Chapter Test: Solutions

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CBSE Class 12 Chapter Test: SOLUTIONS

Time Allowed: 1.5 Hours Maximum Marks: 35

General Instructions:

1. This question paper consists of 14 questions divided into 5 sections: A, B, C, D, and E.
2. Section A comprises 5 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
3. Section B comprises 2 Very Short Answer questions of 2 marks each.
4. Section C comprises 3 Short Answer questions of 3 marks each.
5. Section D comprises 1 Case-Based question of 4 marks.
6. Section E comprises 1 Long Answer question of 5 marks.
7. Use of calculators is not permitted. Log tables may be used if required.

Section A (1 Mark Each)

[1] 1. The value of Henry's law constant (K_H) for four gases Ar, CO₂, HCHO and CH₄ in water are 40.39, 1.67, 1.83×10^-5 and 0.413 kbar respectively at 298 K. The correct order of their increasing solubility in water is:

• Ar < CO₂ < CH₄ < HCHO
• HCHO < CH₄ < CO₂ < Ar
• Ar < CH₄ < CO₂ < HCHO
• HCHO < CO₂ < CH₄ < Ar

[1] 2. Which of the following aqueous solutions should have the highest boiling point? (Assume complete dissociation)

• 1.0 M NaOH
• 1.0 M Na₂SO₄
• 1.0 M NH₄NO₃
• 1.0 M KNO₃

[1] 3. A mixture of ethanol and acetone shows positive deviation from Raoult's Law. This is because:

• In pure ethanol, molecules are hydrogen bonded. Acetone molecules break these hydrogen bonds.
• Acetone and ethanol form new stronger hydrogen bonds.
• Acetone acts as a catalyst.
• The volume of the mixture decreases upon mixing.

[1] 4. Which condition is not satisfied by an ideal solution?

• ΔH_mix = 0
• ΔV_mix = 0
• Raoult's law is obeyed over the entire range of concentration
• ΔS_mix = 0

[1] 5. What is the van't Hoff factor (i) for a dilute aqueous solution of Potassium ferricyanide, K₃[Fe(CN)₆], assuming complete ionization?

• 3
• 4
• 5
• 2

Directions for Q6 & Q7: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:

• Both A and R are true and R is the correct explanation of A.
• Both A and R are true but R is not the correct explanation of A.
• A is true but R is false.
• A is false but R is true.

[1] 6. Assertion (A): The boiling point of 0.1 M urea solution is less than that of 0.1 M KCl solution.
Reason (R): Elevation of boiling point is directly proportional to the number of particles (ions or molecules) present in the solution.

[1] 7. Assertion (A): Osmotic pressure method is widely used to determine the molar masses of macromolecules like proteins and polymers.
Reason (R): Osmotic pressure is measured at room temperature and the magnitude is large even for very dilute solutions.

Section B (2 Marks Each)

[2] 8. State Raoult's law for a solution containing volatile components. How does it become a special case of Henry's law?

[2] 9. 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 for acetic acid = 3.9 K kg mol^-1, Atomic masses: C=12, H=1, O=16)

Section C (3 Marks Each)

[3] 10. A solution containing 15 g of urea (molar mass = 60 g mol^-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol^-1) in water at the same temperature. Calculate the mass of glucose present in one litre of its solution.

[3] 11. Explain the following with suitable examples:
(a) Minimum boiling azeotrope.
(b) What type of deviation from Raoult's law is shown by a mixture of chloroform and acetone? Give reason.

[3] 12. 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol^-1. Find the molar mass of the solute.

Section D: Case-Based (4 Marks)

[4] 13. Read the passage carefully and answer the questions that follow:

The solubility of gases in liquids is greatly affected by pressure and temperature. The relationship between pressure and solubility of a gas in a liquid was given by an English chemist. This law has several biological and industrial applications. For instance, aquatic species are more comfortable in cold water rather than in warm water. Furthermore, to avoid a dangerous medical condition known as 'bends', scuba divers carry cylinders containing breathing air diluted with a specific inert gas. Similarly, people living at high altitudes experience a condition called anoxia due to the low concentration of oxygen in their blood and tissues.

(a) Name the scientific law that governs the phenomena described in the passage. Write its mathematical expression. [1 Mark]
(b) Why are aquatic species more comfortable in cold water rather than warm water? [1 Mark]
(c) Name the inert gas used to dilute oxygen in scuba diver's tanks and explain why it is used instead of normal compressed air. [2 Marks]

Section E: Long Answer (5 Marks)

[5] 14. (a) Define Ebullioscopic constant (K_b) and state its SI unit. [1.5 Marks]

(b) 2 g of benzoic acid (C₆H₅COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol^-1. What is the percentage association of acid if it forms a dimer in solution? [3.5 Marks]

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SOLUTIONS & MARKING SCHEME

1. Answer: (a) Ar < CO₂ < CH₄ < HCHO

Reasoning: According to Henry's law, solubility of a gas is inversely proportional to its Henry's law constant (K_H). The K_H values are highest for Ar (40.39) and lowest for HCHO (1.83×10^-5). Therefore, Ar is least soluble and HCHO is most soluble.

2. Answer: (b) 1.0 M Na₂SO₄

Reasoning: Elevation in boiling point is a colligative property, meaning it depends on the number of particles (ions). Na₂SO₄ dissociates into 3 ions (2 Na⁺ and 1 SO₄^2-), giving it a van't Hoff factor (i) = 3. NaOH gives 2, NH₄NO₃ gives 2, and KNO₃ gives 2. Highest 'i' yields highest boiling point.

3. Answer: (a) In pure ethanol, molecules are hydrogen bonded. Acetone molecules break these hydrogen bonds.

Reasoning: The addition of acetone breaks some of the strong hydrogen bonds between ethanol molecules. As a result, the A-B interactions become weaker than A-A and B-B interactions, causing molecules to escape more easily and showing a positive deviation.

4. Answer: (d) ΔS_mix = 0

Reasoning: For any solution (ideal or non-ideal), mixing is a spontaneous process resulting in an increase in randomness. Therefore, the entropy of mixing (ΔS_mix) is always greater than zero, never zero.

5. Answer: (b) 4

Reasoning: K₃[Fe(CN)₆] dissociates completely into 3 K⁺ ions and 1 [Fe(CN)₆]^3- complex ion. Total number of ions produced from one molecule = 3 + 1 = 4. Hence, i = 4.

6. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: KCl dissociates into 2 ions (i=2), while urea is a non-electrolyte and does not dissociate (i=1). Because boiling point elevation depends on the number of particles, KCl will elevate the boiling point more than urea. Thus, urea's BP is less than KCl's BP.

7. Answer: (a) Both A and R are true and R is the correct explanation of A.

Reasoning: Macromolecules like proteins are unstable at high temperatures and have high molar masses (leading to very small freezing/boiling point changes). Osmotic pressure is measured at room temperature, and its magnitude is measurable even for very dilute solutions.

8. Solution:

Raoult's Law: For a solution of volatile liquids, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction present in solution.
p₁ = p₁° × x₁

Special Case of Henry's Law: According to Henry's law, the partial pressure of a gas in the liquid is proportional to its mole fraction: p = K_H × x. If we compare both equations, Raoult's law becomes a special case of Henry's law when the proportionality constant K_H becomes equal to the vapour pressure of the pure component (p₁°).

9. Solution:

Molar mass of Ascorbic acid (M₂) = (12×6) + (1×8) + (16×6) = 72 + 8 + 96 = 176 g mol^-1

Given: ΔT_f = 1.5 K (or °C), K_f = 3.9 K kg mol^-1, Mass of solvent (W₁) = 75 g = 0.075 kg

ΔT_f = K_f × m = K_f × [W₂ / (M₂ × W₁ in kg)]

1.5 = 3.9 × [W₂ / (176 × 0.075)]

W₂ = (1.5 × 176 × 0.075) / 3.9 = 5.077 g

10. Solution:

Isotonic solutions have the same osmotic pressure at the same temperature (π₁ = π₂). Since both are non-electrolytes, C₁ = C₂.

Concentration of Urea (C₁) = n_urea / V = (W_urea / M_urea) / V = (15 / 60) / 1 L = 0.25 M

Concentration of Glucose (C₂) = n_glucose / V = (W_glucose / 180) / 1 L

Equating C₁ and C₂:
0.25 = W_glucose / 180
W_glucose = 0.25 × 180 = 45 g

11. Solution:

(a) Minimum boiling azeotropes are solutions that show a large positive deviation from Raoult's law. At a specific composition, they boil at a constant temperature which is lower than the boiling points of both pure components. Example: 95% Ethanol and 5% water by volume.

(b) A mixture of chloroform (CHCl₃) and acetone (CH₃COCH₃) shows a negative deviation from Raoult's law.
Reason: Chloroform forms new, stronger hydrogen bonds with the oxygen atom of acetone molecules. Since A-B interactions (chloroform-acetone) are stronger than A-A and B-B interactions, the escaping tendency of molecules decreases, lowering the vapour pressure.

12. Solution:

Given: W₂ (mass of solute) = 1.00 g, W₁ (mass of benzene) = 50 g = 0.050 kg, ΔT_f = 0.40 K, K_f = 5.12 K kg mol^-1.

Formula: ΔT_f = K_f × (W₂ / M₂) × (1000 / W₁ in g)

0.40 = 5.12 × (1.00 / M₂) × (1000 / 50)

0.40 = (5.12 × 20) / M₂

M₂ = 102.4 / 0.40 = 256 g mol^-1

13. Solution (Case-Based):

(a) The scientific law is Henry's Law. Mathematical expression: p = K_H × x (where p is the partial pressure of the gas, x is its mole fraction in liquid, and K_H is Henry's constant).

(b) The solubility of gases (like oxygen) in liquids decreases with an increase in temperature. Therefore, cold water contains more dissolved oxygen compared to warm water, making aquatic species more comfortable.

(c) Helium is used. When divers breathe normal compressed air at high pressure underwater, large amounts of Nitrogen dissolve in the blood. As they ascend, pressure decreases, and Nitrogen bubbles out, blocking capillaries (causing a painful condition called 'bends'). Helium is used to dilute the oxygen because Helium is much less soluble in blood even at high pressures, preventing the bends.

14. Solution:

(a) Ebullioscopic constant (K_b): It is defined as the elevation in boiling point produced when 1 mole of a non-volatile solute is dissolved in 1 kg of the solvent (i.e., when molality is 1 m).
SI Unit: K kg mol^-1.

(b) Numerical:

Given: W₂ (benzoic acid) = 2 g, M₂ (benzoic acid) = 122 g mol^-1, W₁ (benzene) = 25 g, ΔT_f = 1.62 K, K_f = 4.9 K kg mol^-1.

Formula for abnormal mass: ΔT_f = i × K_f × m
where m = (W₂ × 1000) / (M₂ × W₁)

1.62 = i × 4.9 × [(2 × 1000) / (122 × 25)]

1.62 = i × 4.9 × 0.6557

1.62 = i × 3.213

i (van't Hoff factor) = 0.504

Benzoic acid forms a dimer in benzene. The reaction is:
2 C₆H₅COOH ⇔ (C₆H₅COOH)₂

Let initial moles be 1. After association, moles left = 1 - α, moles of dimer formed = α/2.
Total moles = 1 - α + α/2 = 1 - α/2

i = Total moles at equilibrium / Initial moles = 1 - α/2

0.504 = 1 - α/2
α/2 = 1 - 0.504 = 0.496
α = 0.992

Percentage Association = 99.2%

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