NCERT Solutions for Class 12 Chemistry Chapter 1: Solutions (Updated 2025-26)

NCERT Solutions for Class 12 Chemistry Chapter 1 – Solutions provide detailed, step-by-step answers to all textbook questions, including in-text exercises, numericals, and conceptual problems. This chapter, titled “Solutions”, is one of the most important topics in Class 12 Chemistry, as it explains fundamental concepts like molarity, molality, mole fraction, colligative properties, ideal and non-ideal solutions, Raoult’s Law, Henry’s Law, and the Van’t Hoff factor.

For exam preparation, especially in CBSE Board Class 12 exams, JEE, and NEET, students often search for Class 12 Chemistry Chapter 1 solutions PDF, numerical problems with solutions, and easy explanations of concentration terms. This page provides accurate NCERT answers, solved numericals with step-by-step calculation methods, and simplified notes to help you understand formulas, conversions, and applications quickly.

NCERT Solutions for Class 12 Chemistry Chapter 1 Overview

The NCERT Class 12 Chemistry Chapter 1 Solutions PDF included here follows the complete Class 12 chemistry syllabus and exam pattern, making it useful for quick revision, homework help, and competitive exam practice. Along with detailed NCERT Solutions for Class 12 Chemistry, you’ll find comparison tables, real-life examples, previous year questions, and exam-focused tips to help you score better.

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions PDF

A solution is a homogeneous mixture of two or more substances. In this chapter, you’ll learn about different types of solutions (solid, liquid, gaseous), various ways to express concentration (molarity, molality, mole fraction, mass percent), and the key laws governing solutions — like Raoult’s Law and Henry’s Law. The chapter also covers colligative properties, which explain phenomena such as boiling point elevation, freezing point depression, osmotic pressure, and their applications.

Detailed NCERT Solutions for Class 12 Chemistry Chapter 1

Chapter 1 "Solutions" is a high-weightage chapter for CBSE Class 12 Boards (approx. 7 Marks) and essential for JEE/NEET. This guide provides step-by-step NCERT solutions, simplified formula sheets, and visual graphs for Raoult's Law and Henry's Law.

Question 1.1 Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

Solution:

Given:

• Mass of benzene (C₆H₆) = 22 g
• Mass of carbon tetrachloride (CCl₄) = 122 g

Total mass of solution = 22 + 122 = 144 g

Mass percentage of benzene = (Mass of benzene / Total mass) × 100

= (22 / 144) × 100

= 15.28%

Mass percentage of CCl₄ = (Mass of CCl₄ / Total mass) × 100

= (122 / 144) × 100

= 84.72%

Answer: Mass % of C₆H₆ = 15.28%, Mass % of CCl₄ = 84.72%

Question 1.2

Question: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

Solution:

Given: 30% benzene by mass

Assume total mass of solution = 100 g

Therefore:

• Mass of benzene = 30 g
• Mass of CCl₄ = 70 g

Molar mass of benzene (C₆H₆) = 6 × 12 + 6 × 1 = 78 g mol^-1

Molar mass of CCl₄ = 12 + 4 × 35.5 = 154 g mol^-1

Moles of benzene = 30 / 78 = 0.385 mol

Moles of CCl₄ = 70 / 154 = 0.455 mol

Total moles = 0.385 + 0.455 = 0.840 mol

Mole fraction of benzene = 0.385 / 0.840 = 0.458

Mole fraction of CCl₄ = 0.455 / 0.840 = 0.542

Answer: Mole fraction of benzene = 0.458 (or 0.459)

Question 1.3 Calculate the molarity of each of the following solutions:

(a) 30 g of Co(NO₃)₂·6H₂O in 4.3 L of solution

(b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Solution:

(a)

Molar mass of Co(NO₃)₂·6H₂O:

= 59 + 2(14 + 3×16) + 6(2 + 16)

= 59 + 2(14 + 48) + 6(18)

= 59 + 124 + 108

= 291 g mol^-1

Moles of Co(NO₃)₂·6H₂O = 30 / 291 = 0.103 mol

Molarity = Moles / Volume in litres

= 0.103 / 4.3

= 0.024 M

(b)

Using dilution formula: M₁V₁ = M₂V₂

0.5 × 30 = M₂ × 500

M₂ = (0.5 × 30) / 500

M₂ = 15 / 500

M₂ = 0.03 M

Answer: (a) 0.024 M (b) 0.03 M

Question 1.4 Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.

Solution:

Given:

• Total mass of solution = 2.5 kg = 2500 g
• Molality = 0.25 mol kg^-1

Molality = (Moles of solute) / (Mass of solvent in kg)

Let mass of urea = w g

Then mass of water = (2500 - w) g = (2.5 - w/1000) kg

Molar mass of urea (NH₂CONH₂) = 14 + 2 + 12 + 16 + 14 + 2 = 60 g mol^-1

Moles of urea = w / 60

0.25 = (w/60) / (2.5 - w/1000)

0.25(2.5 - w/1000) = w/60

0.625 - 0.25w/1000 = w/60

0.625 = w/60 + w/4000

0.625 = w(1/60 + 1/4000)

0.625 = w(66.67 + 0.25)/4000

0.625 = w(66.92/4000)

w = (0.625 × 4000) / 66.92

w = 37.35 g

Alternatively, assuming mass of solvent ≈ 2.5 kg:

0.25 = (w/60) / 2.5

w/60 = 0.25 × 2.5 = 0.625

w = 0.625 × 60 = 37.5 g

Answer: Mass of urea required = 37.5 g (approximately)

Question 1.5 Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^-1.

Solution:

Assume 100 g of solution

• Mass of KI = 20 g
• Mass of water = 80 g = 0.080 kg
• Density = 1.202 g mL^-1

Molar mass of KI = 39 + 127 = 166 g mol^-1

Molar mass of water = 18 g mol^-1

(a) Molality:

Moles of KI = 20 / 166 = 0.120 mol

Molality = 0.120 / 0.080 = 1.5 mol kg^-1

(b) Molarity:

Volume of solution = Mass / Density = 100 / 1.202 = 83.19 mL = 0.08319 L

Molarity = 0.120 / 0.08319 = 1.45 M (approximately)

(c) Mole fraction:

Moles of water = 80 / 18 = 4.444 mol

Total moles = 0.120 + 4.444 = 4.564 mol

Mole fraction of KI = 0.120 / 4.564 = 0.0263

Answer: (a) 1.5 mol kg^-1 (b) 1.45 M (c) 0.0263

Question 1.6 H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's law constant.

Solution:

Given:

• Molality of H₂S = 0.195 m
• At STP, pressure = 1 bar = 0.987 atm = 750 mm Hg (approximately 1 bar)

For dilute solution, molality ≈ molarity (in mol/L)

Since 1 kg water ≈ 1 L water

Moles of H₂S in 1 kg water = 0.195 mol

Moles of water = 1000 / 18 = 55.55 mol

Mole fraction of H₂S = 0.195 / (0.195 + 55.55)

x = 0.195 / 55.745 = 0.0035

Using Henry's law: p = K_H × x

1 bar = K_H × 0.0035

K_H = 1 / 0.0035 = 285.7 bar

Or in different units:

K_H = 285.7 bar ≈ 286 bar

Answer: K_H ≈ 286 bar (or approximately 2.85 × 10⁵ Pa)

Question 1.7 Henry's law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.

Solution:

Given:

• K_H = 1.67 × 10⁸ Pa
• Pressure = 2.5 atm = 2.5 × 101325 Pa = 253312.5 Pa
• Volume of water = 500 mL

Using Henry's law: p = K_H × x

x = p / K_H

x = 253312.5 / (1.67 × 10⁸)

x = 1.517 × 10^-3

Moles of water in 500 mL = (500 × 1) / 18 = 27.78 mol

(Assuming density of water = 1 g/mL)

Let moles of CO₂ = n

x = n / (n + 27.78)

Since n << 27.78:

1.517 × 10^-3 = n / 27.78

n = 1.517 × 10^-3 × 27.78

n = 0.0421 mol

Mass of CO₂ = 0.0421 × 44 = 1.854 g

Answer: Quantity of CO₂ = 1.85 g (approximately)

Question 1.8 The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Solution:

Given:

• p_A⁰ = 450 mm Hg
• p_B⁰ = 700 mm Hg
• p_total = 600 mm Hg

For liquid phase composition:

Using Raoult's law:

p_total = p_A⁰x_A + p_B⁰x_B

600 = 450 x_A + 700 x_B

Since x_A + x_B = 1

x_B = 1 - x_A

600 = 450 x_A + 700(1 - x_A)

600 = 450 x_A + 700 - 700 x_A

600 = 700 - 250 x_A

250 x_A = 100

x_A = 0.4

x_B = 0.6

For vapour phase composition:

p_A = p_A⁰x_A = 450 × 0.4 = 180 mm Hg

p_B = p_B⁰x_B = 700 × 0.6 = 420 mm Hg

y_A = p_A / p_total = 180 / 600 = 0.3

y_B = p_B / p_total = 420 / 600 = 0.7

Answer: Liquid phase: x_A = 0.4, x_B = 0.6; Vapour phase: y_A = 0.3, y_B = 0.7

Question 1.9 Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Solution:

Given:

• p₁⁰ = 23.8 mm Hg
• Mass of urea = 50 g
• Mass of water = 850 g

Molar mass of urea = 60 g mol^-1

Molar mass of water = 18 g mol^-1

Moles of urea = 50 / 60 = 0.833 mol

Moles of water = 850 / 18 = 47.22 mol

Mole fraction of urea (x₂) = 0.833 / (0.833 + 47.22)

x₂ = 0.833 / 48.053 = 0.0173

Using Raoult's law:

Relative lowering of vapour pressure = (p₁⁰ - p₁) / p₁⁰ = x₂

= 0.0173

p₁⁰ - p₁ = 0.0173 × 23.8

p₁⁰ - p₁ = 0.412 mm Hg

p₁ = 23.8 - 0.412 = 23.388 mm Hg

p₁ ≈ 23.4 mm Hg

Answer: Vapour pressure = 23.4 mm Hg, Relative lowering = 0.0173

Question 1.10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C?

Solution:

Given:

• Boiling point of water at 750 mm Hg = 99.63°C = 372.78 K
• Desired boiling point = 100°C = 373.15 K
• Mass of water = 500 g = 0.5 kg

Elevation in boiling point:

ΔT_b = 373.15 - 372.78 = 0.37 K

For water, K_b = 0.52 K kg mol^-1

Molar mass of sucrose (C₁₂H₂₂O₁₁) = 342 g mol^-1

Using: ΔT_b = K_b × m

0.37 = 0.52 × m

m = 0.37 / 0.52 = 0.712 mol kg^-1

Moles of sucrose needed = 0.712 × 0.5 = 0.356 mol

Mass of sucrose = 0.356 × 342 = 121.75 g

Answer: Mass of sucrose = 121.75 g (approximately 122 g)

Question 1.11 Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol^-1.

Solution:

Given:

• ΔT_f = 1.5 K
• K_f = 3.9 K kg mol^-1
• Mass of acetic acid = 75 g = 0.075 kg

Molar mass of ascorbic acid (C₆H₈O₆):

= 6(12) + 8(1) + 6(16) = 72 + 8 + 96 = 176 g mol^-1

Using: ΔT_f = K_f × m

1.5 = 3.9 × m

m = 1.5 / 3.9 = 0.385 mol kg^-1

Moles of ascorbic acid = 0.385 × 0.075 = 0.0289 mol

Mass of ascorbic acid = 0.0289 × 176 = 5.08 g

Answer: Mass of ascorbic acid = 5.08 g (approximately 5.1 g)

Question 1.12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Solution:

Given:

• Mass of polymer (w₂) = 1.0 g
• Molar mass (M₂) = 185,000 g mol^-1
• Volume = 450 mL = 0.450 L
• Temperature = 37°C = 310 K
• R = 8.314 J K^-1 mol^-1 (or 0.0821 L atm K^-1 mol^-1)

Moles of polymer = 1.0 / 185,000 = 5.405 × 10^-6 mol

Using: π = (n₂RT) / V

π = (5.405 × 10^-6 × 8.314 × 310) / 0.450

π = 0.01393 / 0.450

π = 0.0309 Pa

Or using L atm units and converting:

π = (5.405 × 10^-6 × 0.0821 × 310) / 0.450

π = 1.375 × 10^-4 / 0.450

π = 3.056 × 10^-4 atm

π = 3.056 × 10^-4 × 101325 Pa = 30.96 Pa

Answer: Osmotic pressure = 30.96 Pa (approximately 31 Pa)

END-OF-CHAPTER EXERCISES

Question 1.1 Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.

Solution:

Definition: A solution is a homogeneous mixture of two or more substances. The composition and properties are uniform throughout the mixture.

Types of Solutions:

Based on the physical states of solute and solvent, there are nine types of solutions:

1. Gaseous Solutions:

• Gas in Gas: Mixture of oxygen and nitrogen gases (air)
• Liquid in Gas: Chloroform mixed with nitrogen gas
• Solid in Gas: Camphor in nitrogen gas

2. Liquid Solutions:

• Gas in Liquid: Oxygen dissolved in water (aquatic life depends on it)
• Liquid in Liquid: Ethanol dissolved in water
• Solid in Liquid: Glucose dissolved in water, salt in water

3. Solid Solutions:

• Gas in Solid: Solution of hydrogen in palladium
• Liquid in Solid: Amalgam of mercury with sodium
• Solid in Solid: Copper dissolved in gold (alloys like brass, bronze)

The component present in the largest quantity is called the solvent and determines the physical state of the solution. Other components are called solutes.

Question 1.2 Give an example of a solid solution in which the solute is a gas.

Solution:

An example of a solid solution in which the solute is a gas is:

Hydrogen gas dissolved in palladium metal

Other examples include:

• Hydrogen in platinum
• Hydrogen in nickel

In these solutions, gas molecules occupy the interstitial spaces in the metal lattice.

Question 1.3 Define the following terms:

(i) Mole fraction (ii) Molality (iii) Molarity (iv) Mass percentage.

Solution:

(i) Mole fraction (x):

The mole fraction of a component is defined as the ratio of the number of moles of that component to the total number of moles of all components in the solution.

For component i:

x_i = n_i / (n₁ + n₂ + ... + n_i)

where n_i = number of moles of component i

Sum of all mole fractions = 1

It is a dimensionless quantity.

(ii) Molality (m):

Molality is defined as the number of moles of solute dissolved in 1 kg (1000 g) of solvent.

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

Unit: mol kg^-1 or m

It is independent of temperature.

(iii) Molarity (M):

Molarity is defined as the number of moles of solute dissolved in 1 litre (or 1 dm³) of solution.

Molarity (M) = (Moles of solute) / (Volume of solution in litres)

Unit: mol L^-1 or M or mol dm^-3

It depends on temperature as volume changes with temperature.

(iv) Mass percentage (w/w):

Mass percentage is defined as the mass of a component per 100 g of the solution.

Mass % of component = (Mass of component / Total mass of solution) × 100

It is independent of temperature.

Question 1.4 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:

Given:

• Mass percentage of HNO₃ = 68%
• Density of solution = 1.504 g mL^-1

Assume 100 g of solution

Mass of HNO₃ = 68 g

Mass of water = 32 g

Molar mass of HNO₃ = 1 + 14 + 3(16) = 63 g mol^-1

Moles of HNO₃ = 68 / 63 = 1.079 mol

Volume of solution = Mass / Density

= 100 / 1.504 = 66.49 mL = 0.06649 L

Molarity = Moles / Volume in litres

= 1.079 / 0.06649

= 16.23 M

Answer: Molarity = 16.23 M (approximately 16.2 M)

Question 1.5 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 solution is 1.2 g mL^-1, then what shall be the molarity of the solution?

Solution:

Given:

• 10% w/w glucose solution
• Density = 1.2 g mL^-1

Assume 100 g of solution:

Mass of glucose = 10 g

Mass of water = 90 g = 0.090 kg

Molar mass of glucose (C₆H₁₂O₆) = 180 g mol^-1

Molar mass of water = 18 g mol^-1

Molality:

Moles of glucose = 10 / 180 = 0.0556 mol

Molality = 0.0556 / 0.090 = 0.617 mol kg^-1

Mole fraction:

Moles of water = 90 / 18 = 5 mol

Total moles = 0.0556 + 5 = 5.0556 mol

Mole fraction of glucose = 0.0556 / 5.0556 = 0.011

Mole fraction of water = 5 / 5.0556 = 0.989

Molarity:

Volume of solution = 100 / 1.2 = 83.33 mL = 0.08333 L

Molarity = 0.0556 / 0.08333 = 0.667 M

Answer: Molality = 0.617 mol kg^-1; Mole fractions: glucose = 0.011, water = 0.989; Molarity = 0.667 M

Question 1.6 How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂CO₃ and NaHCO₃ containing equimolar amounts of both?

Solution:

Given: 1 g mixture containing equimolar amounts of Na₂CO₃ and NaHCO₃

Let moles of Na₂CO₃ = moles of NaHCO₃ = n

Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g mol^-1

Molar mass of NaHCO₃ = 23 + 1 + 12 + 3(16) = 84 g mol^-1

Total mass = 106n + 84n = 190n = 1 g

n = 1/190 = 0.00526 mol

Reactions:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

NaHCO₃ + HCl → NaCl + H₂O + CO₂

Moles of HCl required:

For Na₂CO₃ = 2 × 0.00526 = 0.01052 mol

For NaHCO₃ = 1 × 0.00526 = 0.00526 mol

Total = 0.01578 mol

Volume of 0.1 M HCl = Moles / Molarity

= 0.01578 / 0.1 = 0.1578 L = 157.8 mL

Answer: Volume of HCl required = 157.8 mL (approximately 158 mL)

Question 1.7 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:

Given:

• Solution 1: 300 g of 25% solution
• Solution 2: 400 g of 40% solution

Mass of solute in solution 1 = (25/100) × 300 = 75 g

Mass of solute in solution 2 = (40/100) × 400 = 160 g

Total mass of solute = 75 + 160 = 235 g

Total mass of solution = 300 + 400 = 700 g

Mass percentage = (235 / 700) × 100 = 33.57%

Answer: Mass percentage of resulting solution = 33.57% (approximately 33.6%)

Question 1.8 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:

Given:

• Mass of ethylene glycol = 222.6 g
• Mass of water = 200 g = 0.200 kg
• Density of solution = 1.072 g mL^-1

Molar mass of C₂H₆O₂ = 2(12) + 6(1) + 2(16) = 62 g mol^-1

Molality:

Moles of ethylene glycol = 222.6 / 62 = 3.59 mol

Molality = 3.59 / 0.200 = 17.95 mol kg^-1

Molarity:

Total mass of solution = 222.6 + 200 = 422.6 g

Volume of solution = 422.6 / 1.072 = 394.22 mL = 0.39422 L

Molarity = 3.59 / 0.39422 = 9.11 M

Answer: Molality = 17.95 mol kg^-1, Molarity = 9.11 M

Question 1.9 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 molality of chloroform in the water sample.

Solution:

Given: Contamination level = 15 ppm

(i) Percent by mass:

15 ppm means 15 parts per million parts

= 15 g chloroform per 10⁶ g solution

= (15 / 10⁶) × 100 %

= 1.5 × 10^-3 %

= 0.0015%

(ii) Molality:

Consider 10⁶ g (1000 kg) of solution

Mass of CHCl₃ = 15 g

Mass of water ≈ 10⁶ g = 1000 kg (since CHCl₃ is very small)

Molar mass of CHCl₃ = 12 + 1 + 3(35.5) = 119.5 g mol^-1

Moles of CHCl₃ = 15 / 119.5 = 0.1255 mol

Molality = 0.1255 / 1000 = 1.255 × 10^-4 mol kg^-1

Answer: (i) 0.0015% (ii) 1.255 × 10^-4 mol kg^-1

Question 1.10 What role does the molecular interaction play in a solution of alcohol and water?

Solution:

In a solution of alcohol and water, molecular interactions play a very important role:

1. Hydrogen Bonding:

• In pure water, water molecules are extensively hydrogen bonded to each other (H₂O···H-O-H)
• In pure alcohol, alcohol molecules form hydrogen bonds with each other (R-O-H···O-H-R)
• When alcohol and water are mixed, new hydrogen bonds form between alcohol and water molecules (R-O-H···O-H₂)

2. Breaking and Formation of H-bonds:

• Some H-bonds between water molecules break
• Some H-bonds between alcohol molecules break
• New H-bonds form between alcohol and water molecules

3. Effect on Properties:

• The new alcohol-water H-bonds are weaker than the original water-water H-bonds
• This leads to positive deviation from Raoult's law
• The vapor pressure of solution is higher than expected
• Heat is absorbed during mixing (endothermic, ΔH_mix > 0)
• Volume of solution is slightly greater than sum of individual volumes (ΔV_mix > 0)

4. Solubility:

Both alcohol and water are polar and can form H-bonds, following the principle "like dissolves like," making them completely miscible in all proportions.

Question 1.11 Why do gases always tend to be less soluble in liquids as the temperature is raised?

Solution:

Gases become less soluble in liquids as temperature increases due to the following reasons:

1. Exothermic Process:

The dissolution of a gas in a liquid is an exothermic process:

Gas + Liquid → Solution + Heat

ΔH_sol < 0

2. Le Chatelier's Principle:

When a gas dissolves in liquid:

Gas (less ordered) → Dissolved gas (more ordered)

This process releases heat (exothermic)

According to Le Chatelier's Principle:

• When temperature increases, the equilibrium shifts in the direction that absorbs heat
• This means the equilibrium shifts backward (toward gas phase)
• Therefore, solubility decreases

3. Kinetic Energy:

• At higher temperatures, gas molecules have more kinetic energy
• They can more easily escape from the solution back to the gas phase
• This reduces the amount of dissolved gas

4. Practical Examples:

• Cold water can hold more dissolved oxygen than warm water
• Aquatic animals are more comfortable in cold water (more O₂ available)
• Carbonated drinks release CO₂ faster when warm
• Fish tanks need cooling in summer to maintain oxygen levels

Question 1.12 State Henry's law and mention some important applications.

Solution:

Henry's Law Statement:

"At a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of the liquid or solution."

Mathematically: p = K_H × x

where:

• p = partial pressure of the gas
• x = mole fraction of gas in solution
• K_H = Henry's law constant (depends on nature of gas and solvent)

Important Applications:

1. Soft Drinks and Soda Water:

• CO₂ is dissolved in water under high pressure to increase its solubility
• When bottle is opened, pressure decreases and CO₂ escapes, causing fizz

2. Scuba Diving (The Bends):

• At high underwater pressure, more nitrogen dissolves in blood
• If divers surface too quickly, pressure drops suddenly
• Nitrogen forms bubbles in blood, blocking capillaries
• This causes painful and dangerous condition called "bends" or decompression sickness
• To avoid this, divers use helium-oxygen mixture instead of air, or ascend slowly

3. High Altitude Sickness (Anoxia):

• At high altitudes, partial pressure of O₂ is low
• Less oxygen dissolves in blood
• Causes weakness, inability to think clearly, and breathing difficulty
• Climbers carry oxygen cylinders to compensate

4. Lungs and Respiration:

• Oxygen from air dissolves in blood in lungs where its partial pressure is high
• Carbon dioxide from blood is released in lungs where its partial pressure is low

5. Aquatic Life:

• Dissolved oxygen in water bodies sustains aquatic life
• Warm water holds less oxygen, making it difficult for fish to survive

Question 1.13 The partial pressure of ethane over a solution containing 6.56 × 10^-3 g of ethane is 1 bar. If the solution contains 5.00 × 10^-2 g of ethane, then what shall be the partial pressure of the gas?

Solution:

Given:

• When mass of ethane = 6.56 × 10^-3 g, pressure p₁ = 1 bar
• When mass of ethane = 5.00 × 10^-2 g, pressure p₂ = ?

According to Henry's law: p = K_H × x

where x is the mole fraction of gas

For dilute solutions, mole fraction is proportional to mass (or moles) of solute

x ∝ n ∝ mass (when molar mass is same)

Therefore: p ∝ mass of gas dissolved

So: p₁ / p₂ = mass₁ / mass₂

1 / p₂ = (6.56 × 10^-3) / (5.00 × 10^-2)

p₂ = (5.00 × 10^-2) / (6.56 × 10^-3)

p₂ = 50 / 6.56

p₂ = 7.62 bar

Answer: Partial pressure of ethane = 7.62 bar

Question 1.14 What is meant by positive and negative deviations from Raoult's law and how is the sign of ΔH_mix related to positive and negative deviations from Raoult's law?

Solution:

Positive Deviation from Raoult's Law:

Definition:

When the vapor pressure of the solution is higher than that predicted by Raoult's law, it shows positive deviation.

p_total > p_A⁰x_A + p_B⁰x_B

Cause:

• A-B interactions are weaker than A-A and B-B interactions
• Solute-solvent attractive forces are weaker than solute-solute and solvent-solvent forces
• Molecules escape more easily from solution

Examples:

• Ethanol + Acetone
• Carbon disulfide + Acetone
• Ethanol + Water
• Cyclohexane + Ethanol

Sign of ΔH_mix:

ΔH_mix > 0 (Positive, Endothermic)

Heat is absorbed during mixing because energy is required to break the stronger A-A and B-B bonds

Also: ΔV_mix > 0 (Volume increases)

Negative Deviation from Raoult's Law:

Definition:

When the vapor pressure of the solution is lower than that predicted by Raoult's law, it shows negative deviation.

p_total < p_A⁰x_A + p_B⁰x_B

Cause:

• A-B interactions are stronger than A-A and B-B interactions
• Solute-solvent attractive forces are stronger than individual forces
• Molecules escape less easily from solution
• Often involves formation of new hydrogen bonds

Examples:

• Chloroform + Acetone (new H-bonds form)
• Phenol + Aniline (H-bonding between phenolic H and N of aniline)
• Chloroform + Diethyl ether
• Acetic acid + Pyridine

Sign of ΔH_mix:

ΔH_mix < 0 (Negative, Exothermic)

Heat is released during mixing because stronger A-B bonds are formed

Also: ΔV_mix < 0 (Volume decreases)

Question 1.15 An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution:

Given:

• Mass percentage of solute = 2%
• Vapor pressure of solution = 1.004 bar
• At normal boiling point of pure water, p₀ = 1.013 bar

Assume 100 g of solution:

Mass of solute (w₂) = 2 g

Mass of solvent (water) (w₁) = 98 g

Molar mass of water (M₁) = 18 g mol^-1

Using Raoult's law for relative lowering of vapor pressure:

(p₀ - p) / p₀ = (w₂ × M₁) / (M₂ × w₁)

(1.013 - 1.004) / 1.013 = (2 × 18) / (M₂ × 98)

0.009 / 1.013 = 36 / (98 × M₂)

0.00889 = 36 / (98 × M₂)

M₂ = 36 / (0.00889 × 98)

M₂ = 36 / 0.871

M₂ = 41.33 g mol^-1

Answer: Molar mass of solute = 41.33 g mol^-1 (approximately 41 g mol^-1)

Question 1.16 Heptane and octane form an ideal solution. At 373 K, 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?

Solution:

Given:

• p_heptane⁰ = 105.2 kPa
• p_octane⁰ = 46.8 kPa
• Mass of heptane (C₇H₁₆) = 26.0 g
• Mass of octane (C₈H₁₈) = 35 g

Molar mass of heptane (C₇H₁₆) = 7(12) + 16(1) = 100 g mol^-1

Molar mass of octane (C₈H₁₈) = 8(12) + 18(1) = 114 g mol^-1

Moles of heptane = 26.0 / 100 = 0.26 mol

Moles of octane = 35 / 114 = 0.307 mol

Total moles = 0.26 + 0.307 = 0.567 mol

Mole fraction of heptane = 0.26 / 0.567 = 0.459

Mole fraction of octane = 0.307 / 0.567 = 0.541

Using Raoult's law for ideal solution:

p_total = p_heptane⁰ × x_heptane + p_octane⁰ × x_octane

p_total = 105.2 × 0.459 + 46.8 × 0.541

p_total = 48.29 + 25.32

p_total = 73.61 kPa

Answer: Vapor pressure of mixture = 73.61 kPa (approximately 73.6 kPa)

Question 1.17 The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it.

Solution:

Given:

• p₀ = 12.3 kPa
• Molality = 1 m = 1 mol kg^-1

1 molal solution means 1 mole of solute is dissolved in 1 kg (1000 g) of water

Moles of water = 1000 / 18 = 55.56 mol

Moles of solute = 1 mol

Mole fraction of solute = 1 / (1 + 55.56) = 1 / 56.56 = 0.0177

Using Raoult's law:

(p₀ - p) / p₀ = x_solute

(12.3 - p) / 12.3 = 0.0177

12.3 - p = 0.0177 × 12.3

12.3 - p = 0.218

p = 12.3 - 0.218 = 12.082 kPa

Answer: Vapor pressure of solution = 12.08 kPa (approximately 12.1 kPa)

Question 1.18 Calculate the mass of a non-volatile solute (molar mass 40 g mol^-1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.

Solution:

Given:

• Molar mass of solute (M₂) = 40 g mol^-1
• Mass of octane (w₁) = 114 g
• Vapor pressure of solution = 80% of pure octane

This means: p = 0.80 × p₀

Relative lowering of vapor pressure:

(p₀ - p) / p₀ = (p₀ - 0.80p₀) / p₀ = 0.20

Molar mass of octane (C₈H₁₈) (M₁) = 114 g mol^-1

Let mass of solute = w₂

Using Raoult's law:

(p₀ - p) / p₀ = (w₂ × M₁) / (M₂ × w₁)

0.20 = (w₂ × 114) / (40 × 114)

0.20 = w₂ / 40

w₂ = 0.20 × 40 = 8 g

Answer: Mass of solute required = 8 g

Question 1.19 A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

(i) molar mass of the solute

(ii) vapour pressure of water at 298 K.

Solution:

Initial Solution:

• Mass of solute (w₂) = 30 g
• Mass of water (w₁) = 90 g
• Vapor pressure p₁ = 2.8 kPa

After adding water:

• Mass of solute = 30 g (same)
• Mass of water = 90 + 18 = 108 g
• Vapor pressure p₂ = 2.9 kPa

Let molar mass of solute = M g mol^-1

Let vapor pressure of pure water = p₀

Molar mass of water = 18 g mol^-1

For initial solution:

Moles of solute = 30/M

Moles of water = 90/18 = 5

Mole fraction of solute = (30/M) / (30/M + 5)

Using Raoult's law:

(p₀ - 2.8) / p₀ = (30/M) / (30/M + 5)

... (1)

For final solution:

Moles of solute = 30/M

Moles of water = 108/18 = 6

Mole fraction of solute = (30/M) / (30/M + 6)

(p₀ - 2.9) / p₀ = (30/M) / (30/M + 6)

... (2)

From equation (1):

(p₀ - 2.8) / p₀ = (30/M) / (30/M + 5)

p₀(30/M) = (p₀ - 2.8)(30/M + 5)

30p₀/M = 30p₀/M + 5p₀ - 84/M - 14

5p₀ - 14 = 84/M

... (3)

From equation (2):

(p₀ - 2.9) / p₀ = (30/M) / (30/M + 6)

30p₀/M = 30p₀/M + 6p₀ - 87/M - 17.4

6p₀ - 17.4 = 87/M

... (4)

Dividing equation (4) by equation (3):

(6p₀ - 17.4) / (5p₀ - 14) = 87/84

84(6p₀ - 17.4) = 87(5p₀ - 14)

504p₀ - 1461.6 = 435p₀ - 1218

69p₀ = 243.6

p₀ = 3.53 kPa

Substituting in equation (3):

5(3.53) - 14 = 84/M

17.65 - 14 = 84/M

3.65 = 84/M

M = 84 / 3.65 = 23 g mol^-1

Answer: (i) Molar mass of solute = 23 g mol^-1

(ii) Vapor pressure of water at 298 K = 3.53 kPa

Question 1.20 A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Solution:

Given:

• For cane sugar solution: Mass % = 5%, T_f = 271 K
• For glucose solution: Mass % = 5%
• T_f⁰ (water) = 273.15 K

For cane sugar (C₁₂H₂₂O₁₁) solution:

Depression in freezing point = 273.15 - 271 = 2.15 K

Assume 100 g solution:

Mass of cane sugar = 5 g

Mass of water = 95 g = 0.095 kg

Molar mass of cane sugar = 342 g mol^-1

Moles of cane sugar = 5 / 342 = 0.0146 mol

Molality = 0.0146 / 0.095 = 0.154 mol kg^-1

Using: ΔT_f = K_f × m

2.15 = K_f × 0.154

K_f = 2.15 / 0.154 = 13.96 K kg mol^-1

Or we can use the standard value: K_f (water) = 1.86 K kg mol^-1

For glucose (C₆H₁₂O₆) solution:

Mass of glucose = 5 g

Mass of water = 95 g = 0.095 kg

Molar mass of glucose = 180 g mol^-1

Moles of glucose = 5 / 180 = 0.0278 mol

Molality = 0.0278 / 0.095 = 0.293 mol kg^-1

Using K_f = 1.86 K kg mol^-1:

ΔT_f = 1.86 × 0.293 = 0.545 K

Freezing point of glucose solution = 273.15 - 0.545 = 272.605 K

Alternatively, using ratio method:

ΔT_f is proportional to (mass/molar mass) for same mass of solvent

ΔT_f(glucose) / ΔT_f(sugar) = (5/180) / (5/342)

ΔT_f(glucose) / 2.15 = 342/180 = 1.9

ΔT_f(glucose) = 2.15 × 1.9 = 4.085 K

Freezing point = 273.15 - 4.085 = 269.065 K

Note: The answer depends on whether we use the calculated K_f from cane sugar data or standard K_f value.

Answer: Using standard K_f: Freezing point ≈ 272.6 K

Question 1.21 Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.

Solution:

Given:

• K_f = 5.1 K kg mol^-1
• Mass of benzene = 20 g = 0.020 kg
• For AB₂: mass = 1 g, ΔT_f = 2.3 K
• For AB₄: mass = 1 g, ΔT_f = 1.3 K

For AB₂:

Using: ΔT_f = (K_f × w₂ × 1000) / (M₂ × w₁)

2.3 = (5.1 × 1 × 1000) / (M_AB₂ × 20)

M_AB₂ = (5.1 × 1000) / (2.3 × 20)

M_AB₂ = 5100 / 46 = 110.87 g mol^-1

For AB₄:

1.3 = (5.1 × 1 × 1000) / (M_AB₄ × 20)

M_AB₄ = (5.1 × 1000) / (1.3 × 20)

M_AB₄ = 5100 / 26 = 196.15 g mol^-1

Let atomic mass of A = a, atomic mass of B = b

For AB₂: a + 2b = 110.87 ... (1)

For AB₄: a + 4b = 196.15 ... (2)

Subtracting equation (1) from equation (2):

2b = 196.15 - 110.87 = 85.28

b = 42.64 ≈ 43

Substituting in equation (1):

a + 2(42.64) = 110.87

a = 110.87 - 85.28 = 25.59 ≈ 26

Answer: Atomic mass of A = 26 u, Atomic mass of B = 43 u

(Element A could be Mg or Fe, Element B could be related to a group element)

Question 1.22 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 bars at the same temperature, what would be its concentration?

Solution:

Given:

• Temperature T = 300 K
• When mass = 36 g in 1 L, π₁ = 4.98 bar
• When π₂ = 1.52 bar, concentration = ?

Using: π = CRT

where C = concentration in mol L^-1

Since T and R are constant:

π ∝ C

Therefore: π₁ / π₂ = C₁ / C₂

First, find C₁:

Molar mass of glucose = 180 g mol^-1

C₁ = 36 / 180 = 0.2 mol L^-1

Now:

4.98 / 1.52 = 0.2 / C₂

C₂ = (0.2 × 1.52) / 4.98

C₂ = 0.304 / 4.98

C₂ = 0.061 mol L^-1

Mass of glucose in 1 L = 0.061 × 180 = 10.98 g

Answer: Concentration = 0.061 M or 10.98 g/L

Question 1.23

Question: Suggest the most important type of intermolecular attractive interaction in the following pairs.

(i) n-hexane and n-octane

(ii) I₂ and CCl₄

(iii) NaClO₄ and water

(iv) methanol and acetone

(v) acetonitrile (CH₃CN) and acetone (C₃H₆O).

Solution:

(i) n-hexane and n-octane:

Type of interaction: London dispersion forces (Van der Waals forces)

Reason: Both are non-polar hydrocarbons. Only weak Van der Waals forces exist between them.

(ii) I₂ and CCl₄:

Type of interaction: London dispersion forces (Van der Waals forces)

Reason: Both are non-polar molecules. I₂ is non-polar and CCl₄ is also non-polar due to symmetrical tetrahedral structure.

(iii) NaClO₄ and water:

Type of interaction: Ion-dipole interaction

Reason: NaClO₄ is an ionic compound (Na⁺ and ClO₄^- ions) and water is a polar molecule with dipole moment. Strong ion-dipole attractions occur between ions and polar water molecules.

(iv) methanol and acetone:

Type of interaction: Hydrogen bonding

Reason: Methanol (CH₃OH) has -OH group which can form hydrogen bonds with the oxygen of acetone (C=O group). The H of -OH bonds with O of C=O.

(v) acetonitrile (CH₃CN) and acetone (C₃H₆O):

Type of interaction: Dipole-dipole interaction

Reason: Both are polar molecules. Acetonitrile has C≡N group (polar) and acetone has C=O group (polar). Strong dipole-dipole attractions exist between them.

Question 1.24 Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.

Solution:

n-Octane (C₈H₁₈) is a non-polar hydrocarbon solvent.

According to the principle "like dissolves like," non-polar solutes dissolve in non-polar solvents.

Analysis of solutes:

1. KCl:

• Ionic compound
• Will have very poor solubility in non-polar n-octane
• Least soluble

2. CH₃CN (Acetonitrile):

• Polar molecule (C≡N group)
• Has significant dipole moment
• Poor solubility in non-polar solvent

3. CH₃OH (Methanol):

• Polar molecule with -OH group
• Can form hydrogen bonds
• But has small hydrocarbon part (CH₃-)
• Low solubility in non-polar solvent

4. Cyclohexane (C₆H₁₂):

• Non-polar hydrocarbon
• Similar to n-octane in nature
• Will be highly soluble
• Most soluble

Order of increasing solubility in n-octane:

KCl < CH₃CN < CH₃OH < Cyclohexane

Or considering that methanol might be slightly less soluble than acetonitrile due to stronger H-bonding:

KCl < CH₃OH < CH₃CN < Cyclohexane

Explanation: Non-polar cyclohexane has maximum solubility in non-polar n-octane due to similar London dispersion forces. Polar molecules (methanol and acetonitrile) have poor solubility. Ionic KCl has negligible solubility.

Question 1.25 Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?

(i) phenol (ii) toluene (iii) formic acid

(iv) ethylene glycol (v) chloroform (vi) pentanol.

Solution:

Water is a polar solvent. Polar and ionic substances dissolve well in water due to hydrogen bonding and dipole interactions.

(i) Phenol (C₆H₅OH):

Partially soluble

Reason: Has -OH group which can form H-bonds with water, but large non-polar benzene ring reduces solubility. About 8-9 g dissolves in 100 mL water.

(ii) Toluene (C₆H₅CH₃):

Insoluble (or very slightly soluble)

Reason: Non-polar hydrocarbon with no H-bonding capability. Does not dissolve in polar water.

(iii) Formic acid (HCOOH):

Highly soluble

Reason: Small molecule with both -OH and C=O groups. Forms extensive H-bonds with water. Miscible in all proportions.

(iv) Ethylene glycol (HO-CH₂-CH₂-OH):

Highly soluble

Reason: Has two -OH groups which form strong H-bonds with water. Small hydrocarbon chain. Miscible in all proportions.

(v) Chloroform (CHCl₃):

Insoluble (or slightly soluble)

Reason: Non-polar molecule with weak polarity. Cannot form H-bonds. About 0.8 g/100 mL water.

(vi) Pentanol (C₅H₁₁OH):

Partially soluble

Reason: Has -OH group for H-bonding but large non-polar hydrocarbon chain (5 carbons) reduces solubility. About 2-3 g/100 mL water.

Summary:

• Highly soluble: Formic acid, Ethylene glycol
• Partially soluble: Phenol, Pentanol
• Insoluble: Toluene, Chloroform

Question 1.26 If the density of some lake water is 1.25 g mL^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molarity of Na⁺ ions in the lake.

Solution:

Given:

• Density of lake water = 1.25 g mL^-1
• 92 g Na⁺ per kg of water

Atomic mass of Na = 23 g mol^-1

Consider 1 kg of water:

Mass of water = 1000 g

Mass of Na⁺ = 92 g

Total mass of solution = 1000 + 92 = 1092 g

Volume of solution = Mass / Density

= 1092 / 1.25 = 873.6 mL = 0.8736 L

Moles of Na⁺ = 92 / 23 = 4 mol

Molarity = Moles / Volume in litres

= 4 / 0.8736

= 4.58 M

Answer: Molarity of Na⁺ ions = 4.58 M

Question 1.27 If the solubility product of CuS is 6 × 10^-16, calculate the maximum molarity of CuS in aqueous solution.

Solution:

Given: K_sp of CuS = 6 × 10^-16

CuS dissociates as:

CuS(s) ⇌ Cu²⁺(aq) + S^2-(aq)

Let solubility of CuS = s mol L^-1

Then: [Cu²⁺] = s and [S^2-] = s

Solubility product:

K_sp = [Cu²⁺][S^2-]

6 × 10^-16 = s × s = s²

s = √(6 × 10^-16)

s = 2.45 × 10^-8 mol L^-1

Answer: Maximum molarity of CuS = 2.45 × 10^-8 M

Question 1.28 Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.

Solution:

Given:

• Mass of aspirin = 6.5 g
• Mass of acetonitrile = 450 g

Total mass of solution = 6.5 + 450 = 456.5 g

Mass percentage of aspirin = (Mass of aspirin / Total mass) × 100

= (6.5 / 456.5) × 100

= 1.424%

Answer: Mass percentage of aspirin = 1.42% (approximately)

Question 1.29

Question: Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10^-3 m aqueous solution required for the above dose.

Solution:

Given:

• Dose required = 1.5 mg = 1.5 × 10^-3 g
• Molality of solution = 1.5 × 10^-3 m = 1.5 × 10^-3 mol kg^-1

Molar mass of nalorphene (C₁₉H₂₁NO₃):

= 19(12) + 21(1) + 14 + 3(16)

= 228 + 21 + 14 + 48

= 311 g mol^-1

Molality = 1.5 × 10^-3 mol kg^-1 means:

1.5 × 10^-3 mol nalorphene is present in 1 kg of water

Mass of nalorphene in 1 kg water = 1.5 × 10^-3 × 311 = 0.4665 g = 466.5 mg

Total mass of solution containing 466.5 mg nalorphene = 1000 + 0.4665 = 1000.4665 g

For 1.5 mg dose:

Mass of solution = (1000.4665 / 466.5) × 1.5

= 1000.4665 × 1.5 / 466.5

= 3.22 g

Answer: Mass of solution required = 3.22 g (approximately 3.2 g)

Question 1.30 Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.

Solution:

Given:

• Volume of solution = 250 mL = 0.250 L
• Molarity = 0.15 M

Molar mass of benzoic acid (C₆H₅COOH):

= 7(12) + 6(1) + 2(16)

= 84 + 6 + 32

= 122 g mol^-1

Moles of benzoic acid = Molarity × Volume in litres

= 0.15 × 0.250

= 0.0375 mol

Mass of benzoic acid = Moles × Molar mass

= 0.0375 × 122

= 4.575 g

Answer: Amount of benzoic acid required = 4.58 g (approximately 4.6 g)

Question 1.31 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.

Solution:

The three acids are:

• Acetic acid: CH₃COOH
• Trichloroacetic acid: CCl₃COOH
• Trifluoroacetic acid: CF₃COOH

Depression in freezing point order:

CH₃COOH < CCl₃COOH < CF₃COOH

Explanation:

Depression in freezing point (ΔT_f) is a colligative property that depends on the number of particles in solution:

ΔT_f = i × K_f × m

where i = van't Hoff factor (number of particles per molecule)

1. Acetic acid (CH₃COOH):

• Weak acid (K_a = 1.8 × 10^-5)
• Partially ionizes: CH₃COOH ⇌ H⁺ + CH₃COO^-
• Degree of dissociation is low
• i is slightly greater than 1 (low ionization)
• Produces fewer particles

2. Trichloroacetic acid (CCl₃COOH):

• Stronger acid than acetic acid
• Electron-withdrawing Cl atoms increase acidity
• Greater degree of dissociation
• i is larger (more ionization)
• Produces more particles than acetic acid

3. Trifluoroacetic acid (CF₃COOH):

• Strongest acid among the three
• F is most electronegative, maximum electron-withdrawing effect
• Nearly complete dissociation (strong acid)
• i is close to 2 (almost complete ionization)
• Produces maximum number of particles

Acid strength order:

CH₃COOH < CCl₃COOH < CF₃COOH

Degree of dissociation order:

CH₃COOH < CCl₃COOH < CF₃COOH

Number of particles order:

CH₃COOH < CCl₃COOH < CF₃COOH

Therefore, depression in freezing point order:

CH₃COOH < CCl₃COOH < CF₃COOH

Conclusion: The more acidic the compound, the greater the dissociation, the more particles formed, and hence greater the depression in freezing point.

Question 1.32 Calculate the depression in the freezing point of water when 10 g of CH₃CH₂CHClCOOH is added to 250 g of water. K_a = 1.4 × 10^-3, K_f = 1.86 K kg mol^-1.

Solution:

Given:

• Mass of CH₃CH₂CHClCOOH = 10 g
• Mass of water = 250 g = 0.250 kg
• K_a = 1.4 × 10^-3
• K_f = 1.86 K kg mol^-1

Molar mass of CH₃CH₂CHClCOOH:

= 3(12) + 6(1) + 35.5 + 12 + 2(16) + 1

= 36 + 6 + 35.5 + 12 + 32 + 1

= 122.5 g mol^-1

Moles of acid = 10 / 122.5 = 0.0816 mol

Molality = 0.0816 / 0.250 = 0.327 mol kg^-1

Since the acid is weak and partially dissociates:

CH₃CH₂CHClCOOH ⇌ H⁺ + CH₃CH₂CHClCOO^-

For weak acid:

α = √(K_a/C)

where α = degree of dissociation, C = concentration

α = √(1.4 × 10^-3 / 0.327)

α = √(4.28 × 10^-3)

α = 0.0654

van't Hoff factor (i) = 1 + α

i = 1 + 0.0654 = 1.0654

Depression in freezing point:

ΔT_f = i × K_f × m

ΔT_f = 1.0654 × 1.86 × 0.327

ΔT_f = 0.648 K

Answer: Depression in freezing point = 0.648 K (approximately 0.65 K or 0.65°C)

Question 1.33 19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.

Solution:

Given:

• Mass of CH₂FCOOH = 19.5 g
• Mass of water = 500 g = 0.500 kg
• ΔT_f (observed) = 1.0 K
• K_f for water = 1.86 K kg mol^-1

Molar mass of CH₂FCOOH:

= 12 + 2(1) + 19 + 12 + 2(16) + 1

= 12 + 2 + 19 + 12 + 32 + 1

= 78 g mol^-1

Moles of acid = 19.5 / 78 = 0.25 mol

Molality = 0.25 / 0.500 = 0.50 mol kg^-1

Calculate theoretical ΔT_f (without dissociation):

ΔT_f (calculated) = K_f × m

= 1.86 × 0.50 = 0.93 K

van't Hoff factor (i):

i = ΔT_f (observed) / ΔT_f (calculated)

i = 1.0 / 0.93 = 1.075

Calculate degree of dissociation (α):

For weak acid dissociating into 2 particles:

CH₂FCOOH ⇌ H⁺ + CH₂FCOO^-

i = 1 + α

1.075 = 1 + α

α = 0.075

Calculate dissociation constant (K_a):

Initial concentration C = 0.50 M (approximately equal to molality for dilute solution)

At equilibrium:

[CH₂FCOOH] = C(1 - α) = 0.50(1 - 0.075) = 0.4625 M

[H⁺] = Cα = 0.50 × 0.075 = 0.0375 M

[CH₂FCOO^-] = Cα = 0.0375 M

K_a = [H⁺][CH₂FCOO^-] / [CH₂FCOOH]

K_a = (0.0375 × 0.0375) / 0.4625

K_a = 0.001406 / 0.4625

K_a = 3.04 × 10^-3

Or using simpler formula:

K_a = Cα² / (1 - α)

K_a = 0.50 × (0.075)² / (1 - 0.075)

K_a = 0.50 × 0.005625 / 0.925

K_a = 3.04 × 10^-3

Answer: van't Hoff factor (i) = 1.075; K_a = 3.04 × 10^-3

Question 1.34 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Solution:

Given:

• p₀ = 17.535 mm Hg
• Mass of glucose = 25 g
• Mass of water = 450 g

Molar mass of glucose (C₆H₁₂O₆) = 180 g mol^-1

Molar mass of water = 18 g mol^-1

Moles of glucose = 25 / 180 = 0.139 mol

Moles of water = 450 / 18 = 25 mol

Mole fraction of glucose = 0.139 / (0.139 + 25)

= 0.139 / 25.139 = 0.00553

Using Raoult's law:

(p₀ - p) / p₀ = x_glucose

(17.535 - p) / 17.535 = 0.00553

17.535 - p = 0.00553 × 17.535

17.535 - p = 0.097

p = 17.535 - 0.097 = 17.438 mm Hg

Answer: Vapour pressure of water = 17.44 mm Hg (approximately)

Question 1.35 Henry's law constant for the molality of methane in benzene at 298 K is 4.27 × 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution:

Given:

• K_H = 4.27 × 10⁵ mm Hg
• Pressure of methane = 760 mm Hg
• Temperature = 298 K

Using Henry's law: p = K_H × x

where x = mole fraction of methane

x = p / K_H

x = 760 / (4.27 × 10⁵)

x = 1.78 × 10^-3

This is the mole fraction of methane in benzene.

To express in molality (as given that K_H is for molality):

Consider 1 kg benzene (C₆H₆):

Molar mass of benzene = 78 g mol^-1

Moles of benzene = 1000 / 78 = 12.82 mol

Let moles of methane = n

x = n / (n + 12.82)

Since n << 12.82:

1.78 × 10^-3 ≈ n / 12.82

n = 1.78 × 10^-3 × 12.82

n = 0.0228 mol

Solubility = 0.0228 mol kg^-1 = 0.0228 m

Answer: Solubility of methane = 1.78 × 10^-3 (mole fraction) or 0.0228 mol kg^-1

Question 1.36 100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.

Solution:

Given:

• Mass of A = 100 g, M_A = 140 g mol^-1
• Mass of B = 1000 g, M_B = 180 g mol^-1
• p_B⁰ = 500 torr
• p_total = 475 torr

Moles of A = 100 / 140 = 0.714 mol

Moles of B = 1000 / 180 = 5.556 mol

Total moles = 0.714 + 5.556 = 6.270 mol

Mole fraction of A = 0.714 / 6.270 = 0.114

Mole fraction of B = 5.556 / 6.270 = 0.886

Using Raoult's law for ideal solution:

p_total = p_A⁰ × x_A + p_B⁰ × x_B

475 = p_A⁰ × 0.114 + 500 × 0.886

475 = 0.114 p_A⁰ + 443

0.114 p_A⁰ = 32

p_A⁰ = 32 / 0.114 = 280.7 torr

Vapour pressure of A in solution:

p_A = p_A⁰ × x_A

p_A = 280.7 × 0.114 = 32 torr

Answer: Vapour pressure of pure A = 280.7 torr (approximately 281 torr)

Vapour pressure of A in solution = 32 torr

Question 1.37 Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot p_total, p_chloroform, and p_acetone as a function of x_acetone. The experimental data observed for different compositions of mixture is given. Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Solution:

Given:

• p_acetone⁰ = 741.8 mm Hg
• p_chloroform⁰ = 632.8 mm Hg

For Ideal Solution (Raoult's Law):

p_acetone = p_acetone⁰ × x_acetone = 741.8 × x_acetone

p_chloroform = p_chloroform⁰ × x_chloroform = 632.8 × (1 - x_acetone)

p_total = 741.8 x_acetone + 632.8(1 - x_acetone)

p_total = 632.8 + 109 x_acetone

Ideal Solution Calculations:

Experimental Data (Given):

Analysis:

Comparing experimental and ideal values:

• At x_acetone = 0.118: p_total(ideal) = 645.6, p_total(expt) = 603.0
• At x_acetone = 0.508: p_total(ideal) = 688.1, p_total(expt) = 580.4

The experimental vapor pressure is lower than the ideal vapor pressure at all compositions.

Conclusion:

The acetone-chloroform solution shows NEGATIVE DEVIATION from Raoult's law.

Reason: Acetone and chloroform form hydrogen bonds with each other:

CH₃-C(=O)-CH₃ ... H-CCl₃

The acetone-chloroform interactions are stronger than acetone-acetone and chloroform-chloroform interactions, leading to lower vapor pressure.

Question 1.38 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Solution:

Given:

• p_benzene⁰ = 50.71 mm Hg
• p_toluene⁰ = 32.06 mm Hg
• Mass of benzene = 80 g
• Mass of toluene = 100 g

Molar mass of benzene (C₆H₆) = 78 g mol^-1

Molar mass of toluene (C₇H₈) = 92 g mol^-1

Moles of benzene = 80 / 78 = 1.026 mol

Moles of toluene = 100 / 92 = 1.087 mol

Total moles = 1.026 + 1.087 = 2.113 mol

In liquid phase:

x_benzene = 1.026 / 2.113 = 0.486

x_toluene = 1.087 / 2.113 = 0.514

Partial pressures:

p_benzene = 50.71 × 0.486 = 24.65 mm Hg

p_toluene = 32.06 × 0.514 = 16.48 mm Hg

p_total = 24.65 + 16.48 = 41.13 mm Hg

In vapour phase:

y_benzene = p_benzene / p_total

y_benzene = 24.65 / 41.13 = 0.599

Answer: Mole fraction of benzene in vapour phase = 0.599 (approximately 0.6)

Question 1.39 The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.

Solution:

Given:

• Total pressure = 10 atm
• Volume % of O₂ = 20%
• Volume % of N₂ = 79%
• K_H(O₂) = 3.30 × 10⁷ mm Hg
• K_H(N₂) = 6.51 × 10⁷ mm Hg

Total pressure = 10 atm = 10 × 760 = 7600 mm Hg

Partial pressure of O₂ = 0.20 × 7600 = 1520 mm Hg

Partial pressure of N₂ = 0.79 × 7600 = 6004 mm Hg

For Oxygen:

Using Henry's law: p = K_H × x

x(O₂) = p(O₂) / K_H(O₂)

x(O₂) = 1520 / (3.30 × 10⁷)

x(O₂) = 4.606 × 10^-5

For Nitrogen:

x(N₂) = p(N₂) / K_H(N₂)

x(N₂) = 6004 / (6.51 × 10⁷)

x(N₂) = 9.222 × 10^-5

Ratio:

x(N₂) : x(O₂) = 9.222 × 10^-5 : 4.606 × 10^-5

= 9.222 : 4.606

= 2.00 : 1

Answer:

Mole fraction of O₂ in water = 4.61 × 10^-5

Mole fraction of N₂ in water = 9.22 × 10^-5

Ratio N₂ : O₂ = 2 : 1

Question 1.40 Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Solution:

Given:

• van't Hoff factor (i) = 2.47
• Volume of solution = 2.5 L
• Osmotic pressure (π) = 0.75 atm
• Temperature = 27°C = 300 K
• R = 0.0821 L atm K^-1 mol^-1

Using: π = i × C × R × T

where C = molarity

0.75 = 2.47 × C × 0.0821 × 300

0.75 = 2.47 × C × 24.63

0.75 = 60.84 × C

C = 0.75 / 60.84 = 0.01233 M

Moles of CaCl₂ = Molarity × Volume

= 0.01233 × 2.5 = 0.03082 mol

Molar mass of CaCl₂ = 40 + 2(35.5) = 111 g mol^-1

Mass of CaCl₂ = 0.03082 × 111 = 3.42 g

Answer: Amount of CaCl₂ = 3.42 g (approximately 3.4 g)

Question 1.41 Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated.

Solution:

Given:

• Mass of K₂SO₄ = 25 mg = 0.025 g
• Volume = 2 L
• Temperature = 25°C = 298 K
• R = 0.0821 L atm K^-1 mol^-1

K₂SO₄ dissociates as:

K₂SO₄ → 2K⁺ + SO₄^2-

Total particles = 3

Therefore, i = 3

Molar mass of K₂SO₄ = 2(39) + 32 + 4(16) = 174 g mol^-1

Moles of K₂SO₄ = 0.025 / 174 = 1.437 × 10^-4 mol

Using: π = i × (n/V) × R × T

π = 3 × (1.437 × 10^-4 / 2) × 0.0821 × 298

π = 3 × 7.185 × 10^-5 × 0.0821 × 298

π = 3 × 7.185 × 10^-5 × 24.47

π = 5.274 × 10^-3 atm

π = 5.27 × 10^-3 atm

Converting to Pascals:

π = 5.27 × 10^-3 × 101325 Pa

π = 534 Pa

Answer: Osmotic pressure = 5.27 × 10^-3 atm or 534 Pa