ChemistryISC Class 12 · Solutions

Osmosis and Osmotic Pressure: ISC Class 12 Notes and Numericals

Updated 15 October 2026 · Fact-checked

Osmosis is the net flow of solvent through a semipermeable membrane from pure solvent, or a dilute solution, into a more concentrated solution. Osmotic pressure is the extra pressure on the solution that just stops this flow. For dilute solutions π = CRT, or π = iCRT for electrolytes, so it gives molar masses of proteins and polymers.

Only water crosses the membrane, moving toward the sugar side; the extra pressure π needed to stop it equals CRT.

Illustration drawn for QuestionBuddy.

Understand Osmosis and Osmotic Pressure

Picture a membrane with pores so fine that water molecules slip through but sugar molecules do not. Such a semipermeable membrane (SPM) may be natural (animal bladder, the membrane of a living cell, parchment) or chemical (a film of copper ferrocyanide deposited in the walls of a porous pot). Put pure water on one side and a sugar solution on the other. Water molecules cross in both directions, but more cross from the pure solvent side, because there the solvent has a higher escaping tendency. The result is a net flow of solvent into the solution. This is osmosis. Only the solvent moves, and it always moves from the side of lower solute concentration to the side of higher concentration.

The flow can be stopped by pressing on the solution. The osmotic pressure $π$ of a solution is the excess pressure that must be applied on the solution side to just prevent osmosis when it is separated from pure solvent by an SPM. It depends on the number of solute particles, not on what they are, so it is a colligative property.

The van't Hoff laws. For dilute solutions van't Hoff found that solutes behave like gases: (i) van't Hoff-Boyle's law: at constant temperature $π \propto C$ ; (ii) van't Hoff-Charles' law: at constant concentration $π \propto T$ ; (iii) van't Hoff-Avogadro's law: isotonic solutions at the same temperature contain equal numbers of solute particles in equal volumes. Combining (i) and (ii) gives $π = C R T = \frac{n}{V} R T$ , where $C$ is molarity, $V$ is in litres and $R = 0.0821$ L atm K⁻¹ mol⁻¹. With $n = w / M$ , the molar mass is $M = \frac{w R T}{π V}$ .

Why it suits big molecules. For a protein of molar mass 50 000 g mol⁻¹, a few grams per litre lower the freezing point by only thousandths of a kelvin, but give an osmotic pressure of a few millimetres of mercury, which can be measured. The measurement is made at room temperature, so heat-sensitive biomolecules are not damaged, and it uses molarity, so no solvent mass is needed.

Isotonic, hypertonic, hypotonic. Two solutions with the same osmotic pressure at the same temperature are isotonic; no osmosis occurs between them. The fluid inside red blood cells is isotonic with 0.9% (mass/volume) NaCl, called normal saline, which is why it is safe for intravenous drips. A solution of higher osmotic pressure (more solute particles) is hypertonic: a cell placed in it loses water (exosmosis) and shrinks. A solution of lower osmotic pressure is hypotonic: a cell placed in it gains water (endosmosis) and swells, and may burst. A raw mango in strong brine shrivels into pickle, salted meat and sugared fruit resist bacteria (the bacteria lose water), and limp carrots or wilted flowers become firm again in fresh water.

Electrolytes. NaCl gives two particles per formula unit, so its solution has nearly twice the osmotic pressure of a glucose solution of the same molarity. The van't Hoff factor corrects for this: $π = i C R T$ , with $i = 2$ for NaCl, $3$ for Na₂SO₄ or CaCl₂ when fully dissociated.

Reverse osmosis. If a pressure larger than $π$ is applied on the solution side, solvent is pushed out of the solution through the membrane. This reverses the natural direction of flow and is used to desalinate sea water: pure water is squeezed out through a film of cellulose acetate on a support, which lets water through but holds back ions and impurities.

Diffusion versus osmosis. Diffusion needs no membrane and both solute and solvent spread out; osmosis needs a semipermeable membrane and only the solvent moves.

Osmosis and Osmotic Pressure: formulas and key facts

Osmotic pressure (van't Hoff equation): $π = C R T = \frac{n}{V} R T$; $C$ in mol L⁻¹, $V$ in litres, $T$ in kelvin, $R = 0.0821$ L atm K⁻¹ mol⁻¹ gives $π$ in atm.
Molar mass from osmotic pressure: $M = \frac{w R T}{π V}$; Convert cm³ (mL) to litres and mm Hg to atm (÷ 760) before substituting.
With the van't Hoff factor: $π = i C R T$; Complete dissociation: $i = 2$ for NaCl, KCl; $i = 3$ for Na₂SO₄, CaCl₂, BaCl₂; $i = 1$ for glucose, urea, sucrose.
Isotonic solutions: $i_{1} C_{1} = i_{2} C_{2} \Rightarrow \frac{w _{1}}{M _{1} V _{1}} = \frac{w _{2}}{M _{2} V _{2}} (non-electrolytes)$; Same temperature; equal osmotic pressure means equal concentration of particles.
van't Hoff-Boyle and van't Hoff-Charles laws: $π \propto C (T constant), π \propto T (C constant)$; Together with van't Hoff-Avogadro's law they give $π V = n R T$ .

How to solve Osmosis and Osmotic Pressure questions

ISC asks osmotic pressure as direct calculations, molar mass of a macromolecule, isotonic comparisons and reasoning on cells and reverse osmosis. One routine covers them.

1Write the formula first: $π = i C R T$ or $M = \frac{w R T}{π V}$ . The formula line carries a mark.
2Convert units: volume to litres (200 cm³ = 0.200 L), temperature to kelvin, pressure to atm (mm Hg ÷ 760) so that $R = 0.0821$ L atm K⁻¹ mol⁻¹ fits.
3Decide $i$ : 1 for a non-electrolyte; the number of ions for a fully dissociated salt; use the given degree of dissociation if one is stated.
4Substitute in one line, then calculate; keep three significant figures and write the unit (atm, g mol⁻¹).
5For isotonic problems, equate particle concentrations: $i_{1} C_{1} = i_{2} C_{2}$ ; masses per litre then give the unknown molar mass.
6For direction-of-flow questions, compare $i C$ on each side: solvent flows towards the larger $i C$ (higher $π$ ).
7For cells, say hypertonic outside means the cell shrinks (exosmosis); hypotonic outside means it swells (endosmosis).

Osmosis and Osmotic Pressure in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A hypertonic syrup preserves fruit by osmosis

Suppose your grandmother preserves amla in a sugar syrup made with 342 g of sucrose in enough water to make 1 litre, kept at 300 K. Assume the cell sap of the fruit, and of any bacteria on it, behaves like a solution with 0.30 mol of solute particles per litre.

Calculate the osmotic pressure of the syrup and of the cell sap, and explain why the fruit shrinks a little and why bacteria cannot grow in the syrup.

Try it first, then show the solutionSolution

Syrup: $n = 342/342 = 1.0$ mol in 1 L, so $C = 1.0$ M and $π = 1.0 \times 0.0821 \times 300 = 24.6$ atm.
Cell sap: $π = 0.30 \times 0.0821 \times 300 = 7.39$ atm.
The syrup has the higher osmotic pressure, so it is hypertonic to the cells.
Water flows out of the fruit cells and out of any bacterial cells into the syrup (exosmosis). The fruit loses some water; the bacteria lose water, shrivel and cannot multiply.

Answer: Syrup about 24.6 atm, cell sap about 7.4 atm. The syrup is hypertonic, so water leaves the fruit and bacterial cells by exosmosis, which preserves the fruit.

Memory hook: Water always moves towards the saltier or sweeter side: that is why sugar and salt preserve food.

Tricks to remember Osmosis and Osmotic Pressure

Short memory aids. Say each one aloud once, then use it on a question the same day.

Solvent follows the solute
Water goes where more particles are. Higher $i C$ means higher $π$ , and the solvent flows towards it.
Use it when: Any 'in which direction will water flow?' question.
Hyper shrinks, hypo swells
HYPERtonic outside: the cell is 'hyper-dried' and shrinks. HYPOtonic outside: think of a hippo, the cell swells up big.
Use it when: Red blood cells, eggs and plant cells placed in different solutions.
Pie = I See RT
$π = i C R T$ : read it as 'pie equals I see RT', and never forget the $i$ for salts.
Use it when: Writing the formula line in electrolyte questions.

Osmosis and Osmotic Pressure: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Which pair is isotonic? · 1 mark · MCQBoard level

Assuming complete dissociation of the salts, which pair of aqueous solutions is isotonic at the same temperature?
(a) 0.1 M glucose and 0.1 M NaCl
(b) 0.1 M urea and 0.05 M NaCl
(c) 0.1 M sucrose and 0.1 M CaCl₂
(d) 0.2 M glucose and 0.1 M Na₂SO₄

Try it first, then show the solutionSolution

Isotonic solutions have equal particle concentration $i C$ .
(a) 0.1 and 0.2; (b) 0.1 and $2 \times 0.05 = 0.1$ ; (c) 0.1 and 0.3; (d) 0.2 and 0.3.
Only (b) has equal values.

Answer: (b) 0.1 M urea and 0.05 M NaCl.

Example 2: Osmotic pressure of a sucrose solution · 2 marks · Short answerBoard level

Calculate the osmotic pressure at 27 °C of a solution made by dissolving 3.42 g of sucrose ( $M = 342$ g mol⁻¹) in water to give 250 mL of solution. ( $R = 0.0821$ L atm K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

$n = \frac{3.42}{342} = 0.0100$ mol; $V = 0.250$ L, so $C = 0.0400$ mol L⁻¹; $T = 300$ K (½ mark).
$π = C R T$ (½ mark).
$π = 0.0400 \times 0.0821 \times 300 = 0.985$ atm (1 mark).

Answer: $π \approx 0.985$ atm.

Example 3: Molar mass of a protein · 3 marks · Short answerHarder board level

0.80 g of a protein is dissolved in water to make 100 mL of solution. Its osmotic pressure at 310 K is 3.10 mm Hg. (a) Calculate the molar mass of the protein. (b) Give one reason why osmotic pressure, and not depression of freezing point, is used for this measurement. ( $R = 0.0821$ L atm K⁻¹ mol⁻¹)

Try it first, then show the solutionSolution

Convert: $π = \frac{3.10}{760} = 4.079 \times 1 0^{- 3}$ atm; $V = 0.100$ L (½ mark).
$M = \frac{w R T}{π V} = \frac{0.80 \times 0.0821 \times 310}{4.079 \times 1 0 ^{- 3} \times 0.100}$ (1 mark).
$M = \frac{20.36}{4.079 \times 1 0 ^{- 4}} \approx 4.99 \times 1 0^{4}$ g mol⁻¹ (½ mark).
(b) The protein would give a freezing-point depression too small to measure, while $π$ is a few mm Hg and is measured at room temperature, so the protein is not denatured (1 mark).

Answer: (a) About $4.99 \times 1 0^{4}$ g mol⁻¹ (49 900 g mol⁻¹). (b) Osmotic pressure is large enough to measure for dilute solutions of macromolecules and is measured at room temperature.

Example 4: Isotonic solutions: unknown molar mass · 3 marks · Short answerHarder board level

A 6.84% (mass/volume) solution of sucrose ( $M = 342$ g mol⁻¹) is isotonic with a 1.8% (mass/volume) solution of a non-electrolyte X at the same temperature. (a) Find the molar mass of X. (b) What percentage (mass/volume) of NaCl, assumed fully dissociated, would be isotonic with the same sucrose solution? (Na = 23, Cl = 35.5)

Try it first, then show the solutionSolution

Sucrose: 68.4 g per litre, $C = \frac{68.4}{342} = 0.200$ M (½ mark).
(a) Isotonic, both $i = 1$ : X has 18 g per litre at 0.200 M, so $M_{X} = \frac{18}{0.200} = 90$ g mol⁻¹ (1 mark).
(b) NaCl has $i = 2$ , so it needs $C = \frac{0.200}{2} = 0.100$ M (½ mark).
Mass per litre $= 0.100 \times 58.5 = 5.85$ g, i.e. 0.585% (mass/volume) (1 mark).

Answer: (a) $M_{X} = 90$ g mol⁻¹. (b) 0.585% (mass/volume) NaCl.

Example 5: How far can reverse osmosis go? · 5 marks · Long answerHOTS / competency

Treat sea water as a 0.60 M solution of NaCl that is fully dissociated, at 298 K. ( $R = 0.0821$ L atm K⁻¹ mol⁻¹)
(a) Calculate its osmotic pressure. [2]
(b) A desalination unit applies 40 atm on the sea-water side of a cellulose acetate membrane. Explain why pure water comes out. [1]
(c) As pure water is removed, the remaining sea water becomes more concentrated. Find the NaCl concentration at which the flow stops, and the percentage of the original water that can be recovered at 40 atm. [2]

Try it first, then show the solutionSolution

(a) $π = i C R T = 2 \times 0.60 \times 0.0821 \times 298$ (1 mark) $= 29.4$ atm (1 mark).
(b) 40 atm is greater than the osmotic pressure, so the natural flow is reversed: water is pushed out of the sea water through the membrane, while the membrane holds back the ions (reverse osmosis) (1 mark).
(c) Flow stops when $π$ of the remaining brine reaches 40 atm: $C = \frac{40}{2 \times 0.0821 \times 298} = 0.817$ M (1 mark).
The amount of NaCl stays fixed, so the volume falls to $\frac{0.60}{0.817} = 0.734$ of the original: about 27% of the water can be recovered (1 mark).

Answer: (a) About 29.4 atm. (b) The applied pressure exceeds π, so solvent flows out of the solution (reverse osmosis). (c) Flow stops at about 0.82 M NaCl, after about 27% of the water has been recovered.

Where students lose marks in Osmosis and Osmotic Pressure

The slips examiners see most often, and what to write instead.

Mistake: Leaving the volume in cm³
ISC examiners reported that many candidates did not convert 200 cm³ to 0.2 L in a protein molar mass question (2024).
Wrong working
$M = \frac{0.80 \times 0.0821 \times 310}{4.079 \times 1 0 ^{- 3} \times 100}$
Right working
$M = \frac{0.80 \times 0.0821 \times 310}{4.079 \times 1 0 ^{- 3} \times 0.100}$
Do this: With $R = 0.0821$ L atm K⁻¹ mol⁻¹, volume must be in litres and pressure in atm.
Usually costs: The final answer mark, and often the substitution mark.
Mistake: Forgetting the van't Hoff factor for a salt
Examiners noted that candidates did not use $i = 3$ for Na₂SO₄ in an osmotic pressure calculation (2025).
Wrong working
$π = C R T$ for Na₂SO₄
Right working
$π = i C R T$ with $i = 3$ (2Na⁺ + SO₄²⁻)
Do this: Before substituting, count the ions a salt gives and write $i$ explicitly.
Usually costs: 1 to 2 marks.
Mistake: Getting the cell's response backwards
Examiners recorded answers saying red blood cells swell in a hypertonic solution, or that an egg shows no change in pure water.
Wrong working
RBCs placed in strong salt solution swell.
Right working
In a hypertonic solution water leaves the cells (exosmosis) and they shrink; in pure water (hypotonic) they swell.
Do this: Find the side with more solute; water moves towards it.
Usually costs: 1 mark per part.
Mistake: Describing reverse osmosis without the direction of flow
Candidates defined reverse osmosis but were unclear which way the solvent moves and where pressure is applied.
Do this: Write: a pressure greater than the osmotic pressure is applied on the solution (sea water) side, so pure solvent flows out of the solution through the membrane.
Usually costs: 1 mark.
Mistake: Giving a numerical answer with no working
In a 2024 case-based question examiners noted that answers given without steps lost marks.
Do this: Write formula, substitution with units, then the answer; each line earns credit.
Usually costs: Up to all but the answer mark.

Exam tips

Use the value and unit of $R$ that the question gives; ISC usually gives $R = 0.0821$ L atm K⁻¹ mol⁻¹, so work in litres and atm.
For proteins and polymers, write the large molar mass with a comma or in standard form (60 377 g mol⁻¹, not 60.377 g mol⁻¹).
In 'explain' questions, use the words endosmosis, exosmosis, hypertonic and hypotonic; they are the marking points.
When comparing solutions, convert every solution to particle concentration $i C$ before deciding isotonic or the direction of flow.
State that osmotic pressure is colligative because it depends on the number of solute particles, not their nature.

Practice questions on Osmosis and Osmotic Pressure

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A red blood cell is placed in a 1.5% (mass/volume) solution of NaCl. The cell will:
(a) swell and burst
(b) shrink
(c) remain unchanged
(d) first swell and then shrink
Hint
The fluid in the cell is isotonic with 0.9% (mass/volume) NaCl.
Answer
(b). 1.5% NaCl is hypertonic to the cell, so water leaves the cell by exosmosis and it shrinks.
Q2 · 2 marks · Short answerBoard level
State van't Hoff-Boyle's law and van't Hoff-Charles' law for dilute solutions, and show how they combine into the van't Hoff equation for osmotic pressure.
Hint
One law fixes temperature, the other fixes concentration.
Answer
Boyle's form: at constant temperature, osmotic pressure is directly proportional to concentration, $π \propto C$ . Charles' form: at constant concentration, $π \propto T$ . Together $π \propto C T$ , so $π = C R T = \frac{n}{V} R T$ , with $R$ the gas constant.
Q3 · 2 marks · Short answerHarder board level
At what temperature will a 0.050 M aqueous solution of glucose have an osmotic pressure of 1.50 atm? ( $R = 0.0821$ L atm K⁻¹ mol⁻¹)
Hint
Rearrange $π = C R T$ for $T$ .
Answer
$T = \frac{π}{C R} = \frac{1.50}{0.050 \times 0.0821} = 365$ K (about 92 °C).
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Osmotic pressure is the preferred colligative property for finding the molar mass of proteins.
Reason (R): Osmotic pressure is measured at about room temperature, and its value is large even for very dilute solutions.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Think about heat-sensitive molecules with very large molar masses.
Answer
(a). Proteins are damaged by heating and give tiny boiling or freezing point changes; osmotic pressure avoids both problems.
Q5 · 3 marks · Short answerHarder board level
A 0.010 M aqueous solution of CaCl₂ has an osmotic pressure of 0.605 atm at 300 K. Calculate the van't Hoff factor and the degree of dissociation of CaCl₂. ( $R = 0.0821$ L atm K⁻¹ mol⁻¹)
Hint
CaCl₂ gives 3 ions, so $α = \frac{i - 1}{3 - 1}$ .
Answer
Calculated $π$ for $i = 1$ : $0.010 \times 0.0821 \times 300 = 0.2463$ atm. $i = \frac{0.605}{0.2463} = 2.46$ . $α = \frac{2.46 - 1}{2} = 0.728$ , i.e. about 73% dissociated.
Q6 · 2 marks · Short answerHarder board level
1.0 g of urea ( $M = 60$ ) and 1.0 g of glucose ( $M = 180$ ) are each dissolved in water to make 1 litre of solution at the same temperature. Which solution has the higher osmotic pressure, and by what factor? Explain.
Hint
Compare moles, not masses.
Answer
Urea: $\frac{1}{60}$ mol; glucose: $\frac{1}{180}$ mol. The urea solution has three times as many particles per litre, so its osmotic pressure is three times larger. Osmotic pressure depends on the number of particles, not their mass.
Q7 · 3 marks · Short answerHOTS / competency
A tube is divided by a semipermeable membrane. The left side holds 0.10 M sucrose and the right side 0.060 M NaCl (fully dissociated), both at 300 K. (a) In which direction does water flow? (b) What extra pressure, and on which side, would stop the flow? ( $R = 0.0821$ L atm K⁻¹ mol⁻¹)
Hint
Compare $i C$ : 0.10 on the left, 0.12 on the right.
Answer
$π_{l e f t} = 0.10 \times 0.0821 \times 300 = 2.46$ atm; $π_{r i g h t} = 2 \times 0.060 \times 0.0821 \times 300 = 2.96$ atm. (a) Water flows from left to right, towards the higher particle concentration. (b) About 0.49 atm must be applied on the right (NaCl) side.
Q8 · 3 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a cook places slices of raw mango in strong brine, and the slices shrivel within a day. She also notices that wilted coriander revives when its stems stand in fresh water, and that salted fish does not spoil for weeks.
(i) Name the process in the mango and say whether the brine is hypertonic or hypotonic to the mango cells. [1]
(ii) Explain why the coriander becomes firm again. [1]
(iii) Why does salting stop the growth of bacteria on fish? [1]
Hint
In each case ask which side has more solute particles.
Answer
(i) Exosmosis: water leaves the mango cells because the brine is hypertonic. (ii) Fresh water is hypotonic to the cell sap, so water enters the cells by endosmosis and they become turgid. (iii) Bacterial cells on the salted fish are surrounded by a hypertonic medium; they lose water by osmosis, shrivel and die.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define osmotic pressure.Show answer
The excess pressure that must be applied on the solution side to just stop osmosis when the solution is separated from pure solvent by a semipermeable membrane.
2.Write the van't Hoff equation for osmotic pressure of an electrolyte.Show answer
$π = i C R T$ .
3.Give one natural and one chemical semipermeable membrane.Show answer
Natural: animal bladder (or a cell membrane, parchment). Chemical: copper ferrocyanide deposited in a porous pot.
4.What concentration of NaCl is isotonic with blood cell fluid?Show answer
0.9% (mass/volume), normal saline.
5.Which membrane is used in reverse osmosis for desalination?Show answer
A film of cellulose acetate on a suitable support.
6.State van't Hoff-Avogadro's law.Show answer
Equal volumes of isotonic solutions at the same temperature contain equal numbers of solute particles.

Osmosis and Osmotic Pressure: frequently asked questions

What is the difference between osmosis and diffusion?

Diffusion needs no membrane, and both solute and solvent particles spread out. Osmosis needs a semipermeable membrane, and only solvent molecules move, from the dilute side to the concentrated side.

Why is osmotic pressure used to find the molar mass of proteins?

It is large enough to measure even for very dilute solutions of macromolecules, it is measured at room temperature so proteins are not denatured, and it uses molarity.

What happens to a red blood cell in a hypotonic solution?

Water enters the cell by endosmosis and it swells; in a strongly hypotonic solution such as pure water it may burst.

What is reverse osmosis?

When a pressure larger than the osmotic pressure is applied on the solution side, pure solvent is forced out of the solution through the semipermeable membrane. It is used to desalinate sea water.

Is osmotic pressure a colligative property?

Yes. For a dilute solution it equals iCRT, so it depends only on the number of solute particles per litre and the temperature, not on what the particles are.