Solutions Notes for NEET: Class 12 NCERT Chemistry Chapter 1

Q: How many questions come from Solutions in NEET each year?

You can expect 3 to 5 questions from Solutions in NEET. Colligative properties (depression in freezing point, elevation in boiling point, osmotic pressure) appear almost every year. van't Hoff factor questions are very frequent in recent NEET papers.

Q: What is the van't Hoff factor and why is it important for NEET?

The van't Hoff factor (i) measures how many particles one formula unit of a solute produces in solution. For NaCl it is 2 (Na+ and Cl-), for Na2SO4 it is 3, and for glucose it is 1. You use i to modify colligative property formulas: delta-Tf = i x Kf x m and pi = iMRT. NEET frequently asks you to calculate i from observed molar mass, or to find degree of dissociation from i.

Q: What is the difference between molarity and molality?

Molarity (M) is moles of solute per litre of solution. Molality (m) is moles of solute per kilogram of solvent. Molality does not change with temperature because it depends on mass rather than volume. Colligative property calculations always use molality.

Q: Why do solutions show positive or negative deviations from Raoult's law?

Raoult's law assumes A-B interactions equal A-A and B-B interactions. If A-B forces are weaker (A-B < A-A, B-B), molecules escape more easily, giving a positive deviation with higher vapour pressure than expected. Example: acetone and ethanol. If A-B forces are stronger, molecules are held more tightly, giving a negative deviation with lower vapour pressure. Example: acetone and chloroform (hydrogen bonding). Azeotropes are constant-boiling mixtures formed at the maximum or minimum of the vapour pressure curve.

Q: How do you identify isotonic, hypotonic, and hypertonic solutions?

Isotonic solutions have equal osmotic pressure. When a cell is placed in a hypertonic solution (higher concentration outside), water leaves the cell and it shrinks (plasmolysis in plant cells). In a hypotonic solution (lower concentration outside), water enters the cell and it swells or bursts (haemolysis in red blood cells). Intravenous fluids must be isotonic with blood plasma (about 0.9% NaCl).

Q: What is Henry's law and when does it apply?

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in solution: p = KH x chi. Higher KH means the gas is less soluble. It applies to dilute solutions of sparingly soluble gases. Applications include dissolved oxygen in blood (relevant to scuba diving and altitude sickness) and CO2 in carbonated drinks.

Q: If observed molar mass of acetic acid in benzene is 120 g/mol, what does this mean?

Acetic acid has a normal molar mass of 60 g/mol. The observed value of 120 is twice that, so van't Hoff factor i = 60/120 = 0.5. This means molecules are associating: two molecules join through hydrogen bonding to form a cyclic dimer. In benzene (non-polar solvent), acetic acid forms dimers, reducing particle count and making colligative effects smaller than expected.

Q: What is reverse osmosis and how is the pressure related to osmotic pressure?

In normal osmosis, solvent moves from low concentration to high concentration through a semipermeable membrane. Reverse osmosis happens when you apply external pressure greater than the osmotic pressure on the concentrated side, forcing solvent to move in the opposite direction. This is used for water desalination and drinking water purification. The minimum pressure needed equals the osmotic pressure of the solution.

Introduction: What NEET Tests from Solutions

Solutions is one of the most calculation-heavy chapters in NEET Chemistry Class 12. You can expect 3 to 5 questions from this chapter in NEET 2027. The most frequently tested topics are colligative properties (especially depression in freezing point and osmotic pressure), the van't Hoff factor, and Raoult's law. Concentration unit conversions (molarity to molality) appear almost every year in combination problems.

A solution is a homogeneous mixture of two or more components. The component present in the larger quantity is called the solvent. The component in the smaller quantity is called the solute. When both components are liquids, the one present in larger amount is called the solvent.

NCERT Class 12 Chapter 1 covers nine types of solutions based on the physical states of solute and solvent. The most common type for NEET problems is a solid solute dissolved in a liquid solvent (like NaCl in water or glucose in water). You must be very comfortable calculating concentration in all six units before moving to the colligative property formulas.

Types of Solutions and Concentration Units

Solutions can be classified based on the physical states of the solute and solvent. There are nine types: gas-in-gas (like air), gas-in-liquid (CO₂ in water), gas-in-solid (H₂ in palladium), liquid-in-gas (water vapour in air), liquid-in-liquid (ethanol in water), liquid-in-solid (mercury in silver/amalgam), solid-in-gas (camphor in nitrogen), solid-in-liquid (NaCl in water), and solid-in-solid (brass = copper + zinc).

For NEET, you need to know six ways to express concentration. Each formula has a different use. Molality is used for colligative properties because it does not depend on temperature. Mole fraction is used in Raoult's law. Parts per million (ppm) is used for very dilute solutions such as pollutants in water.

Unit	Symbol	Formula	Key Notes
Mass percent	w/w %	(mass of solute / mass of solution) x 100	Temperature-independent
Volume percent	v/v %	(volume of solute / volume of solution) x 100	Used for liquid-in-liquid solutions
Parts per million	ppm	(mass of solute / mass of solution) x 10⁶	For very dilute solutions
Mole fraction	χ	n_solute / (n_solute + n_solvent)	χ_A + χ_B = 1; dimensionless
Molarity	M	moles of solute / volume of solution in litres	Temperature-dependent (volume changes)
Molality	m	moles of solute / mass of solvent in kg	Temperature-independent; used in colligative properties

The key relationship you need for converting molarity to molality (and vice versa) is this: if you know molarity M, molar mass of solute M₂, and density d of the solution (g/mL), then molality m = (1000 x M) / (1000 x d - M x M₂). You can derive this yourself by working out the masses in 1 litre of solution.

Solute

NaCl

Glucose

Urea

KCl

NaOH

H2SO4

Mass of solute (g)

Molar mass of solute (g/mol)

Mass of solvent (g)

Density of solution (g/mL)

Solvent

Water

Benzene

CCl4

Ethanol

Molar mass of solvent (g/mol)

Mass percent

% (w/w)

(w₂ / (w₁+w₂)) × 100

Parts per million

100000

ppm

(w₂ / (w₁+w₂)) × 10⁶

Mole fraction (solute)

0.01099

dimensionless

n₂ / (n₁ + n₂)

Molality

0.6173

mol/kg

n₂ / (w₁ in kg)

Molarity

0.5833

mol/L

n₂ / V(solution in L)

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Solubility and Henry's Law

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a particular temperature and pressure. A solution that holds the maximum possible amount of dissolved solute is called a saturated solution. A solution that can still dissolve more solute is unsaturated. Asupersaturated solution holds more solute than the equilibrium amount (unstable).

For solid solutes in liquids: solubility generally increases with temperature for endothermic dissolution (like KNO₃ in water). For exothermic dissolution (like CaCl₂in water), solubility decreases with temperature.

Henry's Law for Gas Solubility

For gases dissolved in liquids, the solubility increases with pressure. William Henry quantified this relationship:

p = K_{H} \cdot χ_{gas}

Here p is the partial pressure of the gas above the solution, K_His Henry's law constant (units: atm or bar), and χ is the mole fraction of the gas dissolved in the liquid. A larger K_H means the gas is less soluble (it takes more pressure to dissolve the same amount).

Gas solubility decreases with increasing temperature. This is why fish in warm water can suffer from low dissolved oxygen. Henry's law applies only to dilute solutions of gases that do not react with the solvent.

NEET Applications of Henry's Law

Carbonated drinks: CO₂ is dissolved under high pressure. When you open the bottle, pressure drops and gas escapes (solubility falls).
Scuba diving (bends): At depth, divers breathe compressed air. Nitrogen dissolves in blood at high pressure. If they surface too fast, nitrogen comes out of solution as bubbles in the bloodstream, causing decompression sickness.
Altitude sickness: At high altitudes, atmospheric pressure is lower. Less O₂ dissolves in blood, causing headaches and fatigue.

Vapour Pressure and Raoult's Law

The vapour pressure of a liquid is the pressure exerted by its vapour in equilibrium with the liquid phase at a given temperature. A non-volatile solute dissolved in a liquid lowers its vapour pressure. Why? Solute molecules occupy some surface positions, reducing the fraction of solvent molecules that can escape into the vapour phase.

Raoult's Law

For an ideal solution of two miscible liquids A and B, the partial vapour pressure of each component is proportional to its mole fraction in solution:

P_{A} = χ_{A} \cdot P_{A}^{\circ} P_{B} = χ_{B} \cdot P_{B}^{\circ}

P_{total} = P_{A} + P_{B} = χ_{A} P_{A}^{\circ} + χ_{B} P_{B}^{\circ}

Here P_A° and P_B° are the vapour pressures of the pure components. The total vapour pressure is a linear function of mole fraction (P vs χ graph is a straight line between P_B° and P_A°).

For a dilute solution of a non-volatile solute in a solvent, only the solvent contributes to vapour pressure:

\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = χ_{solute}

This is the relative lowering of vapour pressure, which is a colligative property. The left side is measurable directly from the vapour pressures.

Benzene + Toluene (ideal)

Acetone + Ethanol (+ve)

Acetone + Chloroform (-ve)

Ethanol + Water (+ve)

Vapour pressure of pure A, P°A (mmHg)

Vapour pressure of pure B, P°B (mmHg)

Mole fraction of A: 0.40 (mole fraction of B: 0.60)

Solution type

Partial pressure of A (P°A = 95 mmHg, χA = 0.40)

38 mmHg

Partial pressure of B (P°B = 29 mmHg, χB = 0.60)

17.4 mmHg

Total vapour pressure

55.4 mmHg

pA (partial)

38 mmHg

pB (partial)

17.4 mmHg

P total

55.4 mmHg

Mole fraction A in vapour

0.69

pA / Ptotal

Ideal vs Non-Ideal Solutions

An ideal solution obeys Raoult's law at all concentrations. The intermolecular forces between A-B, A-A, and B-B are all equal. Mixing is accompanied by zero enthalpy change (delta-H_mix = 0) and zero volume change (delta-V_mix = 0). Example: benzene and toluene (similar structure, similar interactions).

Most real solutions show deviations:

Positive deviation:A-B interactions are weaker than A-A and B-B. Molecules escape more easily. Total vapour pressure is higher than predicted by Raoult's law. delta-H_mix> 0 (endothermic). Example: acetone + ethanol, acetone + CS₂, ethanol + water.
Negative deviation: A-B interactions are stronger than A-A and B-B. Molecules are held more tightly. Total vapour pressure is lower than predicted. delta-H_mix< 0 (exothermic). Example: acetone + chloroform (H-bonding between C=O and H-CCl₃), HCl + water, HNO₃ + water.

Azeotropes

At certain compositions, the vapour and liquid phases have exactly the same composition. Such mixtures are called azeotropes or constant-boiling mixtures. They cannot be separated by simple distillation because distillation does not change the composition.

Minimum boiling azeotrope:forms at the composition where positive deviation gives a maximum in the vapour pressure curve. Boils at a lower temperature than either pure component. Example: ethanol-water (95.5% ethanol, bp 78.1 °C).
Maximum boiling azeotrope: forms at the composition where negative deviation gives a minimum in the vapour pressure curve. Boils at a higher temperature than either pure component. Example: HNO₃-water (68% HNO₃, bp 120.5 °C).

Colligative Properties

Colligative properties depend only on the number of solute particles dissolved, not on the nature (identity, size, or charge) of those particles. There are four colligative properties that NEET tests:

Relative lowering of vapour pressure
Elevation of boiling point (ebullioscopy)
Depression of freezing point (cryoscopy)
Osmotic pressure

Solvent

Molality m (mol/kg)

van't Hoff factor i

1 = non-electrolyte, 2 = NaCl, 3 = BaCl2

Water constants

Bp = 100°C

Fp = 0°C

K_b = 0.52K kg mol⁻¹

K_f = 1.86K kg mol⁻¹

delta-Tb

0.052

i × Kb × m = 1 × 0.52 × 0.1

New boiling point

100.052

°C

100 + ΔTb

delta-Tf

0.186

i × Kf × m = 1 × 1.86 × 0.1

New freezing point

-0.186

°C

0 − ΔTf

Osmotic pressure (needs molarity)

Molarity C (mol/L)

Osmotic pressure (at 25°C)

2.447 atm

i × C × 0.0821 × 298

1. Relative Lowering of Vapour Pressure

\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = χ_{solute} = \frac{n _{solute}}{n _{solute} + n _{solvent}}

For dilute solutions (n_solute very small compared to n_solvent):

\frac{p ^{\circ} - p _{s}}{p ^{\circ}} \approx \frac{n _{solute}}{n _{solvent}} = \frac{w _{2} / M _{2}}{w _{1} / M _{1}}

This formula lets you find the molar mass M₂ of the solute from vapour pressure measurements.

2. Elevation of Boiling Point

A solution boils at a higher temperature than pure solvent because the dissolved solute lowers the vapour pressure. The solvent must be heated to a higher temperature to reach atmospheric pressure.

Δ T_{b} = T_{b} (solution) - T_{b}^{\circ} (solvent) = K_{b} \cdot m

Here K_b is the ebullioscopic constant (molal elevation constant) and m is the molality of the solution. K_b is a property of the solvent only, not the solute.

Solvent	Boiling point (°C)	Kb (K kg mol⁻¹)
Water	100.0	0.52
Benzene	80.1	2.53
Ethanol	78.4	1.20
Chloroform	61.2	3.63
CCl4	76.7	5.02

3. Depression of Freezing Point

A solution freezes at a lower temperature than pure solvent. The dissolved solute interferes with the formation of the solid lattice.

Δ T_{f} = T_{f}^{\circ} (solvent) - T_{f} (solution) = K_{f} \cdot m

K_f is the cryoscopic constant (molal depression constant). This formula is extremely commonly tested in NEET for finding molar masses of unknowns.

Solvent	Freezing point (°C)	Kf (K kg mol⁻¹)
Water	0.0	1.86
Benzene	5.5	5.12
Cyclohexane	6.5	20.0
Camphor	179.8	40.0
Acetic acid	16.6	3.90

Camphor has a very large K_f (40 K kg mol^-1), so even small amounts of solute give a large, easily measurable freezing point depression. This makes it useful for determining molar masses of small organic molecules (Rast method).

4. Osmotic Pressure

Osmosis is the spontaneous flow of solvent molecules through a semipermeable membrane from a region of lower solute concentration (lower osmotic pressure) to a region of higher solute concentration (higher osmotic pressure).

The osmotic pressure(π) is the minimum external pressure that must be applied to the solution side to stop osmosis:

π = M R T

where M is the molarity of the solution, R = 0.0821 L atm mol^-1 K^-1, and T is the temperature in Kelvin. For electrolytes, use π = iMRT.

Osmotic pressure is useful for measuring molar masses of large molecules (proteins, polymers) because even a small concentration gives a large osmotic pressure that is easy to measure.

Tonicity and Biological Importance

Isotonic solutions: Same osmotic pressure as body fluids. Normal saline (0.9% NaCl) is isotonic with blood plasma. Intravenous fluids must be isotonic.
Hypertonic solutions: Higher osmotic pressure than cell contents. Water moves out of cells (plasmolysis in plants, crenation in red blood cells). Preserving food in salt or sugar creates hypertonic conditions around bacteria, killing them.
Hypotonic solutions: Lower osmotic pressure than cell contents. Water enters cells; they swell and may burst (turgidity in plants, haemolysis in red blood cells).

Reverse Osmosis

If you apply external pressure greater than the osmotic pressure on the concentrated side, solvent is forced to flow in the reverse direction (from concentrated to dilute). This isreverse osmosis. It is used in desalination of sea water (sea water osmotic pressure is about 27 atm, so the applied pressure must exceed this) and in water purification systems.

van't Hoff Factor and Abnormal Molar Masses

When solutes dissociate (electrolytes) or associate (acetic acid in benzene) in solution, the actual number of particles differs from what you calculate based on the formula unit. The colligative property observed is therefore different from the expected value. This is called an abnormal molar mass.

van't Hoff introduced a correction factor i defined as:

i = \frac{observed colligative effect}{expected colligative effect (if no dissociation/association)}

i = \frac{theoretical molar mass}{observed molar mass}

Modifying the Colligative Property Formulas

Δ T_{b} = i \cdot K_{b} \cdot m

Δ T_{f} = i \cdot K_{f} \cdot m

π = i \cdot M \cdot R \cdot T

\frac{p ^{\circ} - p _{s}}{p ^{\circ}} = i \cdot χ_{solute}

van't Hoff Factor Values

Type of solute	i value	Example
Non-electrolyte (no dissociation)	i = 1	Glucose, urea, sucrose
1:1 electrolyte (2 ions)	i = 2	NaCl, KCl, HCl
1:2 or 2:1 electrolyte (3 ions)	i = 3	BaCl2, CaCl2, Na2SO4
1:3 or 3:1 electrolyte (4 ions)	i = 4	AlCl3, Na3PO4
Associating solute (2 molecules join)	i = 0.5	Acetic acid in benzene (dimers)
K4[Fe(CN)6] (5 ions)	i = 5	4K+ + [Fe(CN)6]4-

Degree of Dissociation from i

For an electrolyte that gives n ions per formula unit (e.g., NaCl gives n = 2), if α is the degree of dissociation:

i = 1 + (n - 1) α

α = \frac{i - 1}{n - 1}

For acetic acid forming dimers in benzene (2 molecules associate to 1 dimer):

i = 1 - \frac{α}{2} \Rightarrow α = 2 (1 - i)

Why i < 1 (Association)

Acetic acid (CH₃COOH) dissolves in benzene and forms cyclic dimers through double hydrogen bonding between two carboxylic acid groups. This reduces the number of solute particles. The observed molar mass is twice the theoretical value (120 g/mol instead of 60 g/mol). So i = 60/120 = 0.5.

In polar solvents like water, acetic acid does not dimerize because water competes for hydrogen bonds. So i is close to 1 (slightly above 1 due to partial ionisation).

Worked NEET Problems

NEET-style problem · Concentration units

Question

The density of a 3 M NaCl solution is 1.25 g/mL. Calculate the molality of the solution. (Molar mass of NaCl = 58.5 g/mol)

Solution

In 1 L of solution: moles of NaCl = 3 mol; mass of NaCl = 3 × 58.5 = 175.5 g. Mass of solution = 1000 mL × 1.25 g/mL = 1250 g. Mass of water = 1250 − 175.5 = 1074.5 g = 1.0745 kg. Molality = 3 / 1.0745 = 2.79 m.

NEET-style problem · Osmotic pressure

Question

An aqueous solution at 25 °C has osmotic pressure 4.92 atm. What is its molarity? (R = 0.0821 L atm mol⁻¹ K⁻¹)

Solution

π = MRT. M = π / (RT) = 4.92 / (0.0821 × 298) = 4.92 / 24.47 = 0.201 M ≈ 0.2 M.

NEET-style problem · Freezing point depression

Question

1.5 g of urea (M = 60 g/mol) is dissolved in 200 g of water. Calculate the depression in freezing point. (K_f for water = 1.86 K kg mol⁻¹)

Solution

Moles of urea = 1.5/60 = 0.025 mol. Molality = 0.025/0.2 = 0.125 mol/kg. ΔT_f = K_f × m = 1.86 × 0.125 = 0.2325 K ≈ 0.233 K.

NEET-style problem · van't Hoff factor

Question

K₄[Fe(CN)₆] dissociates completely in aqueous solution. What is its van't Hoff factor? If a 0.1 mol/kg solution shows a freezing point depression of X K, find X. (K_f for water = 1.86 K kg mol⁻¹)

Solution

K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻. This produces 5 ions, so i = 5. ΔT_f = i × K_f × m = 5 × 1.86 × 0.1 = 0.93 K.

NEET-style problem · Abnormal molar mass

Question

When 2.56 g of sulphur is dissolved in 100 g of CS₂, the freezing point depression is 0.383 K. Find the molar mass of sulphur in CS₂ and identify its molecular formula. (K_f for CS₂ = 3.83 K kg mol⁻¹)

Solution

ΔT_f = K_f × (w₂/M₂) × (1000/w₁). 0.383 = 3.83 × (2.56/M) × (1000/100). M = 3.83 × 25.6 / 0.383 = 256 g/mol. Number of S atoms = 256/32 = 8. Molecular formula: S₈.

Summary Cheat Sheet

Before your NEET exam, make sure you can recall these formulas instantly. Every formula below has appeared in at least one NEET paper in the last 8 years.

Property	Formula	With van't Hoff factor
Relative VP lowering	(p° - p_s) / p° = χ_solute	(p° - p_s) / p° = i · χ_solute
Boiling point elevation	ΔT_b = K_b · m	ΔT_b = i · K_b · m
Freezing point depression	ΔT_f = K_f · m	ΔT_f = i · K_f · m
Osmotic pressure	π = MRT	π = i · MRT
Degree of dissociation	α = (i - 1) / (n - 1)	where n = number of ions produced
Molar mass from i	M_obs = M_theoretical / i	i > 1: dissociation; i < 1: association

Key Values to Memorise

K_f(water) = 1.86 K kg mol^-1
K_b(water) = 0.52 K kg mol^-1
K_f(benzene) = 5.12 K kg mol^-1
K_b(benzene) = 2.53 K kg mol^-1
K_f(camphor) = 40.0 K kg mol^-1 (highest; used in Rast method)
R = 0.0821 L atm K^-1 mol^-1 for osmotic pressure problems
Normal saline = 0.9% NaCl (isotonic with blood plasma)
Acetic acid in benzene: dimers form, i = 0.5, observed M = 120 g/mol
Sulphur in CS₂: S₈ molecule, M = 256 g/mol

Quick Recall: Which Formula Uses Which Concentration Unit?

Raoult's law: mole fraction (χ)
Henry's law: mole fraction (χ)
Boiling point elevation and freezing point depression: molality (m)
Osmotic pressure: molarity (M)

Frequently asked questions

How many questions come from Solutions in NEET each year?

You can expect 3 to 5 questions from Solutions in NEET. Colligative properties (depression in freezing point, elevation in boiling point, osmotic pressure) appear almost every year. van't Hoff factor questions are very frequent in recent NEET papers.

What is the van't Hoff factor and why is it important for NEET?

The van't Hoff factor (i) measures how many particles one formula unit of a solute produces in solution. For NaCl it is 2 (Na+ and Cl-), for Na2SO4 it is 3, and for glucose it is 1. You use i to modify colligative property formulas: delta-Tf = i x Kf x m and pi = iMRT. NEET frequently asks you to calculate i from observed molar mass, or to find degree of dissociation from i.

What is the difference between molarity and molality?

Molarity (M) is moles of solute per litre of solution. Molality (m) is moles of solute per kilogram of solvent. Molality does not change with temperature because it depends on mass rather than volume. Colligative property calculations always use molality.

Why do solutions show positive or negative deviations from Raoult's law?

Raoult's law assumes A-B interactions equal A-A and B-B interactions. If A-B forces are weaker (A-B < A-A, B-B), molecules escape more easily, giving a positive deviation with higher vapour pressure than expected. Example: acetone and ethanol. If A-B forces are stronger, molecules are held more tightly, giving a negative deviation with lower vapour pressure. Example: acetone and chloroform (hydrogen bonding). Azeotropes are constant-boiling mixtures formed at the maximum or minimum of the vapour pressure curve.

How do you identify isotonic, hypotonic, and hypertonic solutions?

Isotonic solutions have equal osmotic pressure. When a cell is placed in a hypertonic solution (higher concentration outside), water leaves the cell and it shrinks (plasmolysis in plant cells). In a hypotonic solution (lower concentration outside), water enters the cell and it swells or bursts (haemolysis in red blood cells). Intravenous fluids must be isotonic with blood plasma (about 0.9% NaCl).

What is Henry's law and when does it apply?

Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in solution: p = KH x chi. Higher KH means the gas is less soluble. It applies to dilute solutions of sparingly soluble gases. Applications include dissolved oxygen in blood (relevant to scuba diving and altitude sickness) and CO2 in carbonated drinks.

If observed molar mass of acetic acid in benzene is 120 g/mol, what does this mean?

Acetic acid has a normal molar mass of 60 g/mol. The observed value of 120 is twice that, so van't Hoff factor i = 60/120 = 0.5. This means molecules are associating: two molecules join through hydrogen bonding to form a cyclic dimer. In benzene (non-polar solvent), acetic acid forms dimers, reducing particle count and making colligative effects smaller than expected.

What is reverse osmosis and how is the pressure related to osmotic pressure?

In normal osmosis, solvent moves from low concentration to high concentration through a semipermeable membrane. Reverse osmosis happens when you apply external pressure greater than the osmotic pressure on the concentrated side, forcing solvent to move in the opposite direction. This is used for water desalination and drinking water purification. The minimum pressure needed equals the osmotic pressure of the solution.

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States of Matter

Electrochemistry

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SolutionsNEET Chemistry · Class 12 · NCERT Chapter 1

Introduction: What NEET Tests from Solutions

Types of Solutions and Concentration Units

Solubility and Henry's Law

Henry's Law for Gas Solubility

NEET Applications of Henry's Law

Vapour Pressure and Raoult's Law

Raoult's Law

Ideal vs Non-Ideal Solutions

Azeotropes

Colligative Properties

1. Relative Lowering of Vapour Pressure

2. Elevation of Boiling Point

3. Depression of Freezing Point

4. Osmotic Pressure

Tonicity and Biological Importance

Reverse Osmosis

van't Hoff Factor and Abnormal Molar Masses

Modifying the Colligative Property Formulas

van't Hoff Factor Values

Degree of Dissociation from i

Why i < 1 (Association)

Worked NEET Problems

Summary Cheat Sheet

Key Values to Memorise

Quick Recall: Which Formula Uses Which Concentration Unit?

Frequently asked questions

How many questions come from Solutions in NEET each year?

What is the van't Hoff factor and why is it important for NEET?

What is the difference between molarity and molality?

Why do solutions show positive or negative deviations from Raoult's law?

How do you identify isotonic, hypotonic, and hypertonic solutions?

What is Henry's law and when does it apply?

If observed molar mass of acetic acid in benzene is 120 g/mol, what does this mean?

What is reverse osmosis and how is the pressure related to osmotic pressure?

Continue with the next chapter notes

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