Solutions - Class 12 Chemistry

Class 12 Chemistry | Chapter 1

Solutions

Concentration Terms • Henry's Law • Raoult's Law • Colligative Properties • van't Hoff Factor

1. Introduction to Solutions

Solution: A homogeneous mixture of two or more substances whose composition can be varied within certain limits.

Solvent: Component present in larger quantity. Determines the physical state of the solution.
Solute: Component present in smaller quantity.
Binary Solution: Contains only two components (one solute, one solvent).

1.1 Expressing Concentration of Solutions

1. Mass Percentage (w/w): (Mass of component in solution / Total mass of solution) × 100
2. Volume Percentage (v/v): (Volume of component / Total volume) × 100
3. Mass by Volume Percentage (w/v): Mass of solute (g) dissolved in 100 mL of solution. Used in pharmacy.
4. Parts Per Million (ppm): (Number of parts of component / Total parts in solution) × 10⁶. Used for traces.
5. Mole Fraction (x): x_A = n_A / (n_A + n_B). Note: x_A + x_B = 1.

6. Molarity (M): Number of moles of solute dissolved in 1 Litre of solution.
M = Moles of solute / Volume of solution (L) = (w_B × 1000) / (M_B × V_mL)
7. Molality (m): Number of moles of solute dissolved in 1 kg of solvent.
m = Moles of solute / Mass of solvent (kg) = (w_B × 1000) / (M_B × W_A)

Temperature Dependence: Molarity (M) and w/v change with temperature because volume varies with temp. Molality (m), mole fraction (x), and w/w are independent of temperature.

2. Solubility

Solubility: The maximum amount of a substance that can be dissolved in a specified amount of solvent at a specified temperature.

2.1 Solubility of a Solid in a Liquid

Like dissolves like: Polar solutes (NaCl) dissolve in polar solvents (water). Non-polar solutes (iodine) dissolve in non-polar solvents (benzene, CCl₄).
Effect of Temperature: If dissolution is endothermic (ΔH > 0), solubility increases with temperature. If exothermic (ΔH < 0), solubility decreases with temperature (Le Chatelier's Principle).
Effect of Pressure: No significant effect (solids and liquids are highly incompressible).

2.2 Solubility of a Gas in a Liquid (Henry's Law)

Henry's Law: The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

p = K_H · x
Where K_H = Henry's law constant.

Higher K_H means lower solubility of the gas in the liquid at a given pressure.
K_H increases with increasing temperature → solubility of gases decreases with temperature. (Aquatic species are more comfortable in cold water).
Applications: CO₂ in soft drinks (sealed under high pressure); Scuba divers (preventing the 'bends' using He-diluted air); High altitudes (anoxia due to low pO₂).

3. Vapour Pressure of Liquid Solutions

3.1 Raoult's Law for Volatile Solutes

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 in solution.

p₁ = p₁° · x₁ and p₂ = p₂° · x₂
Total Pressure (Dalton's Law): P_total = p₁ + p₂ = p₁°x₁ + p₂°x₂

Mole fraction in vapour phase (y₁, y₂): p₁ = y₁P_total and p₂ = y₂P_total

3.2 Raoult's Law as a Special Case of Henry's Law

According to Raoult's law, p_i = x_i p_i°. Henry's law: p = K_Hx. They are identical if K_H becomes equal to p_i°.

4. Ideal and Non-Ideal Solutions

4.1 Ideal Solutions

Obey Raoult's law over the entire range of concentration.
A-B interactions are equal to A-A and B-B interactions.
ΔH_mix = 0, ΔV_mix = 0.
Examples: n-hexane + n-heptane, bromoethane + chloroethane, benzene + toluene.

4.2 Non-Ideal Solutions

Do NOT obey Raoult's law. They show deviations.

Property	Positive Deviation	Negative Deviation
Vapour Pressure	p₁ > p₁°x₁ ; P_total is higher	p₁ < p₁°x₁ ; P_total is lower
Interactions	A-B is WEAKER than A-A/B-B	A-B is STRONGER than A-A/B-B
Thermo Changes	ΔH_mix > 0 (Endothermic), ΔV_mix > 0	ΔH_mix < 0 (Exothermic), ΔV_mix < 0
Azeotrope Formed	Minimum boiling azeotrope	Maximum boiling azeotrope
Examples	Ethanol + Acetone Ethanol + Water CS₂ + Acetone	Phenol + Aniline Chloroform + Acetone (H-bond) HNO₃ + Water

4.3 Azeotropes

Binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. They cannot be separated by fractional distillation.

5. Colligative Properties

Properties of a solution which depend ONLY on the number of solute particles irrespective of their nature relative to the total number of particles present.

5.1 Relative Lowering of Vapour Pressure

Addition of non-volatile solute lowers the vapour pressure of the solvent.

(p₁° − p₁) / p₁° = x₂ = n₂ / (n₁ + n₂)
For dilute solutions (n₂ << n₁): (p₁° − p₁) / p₁° ≈ n₂/n₁ = (w₂M₁) / (M₂w₁)

5.2 Elevation of Boiling Point (ΔT_b)

Solution boils at a higher temp than pure solvent. ΔT_b = T_b(solution) − T_b°(solvent).

ΔT_b = K_b · m (m = molality)
ΔT_b = (K_b · 1000 · w₂) / (M₂ · w₁)
K_b = Molal elevation constant (Ebullioscopic constant). Unit: K kg mol⁻¹.

5.3 Depression of Freezing Point (ΔT_f)

Solution freezes at lower temp than pure solvent. ΔT_f = T_f°(solvent) − T_f(solution).

ΔT_f = K_f · m (m = molality)
ΔT_f = (K_f · 1000 · w₂) / (M₂ · w₁)
K_f = Molal depression constant (Cryoscopic constant). Unit: K kg mol⁻¹.

5.4 Osmosis and Osmotic Pressure (π)

Osmosis: Flow of solvent molecules through a semipermeable membrane (SPM) from pure solvent to solution (or lower to higher concentration).
Osmotic Pressure (π): The excess pressure applied on the solution side to stop osmosis.

π = C R T = (n₂ / V) R T
Where C = molarity, R = gas constant, T = temp in Kelvin.

Isotonic solutions: Two solutions having same osmotic pressure at a given temp. No osmosis occurs between them. (e.g., 0.9% w/v NaCl is isotonic with blood plasma).
Hypertonic: Higher π (cells shrink/plasmolysis). Hypotonic: Lower π (cells swell & burst).
Reverse Osmosis (RO): If pressure applied on solution > π, solvent flows from solution to pure solvent. Used in desalination of sea water.

⚠️ NEET TIP: Osmotic pressure is the best method for determining molar masses of biomolecules (proteins, polymers) because it is measured at room temp, and its magnitude is measurable even for very dilute solutions.

6. Abnormal Molar Masses & van't Hoff Factor (i)

If solute undergoes association or dissociation, the number of particles changes, causing calculated molar mass to differ from normal value.

van't Hoff factor (i) = Normal molar mass / Abnormal molar mass
= Observed colligative property / Calculated colligative property
= Total moles of particles after assoc/dissoc / Total moles before assoc/dissoc

Dissociation: i > 1 (e.g., NaCl → Na⁺ + Cl⁻, i = 2 for 100% ionization). Molar mass observed is LOWER.
Association: i < 1 (e.g., CH₃COOH dimerizes in benzene via H-bonding, i = 0.5 for 100% assoc). Molar mass observed is HIGHER.
No assoc/dissoc: i = 1 (e.g., glucose, urea, sucrose).

6.1 Modified Colligative Property Formulas

ΔP / P₁° = i · x₂
ΔT_b = i · K_b · m
ΔT_f = i · K_f · m
π = i · C R T

6.2 Degree of Dissociation (α) and Association (α)

Degree of Dissociation (α): α = (i − 1) / (n − 1)
n = number of ions produced from 1 molecule

Degree of Association (α): α = (1 − i) / (1 − 1/n)
n = number of molecules associating (e.g., n=2 for dimer)

🎓 NEET Previous Year Questions

Q1. [NEET 2022] Which of the following is independent of temperature?

Answer Molality. Molarity, normality, and formality involve volume, which changes with temperature. Molality involves mass, which is temp-independent.

Q2. [NEET 2021] The freezing point depression constant (K_f) of benzene is 5.12 K kg mol⁻¹. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is:

Answer ΔT_f = K_f × m = 5.12 × 0.078 = 0.399 K ≈ 0.40 K. (Since non-electrolyte, i=1).

Q3. [NEET 2020] Mixture showing positive deviation from Raoult's law is:

Answer Ethanol + Acetone. Acetone molecules get between ethanol molecules, breaking ethanol's H-bonds, expanding volume, increasing vapor pressure.

Q4. [NEET 2018] For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

Answer Highest freezing point means LOWEST depression (ΔT_f). ΔT_f = i · K_f · m. So we need lowest i. A complex like [Co(H₂O)₆]Cl₃ gives 4 ions (i=4). [Co(H₂O)₃Cl₃] gives 1 ion (i=1). Complex with lowest i will have highest FP.

Q5. [NEET 2016] At 100°C, vapour pressure of a solution of 6.5g solute in 100g water is 732 mm. K_b=0.52, BP is:

Answer p° at 100°C = 760 mm. P_s = 732 mm. (760−732)/732 = n/N → 28/732 = n/(100/18). Solving gives solute moles. Then ΔT_b = 0.52 × m → T_b = 101.06°C.

💡 Rapid Revision

Molarity changes with temp, Molality does NOT.
Raoult's law (+): Ethanol+Acetone (min boiling azeotrope).
Raoult's law (−): Chloroform+Acetone (max boiling azeotrope).
Osmotic pressure (π = CRT) is best to find molar mass of proteins.
Dissociation: i > 1 | Association: i < 1 | Normal: i = 1.
Isotonic solutions: Same osmotic pressure (π₁ = π₂) → C₁ = C₂.

Numericals - Solutions - Class 12 Chemistry

CLASS 12 CHEMISTRY | NCERT SOLUTIONS

Chapter 1 — Solutions

22 Solved Numericals — Step-by-Step Breakdown

📝 Concentration & Solubility (Q1 – Q7)

3 MarksQ1. Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.

✓ Solution
Let the total mass of solution be 100 g.
Mass of C₂H₆O₂ (solute) = 20 g.
Mass of water (solvent) = 100 − 20 = 80 g.
Molar mass C₂H₆O₂ = 12×2 + 1×6 + 16×2 = 62 g/mol.
Molar mass H₂O = 18 g/mol.
n(C₂H₆O₂) = 20 / 62 = 0.322 mol.
n(H₂O) = 80 / 18 = 4.444 mol.
Total moles = 0.322 + 4.444 = 4.766 mol.

x(C₂H₆O₂) = 0.322 / 4.766 = 0.068
x(H₂O) = 1 − 0.068 = 0.932

3 MarksQ2. Calculate the molarity of a solution containing 5 g of NaOH in 450 mL solution.

✓ Solution
Mass of NaOH (w₂) = 5 g.
Molar mass of NaOH (M₂) = 23 + 16 + 1 = 40 g/mol.
Volume of solution (V) = 450 mL.

Molarity (M) = (w₂ × 1000) / (M₂ × V)
M = (5 × 1000) / (40 × 450) = 5000 / 18000 = 0.278 M (or mol L⁻¹)

3 MarksQ3. Calculate molality of 2.5 g of ethanoic acid (CH₃COOH) in 75 g of benzene.

✓ Solution
Mass of solute (w₂) = 2.5 g. Mass of solvent (w₁) = 75 g.
Molar mass of CH₃COOH (M₂) = 12 + 3 + 12 + 16 + 16 + 1 = 60 g/mol.
Moles of CH₃COOH = 2.5 / 60 = 0.0417 mol.
Mass of benzene in kg = 75 / 1000 = 0.075 kg.

Molality (m) = Moles of solute / Mass of solvent in kg
m = 0.0417 / 0.075 = 0.556 mol kg⁻¹

3 MarksQ4. If N₂ gas is bubbled through water at 293 K, how many millimoles of N₂ gas would dissolve in 1 litre of water? Assume N₂ exerts a partial pressure of 0.987 bar. Given Henry's law constant for N₂ at 293 K is 76.48 kbar.

✓ Solution
p(N₂) = 0.987 bar.
K_H = 76.48 kbar = 76480 bar.
From Henry's law: p = K_H · x
x(N₂) = p(N₂) / K_H = 0.987 / 76480 = 1.29 × 10⁻⁵.
Also, x(N₂) = n(N₂) / (n(N₂) + n(water)) ≈ n(N₂) / n(water) [since n(N₂) is very small].
Moles of water in 1 L (1000 g) = 1000 / 18 = 55.5 mol.
n(N₂) / 55.5 = 1.29 × 10⁻⁵ ⇒ n(N₂) = 55.5 × 1.29 × 10⁻⁵ = 7.16 × 10⁻⁴ mol.

Millimoles = 7.16 × 10⁻⁴ × 1000 = 0.716 mmol

3 MarksQ5. H₂S, a toxic gas with rotten egg like smell, is used for qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry's law constant.

✓ Solution
Solubility = 0.195 m = 0.195 mol H₂S in 1000 g water.
Moles of water = 1000 / 18 = 55.55 mol.
Mole fraction of H₂S, x = 0.195 / (0.195 + 55.55) = 0.195 / 55.745 = 0.0035.
At STP, pressure p = 1 bar (or 0.987 atm, but NCERT uses 1 bar for standard state now).
Using p = K_H · x ⇒ K_H = p / x

K_H = 1 bar / 0.0035 = 285.7 bar (282 bar if using 0.987 atm = 0.987 bar)

3 MarksQ6. 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 if the density of the solution is 1.504 g mL⁻¹?

✓ Solution
68% by mass means 68 g HNO₃ in 100 g solution.
Molar mass of HNO₃ = 1 + 14 + 48 = 63 g/mol.
Moles of HNO₃ = 68 / 63 = 1.079 mol.
Volume of 100 g solution = mass / density = 100 / 1.504 = 66.5 mL = 0.0665 L.

Molarity (M) = 1.079 mol / 0.0665 L = 16.23 M

3 MarksQ7. 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
10% w/w glucose = 10 g glucose in 90 g water.
M(C₆H₁₂O₆) = 180 g/mol.
n(glucose) = 10 / 180 = 0.055 mol. n(H₂O) = 90 / 18 = 5 mol.
1. Molality (m): m = 0.055 mol / 0.090 kg = 0.617 m
2. Mole fraction: Total moles = 5.055.
x(glucose) = 0.055 / 5.055 = 0.011; x(H₂O) = 1 - 0.011 = 0.989
3. Molarity: Volume of 100g sol = 100 / 1.2 = 83.33 mL = 0.0833 L.

Molarity (M) = 0.055 mol / 0.0833 L = 0.66 M

💡 Raoult's Law & Vapour Pressure (Q8 – Q13)

3 MarksQ8. Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate vapour pressure of solution prepared by mixing 25.5 g of CHCl₃ and 40 g of CH₂Cl₂ at 298 K.

✓ Solution
Molar mass CHCl₃ = 119.5 g/mol; CH₂Cl₂ = 85 g/mol.
n(CHCl₃) = 25.5 / 119.5 = 0.213 mol.
n(CH₂Cl₂) = 40 / 85 = 0.47 mol.
Total moles = 0.213 + 0.47 = 0.683 mol.
Mole fractions: x(CHCl₃) = 0.213 / 0.683 = 0.312; x(CH₂Cl₂) = 0.47 / 0.683 = 0.688.

P_total = p°(CHCl₃)·x(CHCl₃) + p°(CH₂Cl₂)·x(CH₂Cl₂)
P_total = 200(0.312) + 415(0.688) = 62.4 + 285.5 = 347.9 mm Hg

3 MarksQ9. Based on Q8, find the mole fraction of each component in vapour phase.

✓ Solution
Partial pressures calculated above:
p(CHCl₃) = 62.4 mm Hg; p(CH₂Cl₂) = 285.5 mm Hg.
P_total = 347.9 mm Hg.
Mole fraction in vapour (y) = p(component) / P_total.

y(CH₂Cl₂) = 285.5 / 347.9 = 0.82
y(CHCl₃) = 62.4 / 347.9 = 0.18

Vapour is richer in the more volatile component (CH₂Cl₂).

3 MarksQ10. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg at 350 K. Find the composition of the liquid mixture if total vapour pressure is 600 mm Hg.

✓ Solution
p°_A = 450; p°_B = 700. P_total = 600.
P_total = p°_A + (p°_B - p°_A)x_B
600 = 450 + (700 - 450)x_B
150 = 250 x_B

x_B = 150 / 250 = 0.60
x_A = 1 - 0.60 = 0.40

Liquid composition: 40% A, 60% B.

3 MarksQ11. 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
M(urea) = 60 g/mol; n(urea) = 50/60 = 0.833 mol.
M(water) = 18 g/mol; n(water) = 850/18 = 47.22 mol.
x(urea) = 0.833 / (0.833 + 47.22) = 0.0173.
(p° - p) / p° = x(urea) = 0.0173 (Relative lowering).
(23.8 - p) / 23.8 = 0.0173 ⇒ 23.8 - p = 0.41.

p = 23.8 - 0.41 = 23.39 mm Hg (Vapour pressure of solution)

4 MarksQ12. 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 80g of benzene is mixed with 100g of toluene.

✓ Solution
M(Benzene, C₆H₆) = 78 g/mol; M(Toluene, C₇H₈) = 92 g/mol.
Moles benzene (n₁) = 80/78 = 1.026 mol. Moles toluene (n₂) = 100/92 = 1.087 mol.
x(benzene) = 1.026 / (1.026 + 1.087) = 0.486. x(toluene) = 0.514.
p(benzene) = 50.71 × 0.486 = 24.645 mm Hg.
p(toluene) = 32.06 × 0.514 = 16.479 mm Hg.
P_total = 24.645 + 16.479 = 41.124 mm Hg.

y(benzene) = 24.645 / 41.124 = 0.60

3 MarksQ13. 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
Normal boiling point pressure of water (p°) = 1.013 bar (1 atm).
P_solution (p) = 1.004 bar.
Relative lowering: (p° - p) / p° = (1.013 - 1.004) / 1.013 = 0.009 / 1.013 = 0.00888.
Dilute solution approximation: x₂ ≈ n₂ / n₁ = (w₂ × M₁) / (M₂ × w₁).
2% solution means 2g solute in 98g water.
0.00888 = (2 × 18) / (M₂ × 98)

M₂ = 36 / (0.00888 × 98) = 41.35 g/mol

📈 Colligative Properties (No factor i) (Q14 – Q18)

3 MarksQ14. 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. K_b for water is 0.52 K kg mol⁻¹.

✓ Solution
ΔT_b = 100 - 99.63 = 0.37 K.
w₁ = 500 g = 0.5 kg. M(sucrose, C₁₂H₂₂O₁₁) = 342 g/mol.
ΔT_b = (K_b × w₂ × 1000) / (M₂ × w₁(g))
0.37 = (0.52 × w₂ × 1000) / (342 × 500) = (0.52 × w₂ × 2) / 342

w₂ = (0.37 × 342) / 1.04 = 126.54 / 1.04 = 121.67 g

3 MarksQ15. Calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) to be dissolved in 75g of acetic acid to lower its melting point by 1.5°C. K_f = 3.9 K kg mol⁻¹.

✓ Solution
ΔT_f = 1.5 K. M₂(Vitamin C) = 12×6 + 8 + 16×6 = 176 g/mol.
w₁ = 75 g.
ΔT_f = (K_f × w₂ × 1000) / (M₂ × w₁)
1.5 = (3.9 × w₂ × 1000) / (176 × 75)
1.5 = (3900 × w₂) / 13200

w₂ = (1.5 × 13200) / 3900 = 19800 / 3900 = 5.08 g

3 MarksQ16. 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
M = 185,000 g/mol; w = 1 g; V = 450 mL = 0.45 L; T = 37 + 273 = 310 K.
R = 8.314 × 10³ Pa L K⁻¹ mol⁻¹. (Since answer is asked in pascals).
π = CRT = (w / (M·V)) · R · T
π = (1 / (185000 × 0.45)) × (8.314 × 10³) × 310

π = (1 / 83250) × 2577340 = 30.96 Pa

3 MarksQ17. 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.

✓ Solution
π = 2.57 × 10⁻³ bar. V = 200 cm³ = 0.200 L. w = 1.26 g. T = 300 K.
R = 0.083 L bar K⁻¹ mol⁻¹.
M = (w × R × T) / (π × V)
M = (1.26 × 0.083 × 300) / (2.57 × 10⁻³ × 0.200)
M = 31.374 / 0.000514

M = 61,039 g/mol

3 MarksQ18. 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
For cane sugar (M = 342), 5% means 5g in 95g water.
ΔT_f = 273.15 - 271 = 2.15 K.
ΔT_f = (K_f × w₂ × 1000) / (M₂ × w₁) ⇒ 2.15 = (K_f × 5 × 1000) / (342 × 95)
K_f = (2.15 × 342 × 95) / 5000 = 69854.5 / 5000 = 13.97 K kg/mol.
For glucose (M = 180), 5% means 5g in 95g water.
ΔT_f(glucose) = (13.97 × 5 × 1000) / (180 × 95) = 69850 / 17100 = 4.08 K.

Freezing point = 273.15 - 4.08 = 269.07 K

⚠️ van't Hoff Factor & Abnormal Molar Mass (Q19 – Q22)

4 MarksQ19. 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 dimer in solution?

✓ Solution
Experimental (abnormal) Molar mass M_obs:
M_obs = (K_f × w₂ × 1000) / (ΔT_f × w₁) = (4.9 × 2 × 1000) / (1.62 × 25) = 241.98 g/mol.
Normal Molar mass M_cal of C₆H₅COOH = 122 g/mol.
van't Hoff factor, i = M_cal / M_obs = 122 / 241.98 = 0.504.
Because it forms a dimer, n = 2. Association reaction: 2A ⇌ A₂.
Degree of association, α = (1 - i) / (1 - 1/n) = (1 - 0.504) / (1 - 1/2) = 0.496 / 0.5 = 0.992.

Percentage association = 0.992 × 100 = 99.2%

4 MarksQ20. 0.6 mL of acetic acid (CH₃COOH), having density 1.06 g mL⁻¹, is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van't Hoff factor and the dissociation constant of acid. (K_f for water = 1.86 K kg mol⁻¹).

✓ Solution
Mass of acetic acid = volume × density = 0.6 × 1.06 = 0.636 g.
Moles (n) = 0.636 / 60 = 0.0106 mol.
Mass of water = 1 kg (1 L = 1 kg for water). Molality (m) = 0.0106 m.
Calculated ΔT_f = K_f × m = 1.86 × 0.0106 = 0.0197 K.
Observed ΔT_f = 0.0205 K.
1. van't Hoff factor: i = Observed ΔT_f / Calculated ΔT_f = 0.0205 / 0.0197 = 1.041.
2. Dissociation constant: CH₃COOH ⇌ CH₃COO⁻ + H⁺ (n=2 ions).
α = (i - 1)/(n - 1) = (1.041 - 1)/(2 - 1) = 0.041.
Dissociation constant K₁ = (Cα²) / (1 - α) ≈ Cα² (as α is small). Here C ≈ m = 0.0106.
K₁ = 0.0106 × (0.041)² = 0.0106 × 0.001681 = 1.78 × 10⁻⁵.

van't Hoff factor i = 1.041. Dissociation constant K_a = 1.78 × 10⁻⁵

3 MarksQ21. 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
π = 0.75 atm; V = 2.5 L; T = 300 K; i = 2.47; R = 0.0821 L atm K⁻¹ mol⁻¹.
π = i · c · R · T = i · (w/M) / V · R · T
Molar mass of CaCl₂ = 40 + (35.5 × 2) = 111 g/mol.
0.75 = (2.47 × w × 0.0821 × 300) / (111 × 2.5)
0.75 = (w × 60.836) / 277.5

w = (0.75 × 277.5) / 60.836 = 208.125 / 60.836 = 3.42 g

4 MarksQ22. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 liter of water at 25°C, assuming that it is completely dissociated.

✓ Solution
w = 25 mg = 0.025 g. V = 2 L. T = 298 K.
K₂SO₄ completely dissociates as 2K⁺ + SO₄²⁻ → 3 ions ⇒ i = 3 (assuming 100% dissoc).
Molar mass K₂SO₄ = (2×39) + 32 + (4×16) = 174 g/mol.
π = i · (w/M) / V · R · T
π = 3 × (0.025 / 174) / 2 × 0.0821 × 298
π = 3 × 0.0001436 / 2 × 24.4658

π = 0.0002154 × 24.4658 = 5.27 × 10⁻³ atm

✍ Score Guide — 22 Questions
All questions from NCERT Exercise & Exemplar with clear explanations matching CBSE marking scheme.

High-Yield Facts & Formulas: Solutions

Solution Definition
A solution is a homogeneous mixture of two or more non-reacting substances.

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

Volume Percentage
Volume % = (Volume of solute / Total volume of solution) × 100

Parts Per Million (ppm)
ppm = (Number of parts of component / Total number of parts of all components) × 10⁶

Mole Fraction (x)
x_A = n_A / (n_A + n_B). The sum of mole fractions of all components is unity (Σx = 1).

Molarity (M)
M = Moles of solute / Volume of solution in Litres. Unit: mol L^-1.

Molality (m)
m = Moles of solute / Mass of solvent in Kilograms. Unit: mol kg^-1.

Temperature Dependence
Molarity changes with temperature (due to volume expansion), while Molality and Mole Fraction are independent of temperature.

Henry's Law
p = K_H × x. Solubility of a gas in a liquid increases with increase in pressure.

Henry's Constant (K_H)
K_H is a function of the nature of the gas. Higher K_H means lower solubility at the same pressure.

Temperature Effect on Gas Solubility
Solubility of gases in liquids decreases with rise in temperature (exothermic process).

Raoult's Law (Volatile Solute)
p_total = p₁⁰x₁ + p₂⁰x₂

Raoult's Law (Non-Volatile Solute)
p₁ = p₁⁰ × x₁. The vapor pressure of solution is less than pure solvent.

Ideal Solution Conditions
ΔH_mix = 0, ΔV_mix = 0, and A-B interactions ≈ A-A or B-B interactions.

Positive Deviation
p_total > p_A⁰x_A + p_B⁰x_B. ΔH_mix > 0 (Endothermic).

Negative Deviation
p_total < p_A⁰x_A + p_B⁰x_B. ΔH_mix < 0 (Exothermic).

Minimum Boiling Azeotropes
Formed by solutions showing large positive deviation from Raoult's law. (e.g., Ethanol-Water).

Maximum Boiling Azeotropes
Formed by solutions showing large negative deviation from Raoult's law. (e.g., Nitric Acid-Water).

Relative Lowering of Vapor Pressure
(p₁⁰ - p₁) / p₁⁰ = x₂. It is a colligative property.

Elevation of Boiling Point
ΔT_b = K_b × m. T_b = T_b⁰ + ΔT_b.

Ebullioscopic Constant (K_b)
Boiling point elevation produced by 1 molal solution. Unity: K kg mol^-1.

Depression of Freezing Point
ΔT_f = K_f × m. T_f = T_f⁰ - ΔT_f.

Cryoscopic Constant (K_f)
Freezing point depression produced by 1 molal solution. Unity: K kg mol^-1.

Ethylene Glycol in Cars
Used as antifreeze to lower the freezing point of water in car radiators.

Osmosis
Flow of solvent through a semipermiable membrane from pure solvent to solution.

Osmotic Pressure (π)
π = CRT or π = (n/V)RT. It is proportional to molarity.

Biological Application of Osmosis
A cell shrinks in hypertonic solution (high salt) and swells in hypotonic solution (pure water).

Reverse Osmosis (RO)
Applied pressure > osmotic pressure; solvent flows from solution to pure solvent. Used for desalination.

Van't Hoff Factor (i)
i = (Normal molar mass / Abnormal molar mass) = (Observed CP / Calculated CP).

i for Association
i < 1. For dimerization (e.g., acetic acid in benzene), i=0.5.

i for Dissociation
i > 1. For NaCl → Na⁺ + Cl^-, i ≈ 2.

Relationship: i and α (Dissociation)
α = (i - 1) / (n - 1), where n is number of ions.

Relationship: i and α (Association)
α = (1 - i) / (1 - 1/n), where n is number of molecules associating.

Modified Raoult's Law
(p₁⁰ - p₁) / p₁⁰ = i × x₂

Modified Osmotic Pressure
π = i C R T

Binary Solution
A solution containing only two components.

Solvent property
The component that determines the physical state of the solution is the solvent.

Amalgam
A solution of mercury (liquid) in another metal (solid).

Standard density of water
1 g/mL or 1000 kg/m³ at 4 °C.

Bends (Decompression sickness)
Caused by nitrogen bubbles in blood for divers. Prevented by using Helium-diluted air.

Anoxia
Low oxygen in blood/tissues at high altitudes due to low partial pressure of O₂.

Drowning prevention
In salt water drowning, osmosis pulls water out of blood into lungs (pulmonary edema).

Vapor Pressure vs Temperature
V.P. increases exponentially with temperature.

Boiling point definition
Temperature at which vapor pressure of liquid equals atmospheric pressure.

Colligative property usage
Mainly used to determine the molar mass of complex molecules like proteins and polymers.

Ebullioscopic constant for Water
0.52 K kg mol^-1.

Cryoscopic constant for Water
1.86 K kg mol^-1.

Isotonic Saline
0.9% (w/v) NaCl solution is isotonic with human blood.

Van't Hoff equation
PV = nRT (approximated for dilute solutions).

Solubility of solids in liquids
Usually increases with temperature if ΔH_sol > 0.

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