Colligative Properties - Osmotic Pressure and Reverse Osmosis

Colligative Properties - Osmotic Pressure and Reverse Osmosis

A gas has a Henry’s law constant of 300 bar. If its partial pressure is 6 bar, what is the mole fraction in the solution

p = KH · x . x = (6/300) = 0.02 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Colligative Properties - Osmotic Pressure and Reverse Osmosis

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

The osmotic pressure of a solution containing 5 g of a solute in 1 L of solution at 27°C is 2.46 atm. What is the molar

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A solution is prepared by dissolving 15 g of a solute in water to make 250 mL of solution. If the osmotic pressure is 1.

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

What is the osmotic pressure of a 0.02 M solution of a non-electrolyte at 27°C? ( R = 0.0821 L atm mol⁻¹ K⁻¹ )

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A solution contains 9 g of a solute (molar mass = 90 g/mol) in 450 mL of solution. What is the molarity?

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A 0.25 M solution of a solute in 400 mL of water has an osmotic pressure of 1.845 atm at 27°C. If the solute dissociates

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

What is the osmotic pressure of a 0.025 M solution of a non-electrolyte at 27°C? ( R = 0.0821 L atm mol⁻¹ K⁻¹ )

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A 0.2 M solution of Na₂CO₃ (assuming complete dissociation) has an osmotic pressure of 1.968 atm at a certain temperatur

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

What is the osmotic pressure of a 0.015 M solution of a non-electrolyte at 27°C? ( R = 0.0821 L atm mol⁻¹ K⁻¹ )

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A solution of 0.05 mole of a non-volatile solute in 500 g of water has an osmotic pressure of 2.463 atm at 27°C. What is

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A solution of 0.03 mole of a non-volatile solute in 300 mL of water has an osmotic pressure of 1.231 atm at 27°C. What i

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

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Colligative Properties - Osmotic Pressure and Reverse Osmosis

A solution is prepared by dissolving 20 g of a solute in water to make 400 mL of solution with a density of 1.25 g/mL. I

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

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