefore, it is most often used for molecular mass determination of expensive substances.

Disclaimer: Please note that this article provides solved exercises for educational purposes and should be used as a supplementary resource alongside your textbook and classroom teachings.