Facts and Quick Notes from NCERT

Solution: A homogeneous mixture of two or more substances. Binary solutions have two components: a solute and a solvent.

Concentration Terms: Molality (m) is temperature-independent (moles solute/kg solvent), whereas Molarity (M) is temperature-dependent (moles solute/L solution).

Henry’s Law: The partial pressure of a gas above a liquid (p) is directly proportional to its mole fraction (x) in the liquid. p = Kₙ x. A higher Kₙ value means lower solubility.

Raoult’s Law: For a volatile component in a solution, its partial vapour pressure is the product of its mole fraction and its vapour pressure in the pure state (pᵢ = pᵢ° xᵢ).

Ideal Solutions: Obey Raoult’s Law. For them, ΔH_mix = 0 and ΔV_mix = 0. Intermolecular forces (A-B) are similar to pure component forces (A-A, B-B).

Non-Ideal Solutions:

Positive Deviation: Vapour pressure is higher than predicted. A-B interactions are weaker. ΔH_mix > 0. Forms minimum boiling azeotropes (e.g., ethanol + acetone).
Negative Deviation: Vapour pressure is lower than predicted. A-B interactions are stronger. ΔH_mix < 0. Forms maximum boiling azeotropes (e.g., chloroform + acetone).

Colligative Properties: Depend only on the number of solute particles, not their nature. They are:

Relative Lowering of Vapour Pressure: (p₁° – p₁) / p₁° = x₂
Elevation in Boiling Point: ΔTₑ = Kₑ m
Depression in Freezing Point: ΔTₒ = Kₒ m
Osmotic Pressure: Π = CRT

Van’t Hoff Factor (i): Corrects for association or dissociation of solutes. i = (Observed Colligative Property) / (Calculated Colligative Property). For dissociation, i > 1. For association, i < 1.

Reverse Osmosis: Applying a pressure greater than the osmotic pressure to the solution side forces the solvent to move from the concentrated to the dilute side. Used for desalination.

NCERT Numerical Problems with Solutions

1. (NCERT Example 2.1) Calculate the mole fraction of ethylene glycol (C₂H₆O₂) in a solution containing 20% of C₂H₆O₂ by mass.

Assume we have 100 g of solution.
Mass of ethylene glycol = 20 g.
Mass of water = 100 g – 20 g = 80 g.
Molar mass of C₂H₆O₂ = (2×12) + (6×1) + (2×16) = 62 g/mol.
Molar mass of H₂O = 18 g/mol.
Moles of C₂H₆O₂ = 20 g / 62 g/mol = 0.322 mol.
Moles of H₂O = 80 g / 18 g/mol = 4.444 mol.
Total moles = 0.322 + 4.444 = 4.766 mol.
Mole fraction of ethylene glycol (x_glycol) = Moles of glycol / Total moles
x_glycol = 0.322 / 4.766 = 0.068.

2. (NCERT Example 2.4) The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g is added to 39.0 g of benzene (molar mass 78 g/mol). The vapour pressure of the solution is 0.845 bar. What is the molar mass of the solid substance?

Given: p₁° = 0.850 bar, p₁ = 0.845 bar, w₂ = 0.5 g, w₁ = 39.0 g, M₁ = 78 g/mol.
Using Raoult’s Law for relative lowering of vapour pressure:
(p₁° – p₁) / p₁° = x₂ = n₂ / (n₁ + n₂)
For dilute solutions, we can approximate: (p₁° – p₁) / p₁° ≈ n₂ / n₁
n₁ = w₁ / M₁ = 39.0 g / 78 g/mol = 0.5 mol.
(0.850 – 0.845) / 0.850 = (w₂/M₂) / n₁
0.005 / 0.850 = (0.5 / M₂) / 0.5
0.00588 = 1 / M₂
M₂ = 1 / 0.00588 = 170 g/mol.

3. (NCERT Example 2.6) 18 g of glucose, C₆H₁₂O₆, is dissolved in 1 kg of water. At what temperature will water boil at 1.013 bar? Kₑ for water is 0.52 K kg/mol.

Molar mass of glucose (M₂) = 180 g/mol.
Mass of solute (w₂) = 18 g.
Mass of solvent (w₁) = 1 kg.
Molality (m) = Moles of solute / Mass of solvent (kg) = (18 g / 180 g/mol) / 1 kg = 0.1 m.
Elevation in boiling point, ΔTₑ = Kₑ * m
ΔTₑ = 0.52 K kg/mol × 0.1 mol/kg = 0.052 K.
Boiling point of pure water at 1.013 bar is 373.15 K (100°C).
Boiling point of solution = 373.15 K + 0.052 K = 373.202 K.

4. (NCERT Example 2.9) 45 g of ethylene glycol (C₂H₆O₂) is mixed with 600 g of water. Calculate the freezing point depression.

Given: w₂ = 45 g, w₁ = 600 g = 0.6 kg. Kₒ for water = 1.86 K kg/mol.
Molar mass of ethylene glycol (M₂) = 62 g/mol.
Molality (m) = (w₂ / M₂) / w₁ = (45 g / 62 g/mol) / 0.6 kg = 0.725 mol / 0.6 kg = 1.2 m.
Freezing point depression, ΔTₒ = Kₒ * m
ΔTₒ = 1.86 K kg/mol × 1.2 mol/kg = 2.2 K.

5. (NCERT Example 2.10) 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.

Given: V = 200 cm³ = 0.200 L, w₂ = 1.26 g, T = 300 K, Π = 2.57 × 10⁻³ bar.
R = 0.083 L bar K⁻¹ mol⁻¹.
Using the osmotic pressure formula: Π = (w₂/M₂) * (RT/V)
Rearranging for M₂: M₂ = (w₂RT) / (ΠV)
M₂ = (1.26 g × 0.083 L bar K⁻¹ mol⁻¹ × 300 K) / (2.57 × 10⁻³ bar × 0.200 L)
M₂ = 31.374 / 0.000514 = 61039 g/mol.

6. (NCERT Example 2.12) 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 (Kₒ) is 4.9 K kg mol⁻¹. What is the percentage association of the acid if it forms a dimer in solution?

Step 1: Calculate the experimental molar mass (M_obs).
M_obs = (Kₒ × w₂ × 1000) / (ΔTₒ × w₁) = (4.9 × 2 × 1000) / (1.62 × 25) = 9800 / 40.5 = 241.98 g/mol.
Step 2: Calculate the van’t Hoff factor (i).
Normal molar mass of benzoic acid (M_norm) = 122 g/mol.
i = M_norm / M_obs = 122 / 241.98 = 0.504.
Step 3: Calculate the degree of association (α).
For dimerization (2A ⇌ A₂), the number of particles associating (n) is 2.
The formula is i = 1 – α + (α/n).
0.504 = 1 – α + (α/2) = 1 – α/2
α/2 = 1 – 0.504 = 0.496
α = 2 × 0.496 = 0.992.
Percentage association = 99.2%.

7. (NCERT Intext 2.5) Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.

0.25 molal solution means 0.25 moles of urea are dissolved in 1 kg (1000 g) of water.
Molar mass of urea = 14+2+12+16+14+2 = 60 g/mol.
Mass of urea in 1000 g of water = 0.25 mol × 60 g/mol = 15 g.
Total mass of this solution = 1000 g (water) + 15 g (urea) = 1015 g.
Now, we need to find the mass of urea for a 2.5 kg (2500 g) solution.
If 1015 g of solution contains 15 g of urea,
Then, 2500 g of solution will contain (15 / 1015) × 2500 = 36.95 g of urea.

8. (NCERT Exercise 2.17) The vapour pressure of water is 12.3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a non-volatile solute in it.

1 molal solution means 1 mole of solute in 1000 g of water.
Moles of water (n₁) = 1000 g / 18 g/mol = 55.55 mol.
Moles of solute (n₂) = 1 mol.
Mole fraction of solvent (x₁) = n₁ / (n₁ + n₂) = 55.55 / (55.55 + 1) = 55.55 / 56.55 = 0.982.
According to Raoult’s Law, p₁ = p₁° x₁.
p₁ = 12.3 kPa × 0.982 = 12.08 kPa.

9. (NCERT Exercise 2.35) Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litres of water at 25 °C, assuming that it is completely dissociated.

K₂SO₄ dissociates as K₂SO₄ → 2K⁺ + SO₄²⁻. Since it is completely dissociated, the number of particles (n) is 3, so the van’t Hoff factor (i) = 3.
Molar mass of K₂SO₄ = 174 g/mol.
w₂ = 25 mg = 0.025 g.
V = 2 L, T = 25 °C = 298 K, R = 0.0821 L atm K⁻¹ mol⁻¹.
Π = i * (w₂/M₂) * (RT/V)
Π = 3 * (0.025 / 174) * (0.0821 * 298 / 2)
Π = 3 * (0.0001437) * (12.23) = 5.27 × 10⁻³ atm.

10. (NCERT Exercise 2.41) Vapour pressure of pure acetone at 298K is 23.3 kPa. 1.2 g of a non-volatile substance was dissolved in 100 g of acetone. The vapour pressure of the solution was found to be 23.1 kPa. Calculate the molar mass of the substance.

p₁° = 23.3 kPa, p₁ = 23.1 kPa.
w₂ = 1.2 g, w₁ = 100 g.
Molar mass of acetone (C₃H₆O), M₁ = 58 g/mol.
Using Raoult’s Law: (p₁° – p₁) / p₁° = x₂ ≈ n₂/n₁ = (w₂/M₂) / (w₁/M₁)
(23.3 – 23.1) / 23.3 = (1.2 / M₂) / (100 / 58)
0.2 / 23.3 = (1.2 / M₂) / 1.724
0.00858 = 0.696 / M₂
M₂ = 0.696 / 0.00858 = 81.1 g/mol.