Osmotic Pressure: The Life-Sustaining Colligative Property
The Real-Life Hook: Why Salt Kills Slugs and Cells Need Balance
Ever wondered why salt kills garden slugs? Why drinking seawater makes you more dehydrated? Why doctors carefully control IV fluid concentrations? Why your cells would burst in pure water? The answer is osmosis and osmotic pressure - perhaps the most biologically important colligative property!
Daily & Medical Applications:
- Medical saline (0.9% NaCl): Isotonic with blood - safe for IV drips
- Red blood cells: Burst in pure water (hypotonic), shrivel in salt water (hypertonic)
- Reverse osmosis: Desalination of seawater for drinking water
- Food preservation: Salt/sugar prevents bacterial growth by osmosis
- Kidney dialysis: Removes waste products from blood using osmosis
- Plant cell turgidity: Plants stand upright due to osmotic pressure
What is Osmosis?
Definition
Osmosis: The spontaneous movement of solvent molecules from a solution of lower concentration (or pure solvent) to a solution of higher concentration through a semi-permeable membrane.
Semi-Permeable Membrane (SPM): A membrane that allows solvent molecules to pass through but blocks solute molecules.
Examples of SPMs:
- Cellophane paper
- Parchment paper
- Animal bladder
- Cell membranes (biological)
- Thin films of Cu₂[Fe(CN)₆] (copper ferrocyanide)
The Osmosis Setup
flowchart LR
subgraph LEFT["Pure Solvent"]
PS["○ Solvent
molecules"]
end
SPM["SPM
(Semipermeable
Membrane)"]
subgraph RIGHT["Solution"]
SOL["● Solute + ○ Solvent"]
end
LEFT -->|"Solvent flows"| SPM
SPM --> RIGHT
LR["Level rises"] -.-> LEFT
LD["Level drops"] -.-> RIGHTWhat happens:
- Solvent flows from left to right (dilute → concentrated)
- Right side level rises
- Pressure difference builds up
- Eventually, flow stops
Osmotic Pressure: The Equilibrium State
Definition
Osmotic Pressure (π): The external pressure that must be applied to the solution side to just prevent osmosis (to stop the flow of solvent).
Van’t Hoff Equation
$$\boxed{\pi = CRT}$$Where:
- π = Osmotic pressure (atm or Pa)
- C = Molar concentration (molarity) in mol/L
- R = Universal gas constant = 0.0821 L·atm/(mol·K) or 8.314 J/(mol·K)
- T = Absolute temperature (K)
Alternative Forms:
In terms of moles and volume:
$$\boxed{\pi = \frac{n}{V} RT}$$In terms of mass and molar mass:
$$\boxed{\pi = \frac{w}{M \times V} RT}$$Where:
- w = mass of solute (g)
- M = molar mass of solute (g/mol)
- V = volume of solution (L)
For molar mass determination:
$$\boxed{M = \frac{wRT}{\pi V}}$$Similarity to Ideal Gas Equation
Notice the striking similarity:
- Ideal Gas: PV = nRT → P = (n/V)RT
- Osmotic Pressure: π = (n/V)RT
Van’t Hoff’s Insight: Solute particles in solution behave like gas molecules!
Memory Trick:
“π = CRT” - “Pressure Is Clearly Related to Temperature”
Why is Osmotic Pressure Special?
Advantages Over Other Colligative Properties
| Property | Magnitude | Measurement | Best For |
|---|---|---|---|
| RLVP | Very small (~ torr) | Difficult | Small molecules |
| ΔT_b | Very small (~ 0.5 K) | Moderate | Non-polymers |
| ΔT_f | Small (~ 2 K) | Good | Non-polymers |
| π | Large (~ atm) | Easy & Precise | Polymers, proteins |
Example Comparison: For 1% glucose solution at 25°C:
- ΔT_f ≈ 0.1°C (hard to measure precisely)
- π ≈ 2.5 atm (easy to measure with manometer!)
Why π is Superior:
- ✅ Measureable at room temperature (no heating/cooling needed)
- ✅ Large magnitude → more precise measurements
- ✅ Best for high molar mass substances (polymers, proteins)
- ✅ Doesn’t require exact temperature control
- ✅ Works for dilute solutions where other methods fail
Berkeley and Hartley Method: Special apparatus to measure osmotic pressure accurately - most precise method for determining molar masses of macromolecules!
Types of Solutions Based on Osmotic Pressure
When two solutions are separated by SPM:
1. Isotonic Solutions
Definition: Two solutions with equal osmotic pressure.
$$\boxed{\pi_1 = \pi_2 \implies C_1 = C_2 \text{ (at same temperature)}}$$Characteristics:
- No net osmosis
- No change in volume
- Cells maintain normal shape
Examples:
- 0.9% NaCl (normal saline) is isotonic with blood
- 5% glucose solution is isotonic with blood
- Red blood cells in these solutions remain normal
Medical Importance: All IV fluids MUST be isotonic to avoid cell damage!
2. Hypertonic Solution
Definition: Solution with higher osmotic pressure (higher concentration).
Effect on Cells:
- Water flows OUT of cells
- Cells shrink (crenation in RBCs)
- Cell becomes dehydrated
Examples:
- Seawater (3.5% salt) is hypertonic compared to blood
- 10% NaCl solution
- Concentrated sugar syrup
Real-Life Examples:
- 🧂 Salt on slugs: Salt solution is hypertonic → water leaves slug → slug dehydrates and dies
- 🌊 Drinking seawater: Makes you MORE dehydrated (water leaves cells)
- 🥒 Pickling: High salt/vinegar concentration preserves vegetables
3. Hypotonic Solution
Definition: Solution with lower osmotic pressure (lower concentration).
Effect on Cells:
- Water flows INTO cells
- Cells swell and may burst (hemolysis in RBCs)
- Cell becomes over-hydrated
Examples:
- Pure water (0% solute)
- 0.1% NaCl solution
- Dilute sugar solution
Real-Life Examples:
- 💧 RBCs in pure water: Burst due to osmotic pressure
- 🌱 Plant cells in rain: Swell but don’t burst (cell wall protection)
- 🏥 Wrong IV fluid: Hypotonic saline would burst RBCs!
Visual Summary: Cell Behavior
| Condition | Solution | Water Flow | Cell Response |
|---|---|---|---|
| Hypotonic | < 0.9% NaCl | Water IN | Cell swells (may burst) |
| Isotonic | 0.9% NaCl | No net flow | Normal cell |
| Hypertonic | > 0.9% NaCl | Water OUT | Cell shrinks (crenation) |
Reverse Osmosis: Osmosis in Reverse!
Concept
Normal Osmosis: Solvent flows from dilute → concentrated (natural direction)
Reverse Osmosis: Apply pressure GREATER than π to force solvent from concentrated → dilute
Condition:
$$\boxed{P_{\text{applied}} > \pi}$$The solvent is forced to flow “uphill” against its natural tendency!
Applications
1. Desalination of Seawater
Problem: Seawater has ~3.5% salt (π ≈ 25-30 atm)
Solution:
- Apply pressure > 30 atm to seawater
- Pure water forced through SPM
- Salt and minerals left behind
- Result: Drinking water from ocean!
Process:
flowchart LR
P["High Pressure
(>30 atm)"] --> SW["Seawater
(3.5% salt)"]
SW --> SPM["SPM"]
SPM --> PW["Pure Water"]
SW --> CB["Concentrated Brine
(rejected)"]Real-World Use:
- Middle East countries (UAE, Saudi Arabia)
- Israel (50% of drinking water from RO)
- California desalination plants
- Ships and submarines
2. Water Purification
Home RO Systems:
- Remove dissolved salts, heavy metals, bacteria
- Apply 3-6 atm pressure
- 99% pure water output
Removes:
- Dissolved salts (NaCl, CaCl₂)
- Heavy metals (Pb²⁺, Hg²⁺, As³⁺)
- Fluoride, nitrates
- Microorganisms
3. Waste Water Treatment
- Industrial effluent treatment
- Recovering valuable materials
- Producing ultra-pure water for semiconductors
Interactive Demo: Complete Calculation
Problem 1: Basic Osmotic Pressure
Question: Calculate the osmotic pressure of a solution containing 6 g urea (MW = 60) in 1 L solution at 27°C. (R = 0.0821 L·atm/(mol·K))
Solution:
Step 1: Calculate molarity Moles of urea = 6/60 = 0.1 mol Volume = 1 L C = 0.1/1 = 0.1 M
Step 2: Convert temperature T = 27 + 273 = 300 K
Step 3: Calculate π π = CRT π = 0.1 × 0.0821 × 300 π = 2.463 atm
Conversion to other units:
- In mmHg: 2.463 × 760 = 1,872 mmHg
- In Pa: 2.463 × 101,325 = 249,600 Pa ≈ 250 kPa
Problem 2: Molar Mass Determination
Question: 2.5 g of a protein dissolved in 100 mL solution shows osmotic pressure of 0.164 atm at 27°C. Calculate the molar mass of protein.
Solution:
Given:
- w = 2.5 g
- V = 100 mL = 0.1 L
- π = 0.164 atm
- T = 300 K
Using formula: M = wRT / (πV) M = (2.5 × 0.0821 × 300) / (0.164 × 0.1) M = 61.575 / 0.0164 M = 3,755 g/mol
Interpretation: This is a protein with molar mass ~3,750 g/mol - typical for small proteins!
Why Osmotic Pressure? If we used freezing point depression: ΔT_f = K_f × m = 1.86 × (2.5/3755) / 0.1 ≈ 0.012°C
This is too small to measure accurately! But π = 0.164 atm is easy to measure!
Problem 3: Isotonic Solutions
Question: A 5% (w/v) glucose solution is isotonic with blood. What concentration of NaCl solution would be isotonic? (MW: glucose = 180, NaCl = 58.5) Assume NaCl fully ionizes (i = 2).
Solution:
Step 1: Calculate π for glucose solution 5% w/v = 5 g per 100 mL = 50 g/L C_glucose = 50/180 = 0.278 M
π_glucose = 0.278 × RT
Step 2: For isotonic NaCl solution π_NaCl = π_glucose But NaCl ionizes: Na⁺ + Cl⁻ (2 particles)
i × C_NaCl × RT = C_glucose × RT 2 × C_NaCl = 0.278 C_NaCl = 0.139 M
Step 3: Convert to % (w/v) Mass = 0.139 × 58.5 = 8.13 g/L = 0.813 g per 100 mL
Answer: 0.813% NaCl ≈ 0.9% NaCl (normal saline!)
Note: This is why physiological saline is 0.9% NaCl, not 5% like glucose!
Abnormal Osmotic Pressure: Association and Dissociation
Case 1: Dissociation (Electrolytes)
For electrolytes that ionize:
$$\boxed{\pi = iCRT}$$Where i = van’t Hoff factor (number of particles per formula unit)
Examples:
- NaCl → Na⁺ + Cl⁻ (i = 2)
- CaCl₂ → Ca²⁺ + 2Cl⁻ (i = 3)
- Glucose (non-electrolyte, i = 1)
Effect: π_observed = i × π_theoretical
Example: 0.1 M NaCl: π = 2 × 0.1 × RT = 0.2 RT (double the pressure!)
Case 2: Association (Dimerization)
Some molecules associate (stick together) in solution:
Example: Acetic acid in benzene forms dimers 2 CH₃COOH ⇌ (CH₃COOH)₂
Effect:
- Fewer particles than expected
- π_observed < π_theoretical
- i < 1
Example: If 50% dimerization:
- Start with 1 mol (100 particles)
- Form dimers: 50% → 25 dimers + 50 monomers = 75 particles
- i = 75/100 = 0.75
Practical Applications in Detail
1. Medical: IV Fluids
Requirements for IV fluids:
- Must be isotonic (π ≈ 7.8 atm at 37°C)
- Must be sterile
- Must be physiologically compatible
Common IV Solutions:
- 0.9% NaCl (“Normal Saline”)
- 5% Glucose (D5W - Dextrose 5% in Water)
- Ringer’s Lactate (balanced electrolytes)
Why 0.9% NaCl specifically?
- Blood plasma has π ≈ 7.8 atm
- 0.9% NaCl has π ≈ 7.8 atm (isotonic!)
- Safe for direct blood contact
2. Biological: Plant Cell Turgidity
Turgid Cell: Plant cell in hypotonic solution
- Water enters cell by osmosis
- Cell swells against rigid cell wall
- Creates turgor pressure
- Plant stands upright!
Plasmolyzed Cell: Plant cell in hypertonic solution
- Water leaves cell
- Cell membrane shrinks away from cell wall
- Plant wilts
This is why:
- 🌱 Plants need water to stay upright (turgor)
- 🥀 Plants wilt in salty soil (plasmolysis)
- 🌾 Fertilizer “burn” happens with too much fertilizer (hypertonic)
3. Food Preservation
High Salt/Sugar Preserves Food:
Principle:
- Salt/sugar solution is hypertonic
- Bacteria cells lose water by osmosis
- Bacteria dehydrate and die
- Food preserved!
Examples:
- 🥒 Pickles (high salt)
- 🍯 Honey (high sugar, π > 100 atm!)
- 🐟 Salted fish
- 🍇 Dried fruits (concentrated sugar)
4. Kidney Dialysis
Problem: Kidney failure → waste products (urea, creatinine) accumulate in blood
Solution: Hemodialysis
Blood (high waste) → SPM ← Dialysate (clean)
↓
Waste flows to dialysate
Mechanism:
- Blood flows on one side of membrane
- Clean dialysate on other side
- Waste products diffuse out (high → low concentration)
- Essential electrolytes maintained at proper levels
Osmotic Balance: Dialysate is carefully formulated to be isotonic for essential components (Na⁺, K⁺, Ca²⁺) but hypotonic for waste products (urea, creatinine).
Common JEE Mistake Traps
Mistake 1: Using Molality Instead of Molarity
❌ Wrong: π = mRT (using molality) ✅ Correct: π = CRT (using molarity)
Why? π depends on number of particles per unit VOLUME, not per unit mass!
Mistake 2: Forgetting van’t Hoff Factor
❌ Wrong: For NaCl, π = CRT ✅ Correct: For NaCl, π = iCRT = 2CRT
Remember: Electrolytes dissociate!
Mistake 3: Wrong R Value
❌ Wrong: Using R = 8.314 J/(mol·K) with atm ✅ Correct: Use R = 0.0821 L·atm/(mol·K) when π is in atm
Or: Use R = 8.314 J/(mol·K) when π is in Pa
Mistake 4: Celsius vs Kelvin
❌ Wrong: π = CRT with T in °C ✅ Correct: T must be in Kelvin (K = °C + 273)
Mistake 5: Concentration Direction in Osmosis
❌ Wrong: Water flows from concentrated → dilute ✅ Correct: Water flows from dilute → concentrated
Memory Aid: “Water goes where salt is” - it follows the solute!
Practice Problems: Three-Level Mastery
Level 1: JEE Main Foundation
Q1. Calculate osmotic pressure of 0.2 M glucose solution at 27°C. (R = 0.0821 L·atm/(mol·K))
Solution
T = 27 + 273 = 300 K π = CRT = 0.2 × 0.0821 × 300 π = 4.926 atm
Q2. A solution has osmotic pressure of 2.46 atm at 300 K. Calculate its molarity. (R = 0.0821)
Solution
π = CRT 2.46 = C × 0.0821 × 300 C = 2.46 / (0.0821 × 300) C = 2.46 / 24.63 C = 0.1 M
Q3. Which has higher osmotic pressure at same temperature? (a) 0.1 M glucose (b) 0.1 M NaCl (c) 0.1 M CaCl₂
Solution
Calculate effective concentration (C × i):
- Glucose: 0.1 × 1 = 0.1
- NaCl: 0.1 × 2 = 0.2 (Na⁺ + Cl⁻)
- CaCl₂: 0.1 × 3 = 0.3 (Ca²⁺ + 2Cl⁻)
Answer: (c) 0.1 M CaCl₂ has highest osmotic pressure!
Level 2: JEE Main Advanced
Q4. 3 g of glucose (MW = 180) and 5.85 g NaCl (MW = 58.5) are dissolved in 500 mL water. Calculate osmotic pressure at 27°C. Assume complete ionization of NaCl.
Solution
Glucose: Moles = 3/180 = 0.0167 mol Effective moles = 0.0167 (i = 1)
NaCl: Moles = 5.85/58.5 = 0.1 mol Effective moles = 0.1 × 2 = 0.2 mol (i = 2)
Total effective moles = 0.0167 + 0.2 = 0.2167 mol
Concentration: C = 0.2167/0.5 = 0.4334 M
Osmotic Pressure: T = 300 K π = 0.4334 × 0.0821 × 300 π = 10.67 atm
Q5. A 1.8% (w/v) solution of a substance is isotonic with 0.9% (w/v) NaCl solution. Calculate molar mass of the substance. (MW of NaCl = 58.5)
Solution
For isotonic solutions: π₁ = π₂
NaCl solution: C_NaCl = 0.9/58.5 = 0.01538 M Effective C = 0.01538 × 2 = 0.03076 M
Substance solution: C_substance = 1.8/M (where M is molar mass)
Equating: 1.8/M = 0.03076 M = 1.8/0.03076 M = 58.5 g/mol
Note: The substance has same molar mass as NaCl but doesn’t ionize (i = 1), so needs double the concentration!
Q6. 200 cm³ of an aqueous solution containing 0.6 g of a protein shows osmotic pressure of 2.46 × 10⁻³ atm at 300 K. Calculate molar mass of protein.
Solution
Given:
- w = 0.6 g
- V = 200 cm³ = 0.2 L
- π = 2.46 × 10⁻³ atm
- T = 300 K
Formula: M = wRT / (πV) M = (0.6 × 0.0821 × 300) / (2.46 × 10⁻³ × 0.2) M = 14.778 / (4.92 × 10⁻⁴) M = 30,037 g/mol ≈ 30 kDa
Note: This is typical for small proteins. The small π value shows why osmotic pressure is best for macromolecules!
Level 3: JEE Advanced Challenge
Q7. (JEE Advanced Type) Two solutions A and B are separated by semi-permeable membrane:
- Solution A: 0.1 M glucose
- Solution B: 0.05 M CaCl₂
At 27°C, which direction will osmosis occur and what is the effective pressure difference?
Solution
Solution A (glucose): Effective C_A = 0.1 × 1 = 0.1 M (non-electrolyte) π_A = 0.1 × 0.0821 × 300 = 2.463 atm
Solution B (CaCl₂): CaCl₂ → Ca²⁺ + 2Cl⁻ (i = 3) Effective C_B = 0.05 × 3 = 0.15 M π_B = 0.15 × 0.0821 × 300 = 3.695 atm
Comparison: π_B > π_A
Direction: Water flows from A → B (from lower to higher π)
Pressure difference: Δπ = 3.695 - 2.463 = 1.232 atm
Answer: Osmosis occurs from glucose solution to CaCl₂ solution with driving force of 1.23 atm.
Q8. (JEE Advanced 2017 Type) At what concentration will a glucose solution be isotonic with 1% (w/v) NaCl solution at the same temperature? Assume complete dissociation. (MW: glucose = 180, NaCl = 58.5)
Solution
NaCl solution: 1% w/v = 1 g per 100 mL = 10 g/L C_NaCl = 10/58.5 = 0.171 M Effective C = 0.171 × 2 = 0.342 M (i = 2) π_NaCl = 0.342 RT
Glucose solution (isotonic): π_glucose = π_NaCl C_glucose × RT = 0.342 RT C_glucose = 0.342 M
Convert to % (w/v): Mass = 0.342 × 180 = 61.56 g/L = 6.156 g per 100 mL
Answer: 6.16% (w/v) glucose solution
Interpretation: Glucose solution needs ~6× higher concentration than NaCl to be isotonic because glucose doesn’t ionize!
Q9. (Challenging) A solution contains a mixture of glucose and sucrose. 1 L of this solution has osmotic pressure of 4.926 atm at 300 K. If the solution contains 18 g glucose, calculate the amount of sucrose. (MW: glucose = 180, sucrose = 342)
Solution
Total osmotic pressure: π_total = 4.926 atm C_total = π/(RT) = 4.926/(0.0821 × 300) = 0.2 M
From glucose: Moles of glucose = 18/180 = 0.1 mol C_glucose = 0.1/1 = 0.1 M
From sucrose: C_sucrose = C_total - C_glucose = 0.2 - 0.1 = 0.1 M Moles of sucrose = 0.1 × 1 = 0.1 mol Mass of sucrose = 0.1 × 342 = 34.2 g
Answer: 34.2 g sucrose
Q10. (JEE Advanced Level) Calculate the minimum pressure required to desalinate seawater at 25°C. Assume seawater contains 3.5% (w/v) NaCl and complete ionization. (MW of NaCl = 58.5)
Solution
Seawater concentration: 3.5% w/v = 35 g/L C_NaCl = 35/58.5 = 0.598 M
Effective concentration (with ionization): C_eff = 0.598 × 2 = 1.196 M
Osmotic pressure of seawater: T = 25 + 273 = 298 K π = 1.196 × 0.0821 × 298 π = 29.26 atm
Minimum pressure for reverse osmosis: P_min = π = 29.26 atm ≈ 30 atm
In practice:
- Applied pressure = 50-80 atm (to ensure good flow rate)
- Energy intensive process!
- This is why RO plants are expensive to operate
Additional calculation: 30 atm = 30 × 101.325 kPa = 3,040 kPa = 3.04 MPa = 30 × 14.7 psi = 441 psi
Answer: Minimum ~30 atm, practical ~50-80 atm
Interactive Demo: Visualize Osmosis
Watch water molecules move through a semi-permeable membrane from dilute to concentrated solution.
Conceptual Questions for JEE
Q1. Why does osmotic pressure measurement give more accurate molar masses for proteins than freezing point depression?
Answer
Reasons:
- Magnitude: π is measurable in atm, while ΔT_f is tiny (< 0.01°C) for dilute protein solutions
- Room temperature: No heating/cooling required
- Precision: Pressure measured to ±0.001 atm, temperature to ±0.01°C
- Proportionality: For macromolecules, π is still large enough but ΔT_f is too small
Example: 1% protein solution
- ΔT_f ≈ 0.003°C (very hard to measure!)
- π ≈ 0.1 atm (easy to measure!)
Q2. Why do plants wilt when soil is over-fertilized?
Answer
Explanation:
- Excess fertilizer increases salt concentration in soil water
- Soil water becomes hypertonic relative to plant cells
- Water flows OUT of plant cells by osmosis
- Cells lose turgor pressure → plasmolysis
- Plant wilts
This is called “fertilizer burn” - not actual burning, but osmotic dehydration!
Prevention: Use dilute fertilizer solutions, water after fertilizing
Q3. Explain why drinking seawater causes dehydration.
Answer
Process:
- Seawater has ~3.5% salt → hypertonic compared to body fluids (~0.9% salt)
- Drinking seawater → intestines absorb some salt
- Blood becomes hypertonic
- Body cells lose water to blood by osmosis
- Kidneys try to excrete excess salt → need water
- Net result: More water lost than gained → dehydration
Calculation: To excrete salt from 1 L seawater, body needs to produce ~1.5 L urine → net loss of 0.5 L water!
Moral: Never drink seawater for survival!
Summary Table: Osmotic Pressure
| Aspect | Details |
|---|---|
| Formula | π = CRT = (n/V)RT |
| Units | atm, Pa, mmHg |
| Depends on | Molarity (not molality!) |
| Best for | Polymers, proteins (high MW) |
| Temperature | Room temperature measurement |
| Advantage | Large, measurable values |
Comparison with other colligative properties:
| Property | Formula | Advantage |
|---|---|---|
| RLVP | ΔP/P° = χ | Fundamental basis |
| ΔT_b | K_b m | Simple apparatus |
| ΔT_f | K_f m | More sensitive than ΔT_b |
| π | CRT | Best for macromolecules |
Cross-Connections to Other Chapters
🔗 Link to Colligative Properties
- Fourth colligative property
- All stem from vapor pressure lowering
- See: Colligative Properties
🔗 Link to van’t Hoff Factor
- π = iCRT for electrolytes
- Abnormal molar masses from association/dissociation
- See: van’t Hoff Factor
🔗 Link to Chemical Equilibrium
- Osmosis is an equilibrium process
- ΔG = 0 at osmotic equilibrium
- See: Equilibrium
🔗 Link to Thermodynamics
- Osmotic work = πV
- Spontaneous process (ΔG < 0) until equilibrium
- See: Gibbs Free Energy
🔗 Link to Electrochemistry
- Similar to membrane potential in cells
- Nernst equation analogy
- See: Electrochemical Cells
🔗 Link to Biology
- Cell membrane transport
- Active vs passive transport
- Homeostasis maintenance
Formula Sheet: Quick Reference
Basic Formula
$$\boxed{\pi = CRT = \frac{n}{V}RT = \frac{w}{MV}RT}$$Molar Mass Determination
$$\boxed{M = \frac{wRT}{\pi V}}$$For Electrolytes
$$\boxed{\pi = iCRT}$$Isotonic Condition
$$\boxed{\pi_1 = \pi_2 \implies C_1 = C_2}$$Reverse Osmosis
$$\boxed{P_{\text{applied}} > \pi}$$Important Constants
- R = 0.0821 L·atm/(mol·K)
- R = 8.314 J/(mol·K)
- Normal saline: 0.9% NaCl (isotonic with blood)
Final JEE Strategy
✅ Highest Weightage Topics:
- Molar mass from π (60% probability)
- Isotonic solutions (50% probability)
- Direction of osmosis (40% probability)
- Reverse osmosis concept (30% probability in Advanced)
✅ Must-Remember Facts:
- π = CRT (uses molarity, not molality!)
- Normal saline = 0.9% NaCl
- Seawater ≈ 3.5% salt → π ≈ 30 atm
- Isotonic glucose ≈ 5% (w/v)
✅ Common Question Types:
- Calculate π from concentration
- Calculate M from π
- Isotonic solution problems
- Compare π of different solutions
- Effect on cells (hypo/iso/hyper-tonic)
✅ Calculation Checklist:
- Using molarity (C), not molality (m)? ✓
- Temperature in Kelvin? ✓
- Correct R value (0.0821 for atm)? ✓
- Applied van’t Hoff factor for electrolytes? ✓
- Volume in liters? ✓
✅ Conceptual Understanding:
- Water flows from low π to high π
- Reverse osmosis needs P > π
- Cells burst in hypotonic, shrink in hypertonic
- π is best for high MW compounds
Previous Topic: Colligative Properties Next Topic: van’t Hoff Factor - understanding abnormal colligative properties!
Last Updated: June 2025 | For JEE Main & Advanced 2026