NCERT Solutions for Class 12 Chemistry Chapter 2 – Solutions

NCERT Textbook Questions Solved

Question 2.1

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

Answer:
Total mass of solution = 22 g (benzene) + 122 g (CCl₄) = 144 g

• Mass % of benzene = (22 / 144) × 100 = 15.28%

• Mass % of CCl₄ = (122 / 144) × 100 = 84.72%

Question 2.2

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

Answer:
Given:

• 30 g of benzene (C₆H₆), 70 g of CCl₄ (in 100 g solution)

• Molar mass of benzene = 78 g/mol

• Molar mass of CCl₄ = 154 g/mol

Moles of C₆H₆ = 30 / 78 = 0.385 mol
Moles of CCl₄ = 70 / 154 = 0.454 mol

Total moles = 0.385 + 0.454 = 0.839 mol

Mole fraction of benzene = 0.385 / 0.839 = 0.459

Question 2.3

Calculate the molarity 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

Answer:
(a) Molar mass of Co(NO₃)₂·6H₂O = 291 g/mol
Moles = 30 / 291 = 0.103 mol
Molarity = 0.103 / 4.3 = 0.0239 M

(b) Using dilution formula M₁V₁ = M₂V₂
M₂ = (0.5 × 30) / 500 = 0.03 M

Question 2.4

Calculate the mass of urea (NH₂CONH₂) required to prepare 2.5 kg of 0.25 molal aqueous solution.

Answer:
Molality (m) = 0.25 mol/kg solvent
Let mass of water (solvent) = x kg
Then:
0.25 = n / x → n = 0.25 mol
Molar mass of urea = 60 g/mol
Mass of urea = 0.25 × 60 = 15 g
Solution mass = 1000 g (solvent) + 15 g (solute) = 1015 g = 1.015 kg

Now for 2.5 kg of solution:
15 / 1.015 × 2.5 = ~37 g urea

Question 2.5

A 20% (mass/mass) aqueous KI solution has a density of 1.202 g/mL. Calculate:

(a) Molality
(b) Molarity
(c) Mole fraction of KI

Answer:

(a) Molality:
Mass of KI = 20 g
Mass of water = 100 – 20 = 80 g = 0.08 kg
Molar mass of KI = 166 g/mol
Molality = (20 / 166) / 0.08 = 1.51 mol/kg

(b) Molarity:
Mass of solution = 100 g
Volume = 100 / 1.202 = 83.2 mL = 0.0832 L
Moles of KI = 20 / 166 = 0.1205 mol
Molarity = 0.1205 / 0.0832 = 1.45 M

(c) Mole fraction of KI:
Moles of water = 80 / 18 = 4.44 mol
Mole fraction of KI = 0.1205 / (0.1205 + 4.44) = 0.0264

Question 2.6

If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.

Answer:
Using:
$p=KH\cdot x$
At STP: Pressure (p) = 1 atm = 101325 Pa
x (mole fraction) = 0.195 mol / (0.195 + 55.5) mol ≈ 0.0035

KH = p / x = 101325 / 0.0035 = 2.90 × 10⁷ Pa

Question 2.7

Henry’s law constant for CO₂ in water at 298 K is 1.67 × 10⁸ Pa. Calculate the amount of CO₂ dissolved in 500 mL soda water packed under 2.5 atm CO₂ pressure at 298 K.

Answer:
p = 2.5 atm = 2.5 × 101325 = 253312.5 Pa
Using: x=pKH=253312.51.67×108=1.516×10−3x = \frac{p}{K_H} = \frac{253312.5}{1.67 \times 10^8} = 1.516 \times 10^{-3}x=KH​p​=1.67×108253312.5​=1.516×10−3

Moles of water = 500 / 18 = 27.78 mol
Moles of CO₂ = 1.516 × 10⁻³ × 27.78 ≈ 0.0421 mol
Mass of CO₂ = 0.0421 × 44 = 1.85 g

Question 2.8

Vapour pressures of pure liquids A and B at 350 K are 450 mmHg and 700 mmHg, respectively. If total vapour pressure of mixture is 600 mmHg, find the liquid and vapour phase compositions.

Answer:
Let mole fraction of A = x
Using Raoult’s law:
P = x × 450 + (1 − x) × 700 = 600
450x + 700 − 700x = 600 → −250x = −100 → x = 0.4

Mole fraction of A in liquid = 0.4, B = 0.6
In vapour phase:
pA=0.4×450=180 mmHgp_A = 0.4 \times 450 = 180 \text{ mmHg}pA​=0.4×450=180 mmHg
pB=0.6×700=420 mmHgp_B = 0.6 \times 700 = 420 \text{ mmHg}pB​=0.6×700=420 mmHg

Total = 600 mmHg
Mole fraction in vapour:

• A = 180 / 600 = 0.3

• B = 420 / 600 = 0.7

Question 2.9

The vapour pressure of pure water at 298 K is 23.8 mmHg. 50 g of urea is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and relative lowering.

Answer:
Moles of urea = 50 / 60 = 0.833 mol
Moles of water = 850 / 18 = 47.22 mol

Mole fraction of water = 47.22 / (47.22 + 0.833) = 0.9827
Vapour pressure = 0.9827 × 23.8 = 23.38 mmHg
Relative lowering = (23.8 − 23.38) / 23.8 = 0.0176 or 1.76%

Question 2.10

Boiling point of water at 750 mmHg is 99.63°C. How much sucrose is added to 500 g of water so that it boils at 100°C?
Given: Kb for water = 0.52 K kg/mol

Answer:
ΔTb = 100 − 99.63 = 0.37°C
Molality = ΔTb / Kb = 0.37 / 0.52 = 0.711 mol/kg
Moles of sucrose = 0.711 × 0.5 = 0.3555 mol
Mass = 0.3555 × 342 = 121.5 g

Question 2.11

Calculate the mass of ascorbic acid (C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf for CH₃COOH = 3.9 K kg/mol

Answer:
ΔTf = Kf × m → m = ΔTf / Kf = 1.5 / 3.9 = 0.3846 mol/kg
Mass of solvent = 75 g = 0.075 kg
Moles of solute = 0.3846 × 0.075 = 0.0288 mol
Mass = 0.0288 × 176 = 5.07 g

Question 2.12

Calculate the osmotic pressure exerted by a solution made by dissolving 1.0 g of polymer (molar mass = 185,000 g/mol) in 450 mL water at 37°C.

Answer:
Moles of solute = 1 / 185000 = 5.41 × 10⁻⁶ mol
Volume = 0.45 L
T = 37°C = 310 K
Using formula:
Π=nVRT=5.41×10−60.45×0.0821×310=3.06×10−4 atm\Pi = \frac{n}{V}RT = \frac{5.41 \times 10^{-6}}{0.45} \times 0.0821 \times 310 = 3.06 \times 10^{-4} \text{ atm}Π=Vn​RT=0.455.41×10−6​×0.0821×310=3.06×10−4 atm
In Pascals:
= 3.06 × 10⁻⁴ × 101325 = 31 Pa

NCERT Class 12 Chemistry Chapter 2 – Solutions: Exercise Questions with Answers

2.1. 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 chemically non-reacting substances. Based on the physical states of solute and solvent, solutions are classified into the following nine types:

2.2. Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be?

Answer:
Such a solution is likely to be an interstitial solid solution, where smaller particles fit into the interstitial spaces between the larger atoms of the host crystal structure.

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

Answer:
(i) Mole Fraction – It is the ratio of the number of moles of a component to the total moles of all components in the solution.
If n₁ is the moles of solute and n₂ of solvent:
Mole fraction of solute=n1n1+n2\text{Mole fraction of solute} = \frac{n_1}{n_1 + n_2}Mole fraction of solute=n1​+n2​n1​​

(ii) Molality (m) – It is the number of moles of solute present in 1 kg (1000 g) of the solvent.

Molality=Moles of soluteMass of solvent (kg)\text{Molality} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}Molality=Mass of solvent (kg)Moles of solute​

Note: Molality is independent of temperature since it is based on mass.

(iii) Molarity (M) – It is the number of moles of solute present in 1 litre of solution.

Molarity=Moles of soluteVolume of solution (L)\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}Molarity=Volume of solution (L)Moles of solute​

Note: Molarity changes with temperature as volume expands or contracts.

(iv) Mass Percentage – It is the mass of solute in 100 g of the solution.

\text{Mass %} = \left(\frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100

2.4. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504 g mL⁻¹?

Answer:
Given:

• Mass percentage of HNO₃ = 68%

• Density of solution = 1.504 g/mL = 1504 g/L

• Molar mass of HNO₃ = 63 g/mol

Mass of HNO₃ in 1 L solution = 68% of 1504 g = 1022.72 g
Moles of HNO₃ = 1022.72 / 63 = 16.23 mol

∴ Molarity = 16.23 M

2.5. 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?

Answer:

• 10% w/w glucose = 10 g glucose + 90 g water

• Molar mass of glucose (C₆H₁₂O₆) = 180 g/mol

• Moles of glucose = 10 / 180 = 0.0556 mol

• Mass of water = 90 g = 0.09 kg

Molality (m) = 0.0556 / 0.09 = 0.617 mol/kg

Mole fraction of glucose:
Moles of water = 90 / 18 = 5
Mole fraction = 0.0556 / (0.0556 + 5) ≈ 0.011 (glucose)

Molarity:
Total mass of solution = 100 g
Density = 1.2 g/mL ⇒ Volume = 100 / 1.2 = 83.33 mL = 0.0833 L
Molarity = 0.0556 / 0.0833 = 0.667 M

2.6. 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?

Answer:

Let’s assume the mixture contains 0.5 g of Na₂CO₃ and 0.5 g of NaHCO₃ (equal moles).

• Molar mass of Na₂CO₃ = 106 g/mol
  ⇒ Moles of Na₂CO₃ = 0.5 / 106 ≈ 0.00472 mol

• Molar mass of NaHCO₃ = 84 g/mol
  ⇒ Moles of NaHCO₃ = 0.5 / 84 ≈ 0.00595 mol

Balanced reactions:

• Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O

• NaHCO₃ + HCl → NaCl + CO₂ + H₂O

⇒ HCl required = 2 × 0.00472 + 0.00595 = 0.01539 mol

Now, using molarity (M = moles/volume in L):

• Volume = moles / molarity = 0.01539 / 0.1 = 0.1539 L = 153.9 mL

2.7. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.

Answer:

• Mass of solute in 300 g of 25% solution = 25% of 300 = 75 g

• Mass of solute in 400 g of 40% solution = 40% of 400 = 160 g

• Total solute = 75 + 160 = 235 g

• Total solution mass = 300 + 400 = 700 g

% Composition by mass = (235 / 700) × 100 = 33.57%

2.8. 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?

Answer:

Step 1: Molality

• Molar mass of ethylene glycol = 62 g/mol

• Moles of C₂H₆O₂ = 222.6 / 62 ≈ 3.59 mol

• Mass of water = 200 g = 0.2 kg

Molality = 3.59 / 0.2 = 17.95 mol/kg

Step 2: Molarity

• Total mass = 222.6 + 200 = 422.6 g

• Volume of solution = 422.6 / 1.072 ≈ 394 mL = 0.394 L

Molarity = 3.59 / 0.394 ≈ 9.11 M

2.9. 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.**

Answer:

(i) 15 ppm = 15 parts per million = 15 g in 10⁶ g of solution
⇒ % by mass = (15 / 1,000,000) × 100 = 0.0015%

(ii) Molality:

• Mass of CHCl₃ = 15 g

• Mass of water = 10⁶ – 15 ≈ 999,985 g ≈ 1000 kg

• Molar mass of CHCl₃ = 119.5 g/mol

• Moles = 15 / 119.5 ≈ 0.1255 mol

Molality = 0.1255 / 1000 = 1.255 × 10⁻⁴ mol/kg

2.10. What role does the molecular interaction play in solution of alcohol in water?

Answer:

Both alcohol and water form hydrogen bonds due to the presence of –OH groups. However, when alcohol is mixed with water, the new hydrogen bonding between alcohol and water is somewhat weaker than that in pure components. This reduces the overall intermolecular attraction, leading to positive deviation from Raoult’s Law, increased vapour pressure, and lower boiling point of the solution.

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

Answer:

Dissolution of gases in liquids is an exothermic process. When temperature increases, the solubility decreases due to Le Chatelier’s principle—heat addition shifts the equilibrium to the gas phase, reducing solubility.

2.12. State Henry’s law and mention some of its important applications.

Answer:

Henry’s Law: The solubility of a gas in a liquid at a given temperature is directly proportional to the pressure of the gas above the liquid.
Mathematically: p = KH × x,
where p is the partial pressure of the gas, x is mole fraction, and KH is Henry’s constant.

Applications:

1. Carbonated drinks are bottled under high pressure to increase CO₂ solubility.

2. Scuba divers use a helium–oxygen mix to avoid nitrogen narcosis.

3. Oxyhaemoglobin formation in blood is influenced by oxygen pressure, helping in oxygen transport in the body.

2.13. 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?

Answer:

Since pressure is directly proportional to mole fraction, and the number of moles of ethane increases:

Let initial mass = 6.56 × 10⁻³ g,
New mass = 5.00 × 10⁻² g

Ratio = (5.00 × 10⁻²) / (6.56 × 10⁻³) ≈ 7.62

New pressure = 1 × 7.62 = 7.62 bar

2.14. According to Raoult’s law, what is meant by positive and negative deviations and how is the sign of ∆solH related to positive and negative deviations from Raoult’s law?

Answer:

• Positive deviation: Occurs when A–B interactions are weaker than A–A or B–B. Results in increased vapour pressure and ∆solH > 0 (endothermic).

• Negative deviation: A–B interactions are stronger than A–A or B–B. Leads to reduced vapour pressure and ∆solH < 0 (exothermic).

2.15. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molecular mass of the solute?

Answer:

Using Raoult’s law:
Relative lowering of vapour pressure = ∆P / P⁰ = w₂ × 1000 / (M₂ × w₁)

Given:

• ∆P = P⁰ – P = 1.013 – 1.004 = 0.009 bar

• w₂ = 2 g (solute), w₁ = 98 g (solvent)

⇒ 0.009 / 1.013 = 2 × 1000 / (M₂ × 98)
Solving gives: M₂ ≈ 41.35 g/mol

2.16. 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 35.0 g of octane?

Answer:

Molar masses:

• Heptane (C₇H₁₆) = 100 g/mol

• Octane (C₈H₁₈) = 114 g/mol

Moles:

• Heptane = 26 / 100 = 0.26 mol

• Octane = 35 / 114 ≈ 0.307 mol

Mole fractions:

• Heptane = 0.26 / (0.26 + 0.307) ≈ 0.458

• Octane = 1 – 0.458 = 0.542

Using Raoult’s Law:
P_total = (0.458 × 105.2) + (0.542 × 46.8) ≈ 48.19 + 25.37 = 73.56 kPa

2.17. 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.

Answer:

Using Raoult’s Law:

Relative lowering = n_solute / (n_solute + n_solvent)

In 1 molal solution:

• n_solute = 1 mol

• Water = 1000 g = 55.56 mol

⇒ Relative lowering = 1 / (1 + 55.56) = 0.01768

∆P = 12.3 × 0.01768 ≈ 0.217 kPa

P_solution = 12.3 – 0.217 = 12.083 kPa