Class 12 Chemistry: Solutions Chapter Test Paper 2025 (With Answers)
Preparing for CBSE Class 12 Chemistry Board Exam 2025? "Solutions" (Chapter 2) is a high-scoring physical chemistry chapter involving Colligative Properties and Numericals. Below is a comprehensive 40 Marks Practice Test Paper covering Previous Year Questions (PYQs), important Reasoning questions, and Numericals. Click on the dropdowns to check the solutions instantly.
SECTION A (1 Mark Each)
Q1. Which of the following is a colligative property?
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Correct Answer: (c) Osmotic pressure
Note: Relative lowering of vapor pressure, Elevation in boiling point, and Depression in freezing point are colligative properties, not the absolute values.
Note: Relative lowering of vapor pressure, Elevation in boiling point, and Depression in freezing point are colligative properties, not the absolute values.
Q2. The value of Van't Hoff factor (i) for K₂SO₄ assuming complete dissociation is:
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Correct Answer: (c) 3
Reaction: K₂SO₄ → 2K⁺ + SO₄²⁻. Total ions = 2 + 1 = 3.
Reaction: K₂SO₄ → 2K⁺ + SO₄²⁻. Total ions = 2 + 1 = 3.
Q3. Low concentration of oxygen in the blood and tissues of people living at high altitudes is due to:
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Correct Answer: (b) Low atmospheric pressure
Application of Henry's Law (Anoxia).
Application of Henry's Law (Anoxia).
Q4. A solution of acetone and ethanol shows:
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Correct Answer: (a) Positive deviation from Raoult's Law
Acetone breaks the strong hydrogen bonding between ethanol molecules, increasing vapor pressure.
Acetone breaks the strong hydrogen bonding between ethanol molecules, increasing vapor pressure.
Q5. The unit of ebullioscopic constant (Kb) is:
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Correct Answer: (a) K kg mol⁻¹
SECTION B (2 Marks Each)
Q6. State Raoult’s law for a solution of volatile liquids. How does the plot of Vapor Pressure vs Mole Fraction look for an ideal solution?
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Statement: For a solution of volatile liquids, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction present in the solution.
p₁ = p₁°x₁ and p₂ = p₂°x₂.
Graph: It is a linear plot (straight line) passing through the origin if plotted individually.
p₁ = p₁°x₁ and p₂ = p₂°x₂.
Graph: It is a linear plot (straight line) passing through the origin if plotted individually.
Q7. Why is the osmotic pressure method preferred for the determination of molar masses of macromolecules like proteins and polymers?
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1. Osmotic pressure (Ï€) can be measured at room temperature, preventing denaturation of proteins (unlike boiling/freezing methods).
2. Its magnitude is large even for very dilute solutions, making measurement accurate.
2. Its magnitude is large even for very dilute solutions, making measurement accurate.
Q8. 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. (Kf = 3.9 K kg mol⁻¹)
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Formula: ΔTf = (Kf × w₂ × 1000) / (M₂ × w₁)
Given: ΔTf = 1.5, Kf = 3.9, w₁ = 75g.
Molar mass of C₆H₈O₆ (M₂) = 176 g mol⁻¹.
1.5 = (3.9 × w₂ × 1000) / (176 × 75)
w₂ = (1.5 × 176 × 75) / (3.9 × 1000)
w₂ = 5.08 g.
Given: ΔTf = 1.5, Kf = 3.9, w₁ = 75g.
Molar mass of C₆H₈O₆ (M₂) = 176 g mol⁻¹.
1.5 = (3.9 × w₂ × 1000) / (176 × 75)
w₂ = (1.5 × 176 × 75) / (3.9 × 1000)
w₂ = 5.08 g.
Q9. Define Azeotropes. Give one example each of minimum boiling and maximum boiling azeotropes.
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Azeotropes: Binary mixtures having the same composition in liquid and vapor phase and boil at a constant temperature.
1. Minimum Boiling: Ethanol + Water (shows +ve deviation).
2. Maximum Boiling: Nitric Acid + Water (shows -ve deviation).
1. Minimum Boiling: Ethanol + Water (shows +ve deviation).
2. Maximum Boiling: Nitric Acid + Water (shows -ve deviation).
Q10. Calculate the Van't Hoff factor for a solution if the degree of dissociation of MgCl₂ is 80%.
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Reaction: MgCl₂ → Mg²⁺ + 2Cl⁻ (n = 3)
Formula: α = (i - 1) / (n - 1)
0.80 = (i - 1) / (3 - 1)
0.80 × 2 = i - 1
1.6 = i - 1 ⇒ i = 2.6
Formula: α = (i - 1) / (n - 1)
0.80 = (i - 1) / (3 - 1)
0.80 × 2 = i - 1
1.6 = i - 1 ⇒ i = 2.6
SECTION C (3 Marks Each)
Q11. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose (C₁₂H₂₂O₁₁) is to be added to 500 g of water such that it boils at 100°C? (Kb for water = 0.52 K kg mol⁻¹)
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ΔTb = 100 - 99.63 = 0.37°C (or K)
Molar mass of Sucrose (M₂) = 342 g mol⁻¹.
Formula: ΔTb = (Kb × w₂ × 1000) / (M₂ × w₁)
0.37 = (0.52 × w₂ × 1000) / (342 × 500)
w₂ = (0.37 × 342 × 500) / (0.52 × 1000)
w₂ = 63270 / 520 = 121.67 g.
Molar mass of Sucrose (M₂) = 342 g mol⁻¹.
Formula: ΔTb = (Kb × w₂ × 1000) / (M₂ × w₁)
0.37 = (0.52 × w₂ × 1000) / (342 × 500)
w₂ = (0.37 × 342 × 500) / (0.52 × 1000)
w₂ = 63270 / 520 = 121.67 g.
Q12. Give reasons for the following:
(i) Aquatic species are more comfortable in cold water than in warm water.
(ii) Measurement of osmotic pressure is preferred over other colligative properties.
(iii) Salt is sprinkled on roads to clear snow.
(i) Aquatic species are more comfortable in cold water than in warm water.
(ii) Measurement of osmotic pressure is preferred over other colligative properties.
(iii) Salt is sprinkled on roads to clear snow.
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(i) According to Henry’s law, solubility of gases (Oxygen) decreases with increase in temperature. Cold water has more dissolved oxygen.
(ii) It gives accurate results for macromolecules at room temperature.
(iii) Salt acts as a non-volatile solute and depresses the freezing point of water, helping ice to melt at lower temperatures.
(ii) It gives accurate results for macromolecules at room temperature.
(iii) Salt acts as a non-volatile solute and depresses the freezing point of water, helping ice to melt at lower temperatures.
Q13. 200 cm³ of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10⁻³ bar. Calculate the molar mass of the protein. (R = 0.083 L bar mol⁻¹ K⁻¹)
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Formula: Ï€ = CRT = (w₂/M₂) × (1/V) × RT
Or, M₂ = (w₂RT) / (Ï€V)
Given: w₂ = 1.26g, V = 0.2 L, T = 300K, Ï€ = 2.57 × 10⁻³ bar.
M₂ = (1.26 × 0.083 × 300) / (2.57 × 10⁻³ × 0.2)
M₂ = 31.374 / 0.000514
M₂ ≈ 61,038 g mol⁻¹
Or, M₂ = (w₂RT) / (Ï€V)
Given: w₂ = 1.26g, V = 0.2 L, T = 300K, Ï€ = 2.57 × 10⁻³ bar.
M₂ = (1.26 × 0.083 × 300) / (2.57 × 10⁻³ × 0.2)
M₂ = 31.374 / 0.000514
M₂ ≈ 61,038 g mol⁻¹
Q14. What are Non-ideal solutions? Draw the Vapor Pressure vs Mole Fraction graph for a solution showing Negative Deviation. Give one example.
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Non-ideal solutions: Solutions that do not obey Raoult's law over the entire range of concentrations.
Negative Deviation: The vapor pressure is lower than predicted (P < P°X). Intermolecular forces (A-B) are stronger than A-A and B-B.
Example: Chloroform + Acetone.
(Imagine a graph where the curves for Total Pressure and Partial Pressures dip below the straight ideal lines).
Negative Deviation: The vapor pressure is lower than predicted (P < P°X). Intermolecular forces (A-B) are stronger than A-A and B-B.
Example: Chloroform + Acetone.
(Imagine a graph where the curves for Total Pressure and Partial Pressures dip below the straight ideal lines).
Q15. Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 L of water such that its osmotic pressure is 0.75 atm at 27°C.
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Formula: π = iCRT = i (n/V) RT
n = (Ï€V) / (iRT)
Given: π=0.75, V=2.5, i=2.47, R=0.0821, T=300K.
n = (0.75 × 2.5) / (2.47 × 0.0821 × 300)
n = 1.875 / 60.836 = 0.0308 mol.
Mass = n × Molar Mass of CaCl₂ (111 g/mol)
Mass = 0.0308 × 111 = 3.42 g.
n = (Ï€V) / (iRT)
Given: π=0.75, V=2.5, i=2.47, R=0.0821, T=300K.
n = (0.75 × 2.5) / (2.47 × 0.0821 × 300)
n = 1.875 / 60.836 = 0.0308 mol.
Mass = n × Molar Mass of CaCl₂ (111 g/mol)
Mass = 0.0308 × 111 = 3.42 g.
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