#osmotic pressure

Membrane Transportation

Which condition would cause the highest osmotic pressure?

Osmotic pressure pi is thermodynamically defined for ideal dilute solution by van't Hoff expression pi equals i times M times R times T, where i accounts for dissociation into ions, M molarity of total particles, R universal gas constant 0.082 liter atm per mol per K, T absolute temperature. Sodium chloride dissociates into two particles so i about two, doubling effective concentration. Consequently pi scales linearly with total osmolarity. Among listed options, 0.1 molar NaCl yields approximately 0.2 osmolar giving pi about 4.9 atm at 298 K, 0.5 molar yields 1.0 osmolar about 24.5 atm, 1.0 mo

Ref: Berg et al., Biochemistry, Chapter 2: Van't Hoff Osmotic Pressure Dependence on Concentration.

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Membrane Transportation

Which of the following conditions describes osmotic pressure?

Osmotic pressure is a colligative property describing tendency of water to move across semipermeable membranes that allow solvent but restrict solute permeation. When compartment containing dilute solution is separated from concentrated solution, water chemical potential is higher where solute activity is lower. Spontaneous diffusion drives water toward higher solute until chemical potentials equalize, causing volume increase and hydrostatic pressure rise on concentrated side. The additional hydrostatic pressure required to stop net influx and hold volumes constant at equilibrium defines osmot

Ref: Alberts et al., Molecular Biology of the Cell, Chapter 11: Osmotic Pressure Definition and van't Hoff Equation.

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Microbial Physiology

In bacterial growth, water availability is controlled by:

Water availability parameter aw water activity influences growth because intracellular reactions require aqueous medium, turgor pressure drives expansion. aw equals relative humidity 100 percent in equilibrium. Bacteria regulate response to aw change mainly via osmotic pressure sensing osmosensors EnvZ. Hyperosmotic shock water exits cytoplasm plasmolysis outer membrane separates, cytoplasm shrivels, growth arrested. Compatible solutes accumulation potassium glutamate, proline, ectoine, trehalose synthesised or imported via ProP, BetT transporters balances osmolarity without denaturing protein

Ref: Brock Biology of Microorganisms, 16th ed., Chapter 5: Water activity and osmotic pressure.

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Types of Solutions and Expressing Concentration

The van’t Hoff factor of a 0.05 m CaCl₂ solution is 2.7. What is the osmotic pressure at 27°C? ( R = 0.0821 L atm mol⁻¹

Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.05 M . Pi = 2.7 × 0.05 × 0.0821 × 300 ≈ 3.326 atm .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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Types of Solutions and Expressing Concentration

A solution containing 0.1 mole of a non-volatile solute in 1 kg of water has an osmotic pressure of 2.46 atm at 27°C. Wh

Pi = i · M · RT . Molarity ≈ molality for dilute solution, so M = 0.1 M . 2.46 = i × 0.1 × 0.0821 × 300 . i = (2.46/0.1 × 0.0821 × 300) ≈ 1 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration

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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 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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