Focuses on osmotic pressure, the movement of solvent across membranes, and reverse osmosis processes in solutions.
p = KH · x . x = (6/300) = 0.02 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = (n/V) RT , n = (w/M) . 2.46 = (5/M) × 0.0821 × 300 . M = (5 × 0.0821 × 300/2.46) ≈ 50 g/mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = (w/M V) RT . 1.476 = (15/M × 0.25) × 0.0821 × 300 . M = (15 × 0.0821 × 300/1.476 × 0.25) ≈ 100 g/mol .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = MRT . Pi = 0.02 × 0.0821 × 300 ≈ 0.492 atm .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Moles of solute = (9/90) = 0.1 mol . Volume = 450 mL = 0.45 L. Molarity = (0.1/0.45) ≈ 0.2222 M .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = i · M · RT . 1.845 = i × 0.25 × 0.0821 × 300 . i = (1.845/0.25 × 24.63) ≈ 0.3 × 6 = 1.8 . i = 1 + α , 1.8 = 1 + α , α = 0.8 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = MRT . Pi = 0.025 × 0.0821 × 300 ≈ 0.6158 atm .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
For Na₂CO₃, i = 3 (2Na⁺ + CO₃²⁻). Pi = i · M · RT . 1.968 = 3 × 0.2 × 0.0821 × T . T = (1.968/0.6 × 0.0821) ≈ 40 K .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = MRT . Pi = 0.015 × 0.0821 × 300 ≈ 0.3695 atm .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = i · M · RT . Molarity = (0.05/0.5) = 0.1 M . 2.463 = i × 0.1 × 0.0821 × 300 . i = (2.463/0.1 × 24.63) ≈ 1 .
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Pi = i · M · RT . Molarity = (0.03/0.3) = 0.1 M . 1.231 = i × 0.1 × 0.0821 × 300 . i = (1.231/0.1 × 24.63) ≈ 0.5 , recalculate: i = 1 (non-electrolyte).
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
Moles = M × V = 0.5 × 0.4 = 0.2 mol . Molar mass = (Mass/Moles) = (20/0.2) = 100 g/mol . (Density is extra info, not needed here.)
Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis
