Solutions Class 12 Chemistry: Concepts, Formulas and Board Numericals

What Solutions is about

Weightage: This chapter is Unit I, Solutions, which carries 7 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit I Solutions

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper. See the questions

Key concepts

Types of solutions and expressing concentration

NCERT section 1.1 Types of Solutions; 1.2 Expressing Concentration of Solutions

A solution has a solvent (the component present in the largest amount, which fixes the physical state) and one or more solutes. Solutions can be gaseous, liquid or solid; for example, oxygen in nitrogen is a gaseous solution, glucose in water is a liquid solution and copper dissolved in gold is a solid solution.

Concentration can be expressed in several ways:

• Mass percentage (w/w): mass of component ÷ total mass of solution × 100.
• Volume percentage (V/V): used for liquids, for example 35% ethylene glycol in water as antifreeze.
• Mass by volume percentage (w/V): grams of solute in 100 mL of solution, used in medicine and pharmacy.
• Parts per million (ppm): used for traces, such as pollutants or dissolved oxygen in water.
• **Mole fraction ($x$)**: moles of a component ÷ total moles. In a two-component solution $x_1+x_2=1$.
• **Molarity ($M$)**: moles of solute per litre of solution.
• **Molality ($m$)**: moles of solute per kilogram of solvent.

Molarity changes with temperature because volume expands or contracts. Mass percentage, ppm, mole fraction and molality depend only on masses, so they do not change with temperature. That is why molality appears in the boiling and freezing point equations.

Worked example

18 g of glucose ($C_6{H}_{12}{O}_6$, molar mass 180 g mol⁻¹) is dissolved in 250 g of water. Calculate (a) the molality of the solution and (b) the mole fraction of glucose. (Molar mass of water = 18 g mol⁻¹) (2 marks)

Try it first, then show the solutionSolution

1. Moles of glucose = $\frac{18}{180}=0.1$ mol.
2. (a) Molality = $\frac{0.1\text{ mol}}{0.250\text{ kg}}=0.4$ mol kg⁻¹. (1)
3. Moles of water = $\frac{250}{18}=13.89$ mol.
4. (b) $x_{glucose}=\frac{0.1}{0.1+13.89}=\frac{0.1}{13.99}=0.0071$. (1)

Answer: (a) 0.4 mol kg⁻¹ (b) mole fraction of glucose ≈ 0.0071

Solubility of gases and Henry's law

NCERT section 1.3 Solubility (1.3.2 Solubility of a Gas in a Liquid)

Solubility is the maximum amount of a substance that dissolves in a fixed amount of solvent at a given temperature. A solid dissolves until the rate of dissolution equals the rate of crystallisation, giving a saturated solution. Pressure has almost no effect on solids and liquids dissolving in liquids because they are nearly incompressible.

For a gas in a liquid, pressure matters a lot. Henry's law: the partial pressure of a gas in the vapour phase is proportional to its mole fraction in the solution, $p=K_{H}x$. The constant $K_H$ depends on the gas and the temperature.

Two results follow:

• **Higher $K_H$ means lower solubility** at the same pressure. For example, at 293 K, $K_H$ for O₂ (34.86 kbar) is smaller than for N₂ (76.48 kbar), so oxygen is more soluble than nitrogen.
• $K_H$ increases with temperature, so gases are less soluble in warm water. Dissolving a gas is exothermic, so by Le Chatelier's principle solubility falls as temperature rises. Aquatic life is more comfortable in cold water for this reason.

Applications in NCERT: soft drink bottles are sealed under high CO₂ pressure; scuba divers breathing compressed air get bends when dissolved nitrogen forms bubbles in the blood as they rise, so their tanks use air diluted with helium; and at high altitude the low partial pressure of oxygen gives anoxia, a feeling of weakness and inability to think clearly.

Worked example

At 293 K the Henry's law constant for oxygen in water is 34.86 kbar. If the partial pressure of oxygen in air is 0.21 bar, calculate the mole fraction of oxygen in water saturated with air, and the mass of oxygen dissolved in 1 L of water. (Take 1 L of water = 55.5 mol) (3 marks)

Try it first, then show the solutionSolution

1. Henry's law: $x_{O_2}=\frac{p_{O_2}}{K_H}=\frac{0.21\text{ bar}}{34860\text{ bar}}$. (1)
2. $x_{O_2}=6.0\times 10^{-6}$. (1)
3. For a very dilute solution, $n_{O_2}\approx x_{O_2}\times n_{water}=6.0\times 10^{-6}\times 55.5=3.3\times 10^{-4}$ mol.
4. Mass = $3.3\times 10^{-4}\times 32=1.07\times 10^{-2}$ g, about 10.7 mg. (1)

Answer: Mole fraction ≈ 6.0 × 10⁻⁶; about 10.7 mg of oxygen dissolves in 1 L of water.

Vapour pressure of solutions of two volatile liquids (Raoult's law)

NCERT section 1.4.1 Vapour Pressure of Liquid-Liquid Solutions; 1.4.2 Raoult's Law as a special case of Henry's Law

A liquid in a closed container sets up an equilibrium with its vapour; the pressure of that vapour is the vapour pressure. In a mixture of two volatile liquids, each one contributes to the total.

Raoult's law: for a solution of volatile liquids, the partial vapour pressure of each component is proportional to its mole fraction:
$p_1=p_1^0{x}_1$ and $p_2=p_2^0{x}_2$, where $p^0$ is the vapour pressure of the pure liquid.

By Dalton's law of partial pressures, $p_{total}=p_1+p_2=p_1^0+(p_2^0-p_1^0)x_2$. So a graph of total pressure against $x_2$ is a straight line joining the two pure vapour pressures.

The vapour is richer in the more volatile component. Its composition is found from $y_1=\frac{p_1}{p_{total}}$, where $y$ is the mole fraction in the vapour.

Raoult's law is also a special case of Henry's law: both say partial pressure is proportional to mole fraction; only the constant differs ($p_1^0$ instead of $K_H$).

Worked example

Pure liquids P and Q have vapour pressures of 120 mm Hg and 40 mm Hg at 300 K. A solution is made with mole fraction of P equal to 0.25. Assuming ideal behaviour, find the total vapour pressure and the mole fraction of P in the vapour. (3 marks)

Try it first, then show the solutionSolution

1. $p_P=120\times 0.25=30$ mm Hg. (½)
2. $x_Q=1-0.25=0.75$, so $p_Q=40\times 0.75=30$ mm Hg. (½)
3. $p_{total}=30+30=60$ mm Hg. (1)
4. $y_P=\frac{30}{60}=0.5$. (1)
5. Check: P is more volatile, and its share rises from 0.25 in the liquid to 0.5 in the vapour, as expected.

Answer: Total vapour pressure = 60 mm Hg; mole fraction of P in the vapour = 0.5.

Vapour pressure of a solution of a non-volatile solute

NCERT section 1.4.3 Vapour Pressure of Solutions of Solids in Liquids; 1.6.1 Relative Lowering of Vapour Pressure

When a non-volatile solute (such as glucose, urea or common salt) dissolves, some solvent molecules at the surface are replaced by solute particles. Fewer solvent molecules can escape, so the vapour pressure of the solution is lower than that of the pure solvent.

Only the solvent contributes to the vapour, so Raoult's law gives $p_1=x_1{p}_1^0$. Rearranging:
$\frac{p_1^0-p_1}{p_1^0}=x_2$
The relative lowering of vapour pressure equals the mole fraction of the solute. It depends only on how many solute particles are present, not on what they are: this is the first colligative property.

For dilute solutions $n_2\ll n_1$, so $\frac{\Delta p}{p_1^0}\approx \frac{n_2}{n_1}=\frac{w_2{M}_1}{M_2{w}_1}$. This lets you find the molar mass $M_2$ of an unknown solute.

Worked example

The vapour pressure of pure water at 298 K is 23.8 mm Hg. Calculate the vapour pressure of a solution made by dissolving 6 g of urea (molar mass 60 g mol⁻¹) in 90 g of water. (2 marks)

Try it first, then show the solutionSolution

1. Moles of urea = $\frac{6}{60}=0.1$; moles of water = $\frac{90}{18}=5.0$.
2. $x_{urea}=\frac{0.1}{5.0+0.1}=0.0196$. (1)
3. $\Delta p=23.8\times 0.0196=0.47$ mm Hg, so $p=23.8-0.47=23.33$ mm Hg. (1)

Answer: About 23.33 mm Hg.

Ideal and non-ideal solutions; azeotropes

NCERT section 1.5 Ideal and Non-ideal Solutions

An ideal solution obeys Raoult's law at every composition. The A–B attractions are the same as A–A and B–B attractions, so ${\Delta }_{mix}H=0$ and ${\Delta }_{mix}V=0$. Examples close to ideal: n-hexane + n-heptane, bromoethane + chloroethane, benzene + toluene.

Positive deviation: A–B attractions are weaker than A–A and B–B. Molecules escape more easily, so vapour pressure is higher than Raoult's law predicts. ${\Delta }_{mix}H>0$ (heat is absorbed) and ${\Delta }_{mix}V>0$ (volume increases). Examples: ethanol + water (ethanol breaks some hydrogen bonds of water), ethanol + acetone, carbon disulphide + acetone.

Negative deviation: A–B attractions are stronger. Vapour pressure is lower than predicted, ${\Delta }_{mix}H<0$ and ${\Delta }_{mix}V<0$. Examples: chloroform + acetone (a new hydrogen bond forms between them), phenol + aniline, nitric acid + water.

Azeotropes are mixtures that boil at a constant temperature with the liquid and vapour having the same composition, so fractional distillation cannot separate them.

• Large positive deviation gives a minimum boiling azeotrope: ethanol–water at about 95% ethanol by volume.
• Large negative deviation gives a maximum boiling azeotrope: nitric acid–water at about 68% HNO₃ by mass, boiling at 393.5 K.

Worked example

When 50 mL of acetone is mixed with 50 mL of chloroform, the mixture becomes warm. (a) Is the volume of the solution more than, less than or equal to 100 mL? (b) Name the type of deviation and give the reason. (c) What type of azeotrope can such a mixture form? (3 marks)

Try it first, then show the solutionSolution

1. Warming means heat is released, so ${\Delta }_{mix}H<0$. This happens when new, stronger attractions form between the two liquids.
2. (a) Stronger attractions pull molecules closer, so ${\Delta }_{mix}V<0$: the volume is less than 100 mL. (1)
3. (b) Negative deviation from Raoult's law. The hydrogen atom of chloroform forms a hydrogen bond with the oxygen of acetone, so A–B attractions are stronger than A–A and B–B, and fewer molecules escape. (1)
4. (c) A maximum boiling azeotrope. (1)

Answer: Less than 100 mL; negative deviation due to hydrogen bonding between chloroform and acetone; maximum boiling azeotrope.

Elevation of boiling point and depression of freezing point

NCERT section 1.6.2 Elevation of Boiling Point; 1.6.3 Depression of Freezing Point

A liquid boils when its vapour pressure equals the atmospheric pressure. A non-volatile solute lowers the vapour pressure, so the solution must be heated to a higher temperature before it boils. The rise is the elevation of boiling point:
$\Delta T_b=K_{b}m$
$K_b$ is the molal elevation constant (ebullioscopic constant), unit K kg mol⁻¹. For water $K_b=0.52$ K kg mol⁻¹.

A liquid freezes when the vapour pressure of the liquid equals that of the solid. Because the solution has a lower vapour pressure, this happens at a lower temperature. The fall is the depression of freezing point:
$\Delta T_f=K_{f}m$
$K_f$ is the molal depression constant (cryoscopic constant); for water $K_f=1.86$ K kg mol⁻¹.

Writing $m=\frac{w_2\times 1000}{M_2\times w_1}$ (masses in grams) gives the working formula for molar mass:
$M_2=\frac{K_b\times w_2\times 1000}{\Delta T_b\times w_1}$ (and the same with $K_f$ and $\Delta T_f$).

Useful shortcut: for the same solution in the same solvent, $\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}$, because the molality cancels.

Worked example

An aqueous solution of a non-electrolyte boils at 100.13 °C at 1 atm. At what temperature will it freeze? ($K_b$ = 0.52 K kg mol⁻¹, $K_f$ = 1.86 K kg mol⁻¹ for water) (3 marks)

Try it first, then show the solutionSolution

1. $\Delta T_b=100.13-100=0.13$ K. (½)
2. $m=\frac{\Delta T_b}{K_b}=\frac{0.13}{0.52}=0.25$ mol kg⁻¹. (1)
3. $\Delta T_f=K_{f}m=1.86\times 0.25=0.465$ K. (1)
4. Freezing point = $0-0.465=-0.465$ °C (272.685 K). (½)
5. Second route: $\Delta T_f=0.13\times \frac{1.86}{0.52}=0.465$ K, the same answer.

Answer: The solution freezes at −0.465 °C (about 272.69 K).

Osmosis and osmotic pressure

NCERT section 1.6.4 Osmosis and Osmotic Pressure; 1.6.5 Reverse Osmosis and Water Purification

Osmosis is the flow of solvent through a semipermeable membrane from pure solvent (or a dilute solution) into a more concentrated solution. The membrane lets solvent molecules through but not solute particles. Examples: animal bladders, parchment, and synthetic films such as cellophane.

Osmotic pressure ($\pi$) is the extra pressure that must be applied to the solution side to just stop osmosis. For dilute solutions:
$\pi =CRT=\frac{n_2}{V}RT$, so $M_2=\frac{w_{2}RT}{\pi V}$
with R = 0.083 L bar mol⁻¹ K⁻¹.

Osmotic pressure is the method of choice for proteins and polymers: it is measured at room temperature (no heating, so biomolecules are not damaged) and even very dilute solutions give a measurable value, whereas $\Delta T_b$ and $\Delta T_f$ would be tiny.

Isotonic solutions have the same osmotic pressure. Red blood cells are isotonic with 0.9% (w/V) NaCl (normal saline). In a hypertonic solution (more than 0.9%) cells lose water and shrink; in a hypotonic one (less than 0.9%) they gain water and swell. Raw mango shrivels in brine, and wilted flowers revive in fresh water, for the same reason.

Reverse osmosis: push on the solution with more than its osmotic pressure and the flow turns round: pure solvent is pushed out through the membrane. It is used to desalinate sea water, using films of cellulose acetate that let water pass but not ions.

Worked example

2.0 g of a protein is dissolved in water to make 250 mL of solution. At 300 K its osmotic pressure is 3.32 × 10⁻³ bar. (a) Calculate the molar mass of the protein. (b) Taking the solution as 250 g of water, estimate the depression of freezing point it would give, and comment. (R = 0.083 L bar mol⁻¹ K⁻¹, $K_f$ = 1.86 K kg mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. (a) $M_2=\frac{w_{2}RT}{\pi V}$ (1)
2. $=\frac{2.0\times 0.083\times 300}{3.32\times 10^{-3}\times 0.250}=\frac{49.8}{8.30\times 10^{-4}}$
3. $M_2=6.0\times 10^4$ g mol⁻¹. (1)
4. (b) Moles of protein = $\frac{2.0}{60000}=3.33\times 10^{-5}$ mol; molality = $\frac{3.33\times 10^{-5}}{0.250}=1.33\times 10^{-4}$ mol kg⁻¹.
5. $\Delta T_f=1.86\times 1.33\times 10^{-4}=2.5\times 10^{-4}$ K, far too small to measure with an ordinary thermometer, while the osmotic pressure (about 3.3 mbar) can be measured. (1)

Answer: (a) 6.0 × 10⁴ g mol⁻¹ (b) ΔT_f ≈ 2.5 × 10⁻⁴ K, too small to measure, which is why osmotic pressure is used for proteins.

Abnormal molar masses and the van't Hoff factor

NCERT section 1.7 Abnormal Molar Masses

Colligative properties count particles. If the solute splits into ions (dissociation) there are more particles than expected, so the measured property is larger and the calculated molar mass comes out smaller than the true value. If molecules join (association), there are fewer particles, the property is smaller and the molar mass comes out larger. Either way the molar mass is abnormal.

Examples: KCl in water gives K⁺ and Cl⁻, so $i$ is close to 2. Ethanoic acid in benzene forms dimers through hydrogen bonding, so $i$ is close to 0.5.

The van't Hoff factor $i$ corrects for this:
$i=\frac{\text{normal molar mass}}{\text{abnormal molar mass}}=\frac{\text{observed colligative property}}{\text{calculated colligative property}}$
It equals the total moles of particles after dissociation or association ÷ moles before.

The modified equations are $\Delta T_b=i{K}_{b}m$, $\Delta T_f=i{K}_{f}m$, $\pi =iCRT$ and $\frac{\Delta p}{p_1^0}=i\frac{n_2}{n_1}$.

To find the degree of change $\alpha$:

• dissociation into $n$ ions: $i=1+(n-1)\alpha$
• association of $n$ molecules into one: $i=1-\alpha \left(1-\frac{1}{n}\right)$; for dimers $i=1-\frac{\alpha }{2}$.

Worked example

0.6 g of ethanoic acid (molar mass 60 g mol⁻¹) is dissolved in 50 g of benzene. The freezing point falls by 0.62 K. Calculate the van't Hoff factor and the percentage of the acid that is associated into dimers. ($K_f$ for benzene = 5.12 K kg mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. Expected molality without association = $\frac{0.6/60}{0.050}=0.2$ mol kg⁻¹.
2. Calculated $\Delta T_f=5.12\times 0.2=1.024$ K. (1)
3. $i=\frac{0.62}{1.024}=0.605$. (1)
4. For dimerisation, $i=1-\frac{\alpha }{2}$, so $\alpha =2(1-0.605)=0.79$, i.e. about 79% associated. (1)

Answer: i ≈ 0.61; about 79% of the acid is associated into dimers.

Chemistry in real life: Solutions

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.

Topics in this chapter

• Raoult's Law, Ideal and Non-ideal Solutions
• Elevation of Boiling Point and Depression of Freezing Point
• Abnormal Molar Mass and van't Hoff Factor

Solutions formula sheet

Mass percentage

$\text{w/w}\%=\frac{\text{mass of component}}{\text{total mass of solution}}\times 100$

Volume percentage and mass by volume percentage are defined the same way with volumes.

Parts per million

$\text{ppm}=\frac{\text{parts of component}}{\text{total parts of solution}}\times 10^6$

Used for traces, such as pollutants or dissolved gases.

Mole fraction

$x_1=\frac{n_1}{n_1+n_2},x_1+x_2=1$

No units; does not change with temperature.

Molarity and molality

$M=\frac{n_2}{V_{solution}(\text{L})},m=\frac{n_2}{w_1(\text{kg})}=\frac{w_2\times 1000}{M_2\times w_1(\text{g})}$

Molarity changes with temperature; molality does not.

Henry's law

$p=K_{H}x$

Higher $K_H$ means lower solubility; $K_H$ rises with temperature.

Raoult's law (volatile components)

$p_1=p_1^0{x}_1,p_2=p_2^0{x}_2,p_{total}=p_1^0+(p_2^0-p_1^0)x_2$

Total pressure against mole fraction is a straight line for an ideal solution.

Composition of the vapour

$y_1=\frac{p_1}{p_{total}}$

The vapour is richer in the more volatile component.

Relative lowering of vapour pressure

$\frac{p_1^0-p_1}{p_1^0}=x_2\approx \frac{n_2}{n_1}=\frac{w_2{M}_1}{M_2{w}_1}$

The approximation holds for dilute solutions.

Elevation of boiling point

$\Delta T_b=K_{b}m,M_2=\frac{K_b\times w_2\times 1000}{\Delta T_b\times w_1}$

$K_b$ of water = 0.52 K kg mol⁻¹.

Depression of freezing point

$\Delta T_f=K_{f}m,M_2=\frac{K_f\times w_2\times 1000}{\Delta T_f\times w_1}$

$K_f$ of water = 1.86 K kg mol⁻¹; of benzene = 5.12 K kg mol⁻¹.

Osmotic pressure

$\pi =CRT=\frac{n_{2}RT}{V},M_2=\frac{w_{2}RT}{\pi V}$

R = 0.083 L bar mol⁻¹ K⁻¹; V in litres.

van't Hoff factor

$i=\frac{\text{normal molar mass}}{\text{abnormal molar mass}}=\frac{\text{observed colligative property}}{\text{calculated colligative property}}$

Greater than 1 for dissociation, less than 1 for association.

Modified colligative equations

$\Delta T_b=i{K}_{b}m,\Delta T_f=i{K}_{f}m,\pi =iCRT,\frac{\Delta p}{p_1^0}=i\frac{n_2}{n_1}$

Use these whenever the solute is an electrolyte or associates.

Degree of dissociation or association

$i=1+(n-1)\alpha (\text{dissociation}),i=1-\alpha \left(1-\frac{1}{n}\right)(\text{association})$

For dimers ($n$ = 2), $i=1-\alpha /2$.

Solutions: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Solutions

The question patterns to expect, and the method that works for each.

Tricks to remember Solutions

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

• Molality is Mass-based

  Molality has an L in it but uses kilograms of solvent, not litres. Remember: molaLity = per kiLogram of solvent; molaRity = per litRe of solution.

  Use it when: Choosing the right concentration term and explaining why molality does not change with temperature.

• High K_H, Hardly dissolves

  In $p=K_{H}x$, $x=p/K_H$. A big $K_H$ in the denominator means a small mole fraction: High K_H, Hardly soluble.

  Use it when: Comparing gas solubilities and the effect of temperature.

• Positive = Pushes out

  Positive deviation: weak attractions push molecules into the vapour, so vapour pressure rises, ΔH and ΔV are positive, and the azeotrope boils at a minimum. Negative deviation is the mirror image.

  Use it when: 2-mark questions on deviations and azeotropes.

• Ion count for i

  Write the ions out: NaCl → 2, CaCl₂ → 3, Na₂SO₄ → 3, Al₂(SO₄)₃ → 5, K₄[Fe(CN)₆] → 5. Water of crystallisation does not count, so Na₂SO₄·10H₂O is still 3.

  Use it when: MCQs that ask for the van't Hoff factor or the order of boiling or freezing points.

• Proteins Prefer Pressure

  For big molecules like Proteins and Polymers, use osmotic Pressure: room temperature and a measurable value.

  Use it when: MCQs and reasons on the best method for molar masses of biomolecules.

Solutions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Solutions

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

• ✗Mistake: Ignoring the van't Hoff factor for electrolytes

  Students apply $\Delta T_f=K_{f}m$ to NaCl or K₂SO₄ as if they were glucose.

  Wrong working

  0.1 m NaCl: ΔT_f = 1.86 × 0.1 = 0.186 K

  Right working

  0.1 m NaCl: i = 2 (Na⁺ + Cl⁻), ΔT_f = 2 × 1.86 × 0.1 = 0.372 K

  ✓Do this: Before using any colligative formula, ask: does this solute dissociate or associate? If yes, put $i$ in.

  Usually costs: 1 to 2 marks in a 3-mark numerical.

• ✗Mistake: Mass of solvent left in grams in the molality formula

  The 1000 is forgotten when solvent mass is given in grams.

  Wrong working

  m = 0.1 mol ÷ 250 = 0.0004 mol kg⁻¹

  Right working

  m = 0.1 mol ÷ 0.250 kg = 0.4 mol kg⁻¹

  ✓Do this: Write the units at every step; molality is per kilogram of solvent, and the solvent is not the whole solution.

  Usually costs: The final answer mark and often the next step.

• ✗Mistake: Mixing up the signs for positive and negative deviation

  Students remember 'positive' but attach the wrong heat and volume changes.

  Wrong working

  Chloroform + acetone shows positive deviation because it gets warm on mixing.

  Right working

  Chloroform + acetone shows negative deviation: hydrogen bonding between them releases heat (ΔH < 0) and the volume decreases (ΔV < 0).

  ✓Do this: Link heat released to stronger attractions, and stronger attractions to lower vapour pressure: negative deviation.

  Usually costs: 1 mark in a 2-mark question.

• ✗Mistake: Using the wrong volume unit in π = CRT

  Millilitres are substituted directly with R in L bar mol⁻¹ K⁻¹.

  Wrong working

  M = (2.0 × 0.083 × 300) ÷ (3.32 × 10⁻³ × 250)

  Right working

  M = (2.0 × 0.083 × 300) ÷ (3.32 × 10⁻³ × 0.250) = 6.0 × 10⁴ g mol⁻¹

  ✓Do this: Convert volume to litres and temperature to kelvin before substituting.

  Usually costs: The final answer mark; the answer is out by 1000.

• ✗Mistake: Writing that K_H decreases with temperature

  Students know solubility falls as temperature rises and wrongly assume $K_H$ falls too.

  ✓Do this: Solubility falls because $K_H$ rises: $x=p/K_H$. State both facts together.

  Usually costs: The full mark in an MCQ or assertion-reason item.

• ✗Mistake: Comparing freezing points by molality alone

  Options like 0.1 m glucose and 0.1 m KCl look equal.

  ✓Do this: Compare $i\times m$: the larger it is, the lower the freezing point and the higher the boiling point.

  Usually costs: 1 mark in an MCQ.

NCERT exercise guide: Solutions

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Solutions: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7 marks per paper; 33 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Solutions

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Solutions question, by year

Practice questions on Solutions

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

1. Q1 · 1 mark · MCQBoard level

   The SI unit of the molal elevation constant ($K_b$) is:
   (a) K mol kg⁻¹ (b) K kg mol⁻¹ (c) mol kg K⁻¹ (d) K⁻¹ kg mol⁻¹

   Hint

   Rearrange ΔT_b = K_b m.

   Answer

   (b) K kg mol⁻¹, because K_b = ΔT_b ÷ m = K ÷ (mol kg⁻¹).

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): 100% ethanol cannot be obtained by fractional distillation of an ethanol–water mixture.
   Reason (R): Ethanol and water form a minimum boiling azeotrope at about 95% ethanol by volume.
   (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

   What is special about an azeotrope's vapour?

   Answer

   (a) At the azeotropic composition the liquid and vapour have the same composition, so distillation cannot enrich ethanol beyond about 95%.

3. Q3 · 2 marks · Very short answerBoard level

   Define molality. Why is molality preferred to molarity for expressing the concentration in colligative property work? (2 marks)

   Hint

   Which one depends on volume?

   Answer

   Molality is the number of moles of solute per kilogram of solvent (1). It depends only on masses, so it does not change with temperature, whereas molarity depends on volume, which changes with temperature (1).

4. Q4 · 3 marks · Short answerBoard level

   State Henry's law. Give two applications of it in daily life or nature. (3 marks)

   Hint

   Think of soda bottles, divers and mountains.

   Answer

   The partial pressure of a gas in the vapour phase is proportional to its mole fraction in the solution, p = K_H x (1). Applications (any two, 1 each): soft drinks are bottled under high CO₂ pressure to increase solubility; scuba divers get bends when dissolved nitrogen forms bubbles as pressure falls, so tanks use air diluted with helium; at high altitude low oxygen partial pressure lowers dissolved oxygen, causing anoxia.

5. Q5 · 3 marks · Short answerHarder board level

   The vapour pressure of water at 298 K is 23.8 mm Hg. Calculate the vapour pressure of a 0.5 molal aqueous NaCl solution at 298 K, assuming complete dissociation. (3 marks)

   Hint

   Count the ions: 0.5 mol NaCl gives 1.0 mol of particles.

   Answer

   Particles = 2 × 0.5 = 1.0 mol per 1000 g (55.6 mol) of water (1). x(solute particles) = 1.0 ÷ 56.6 = 0.0177 (1). p = 23.8 × (1 − 0.0177) = 23.38 mm Hg (1).

6. Q6 · 3 marks · Short answerHarder board level

   Suppose a driver in Leh adds 310 g of ethylene glycol (62 g mol⁻¹) to 2.5 kg of water in the radiator. (a) Find the freezing and boiling points of the coolant. (b) The forecast says the night temperature will drop to −9.3 °C. How much more ethylene glycol must be added to protect the radiator? ($K_f$ = 1.86, $K_b$ = 0.52 K kg mol⁻¹) (3 marks)

   Hint

   Find the molality now, then the molality needed for ΔT_f = 9.3 K.

   Answer

   (a) Moles = 310 ÷ 62 = 5.0; m = 5.0 ÷ 2.5 = 2.0 mol kg⁻¹; ΔT_f = 1.86 × 2.0 = 3.72 K, so it freezes at −3.72 °C; ΔT_b = 0.52 × 2.0 = 1.04 K, so it boils at 101.04 °C (1½). (b) Needed m = 9.3 ÷ 1.86 = 5.0 mol kg⁻¹, i.e. 5.0 × 2.5 = 12.5 mol in all; extra = 12.5 − 5.0 = 7.5 mol = 7.5 × 62 = 465 g (1½).

7. Q7 · 2 marks · Very short answerHarder board level

   1.0 g of a polymer is dissolved in water to make 250 mL of solution. Its osmotic pressure at 300 K is 4.98 × 10⁻³ bar. Calculate the molar mass of the polymer. (R = 0.083 L bar mol⁻¹ K⁻¹) (2 marks)

   Hint

   M = wRT ÷ (πV), with V in litres.

   Answer

   M = (1.0 × 0.083 × 300) ÷ (4.98 × 10⁻³ × 0.250) (1) = 24.9 ÷ (1.245 × 10⁻³) = 2.0 × 10⁴ g mol⁻¹ (1).

8. Q8 · 3 marks · Short answerHarder board level

   A 5% (w/V) glucose solution (180 g mol⁻¹) is used as an intravenous fluid. (a) What percentage (w/V) of urea (60 g mol⁻¹) would be isotonic with it? (b) What percentage (w/V) of NaCl (58.5 g mol⁻¹), assuming complete dissociation, would be isotonic with it? (3 marks)

   Hint

   Isotonic means equal concentration of particles, i × C, at the same temperature.

   Answer

   Glucose: 50 g L⁻¹ ÷ 180 = 0.278 mol L⁻¹ of particles (1). (a) Urea (i = 1): 0.278 × 60 = 16.7 g L⁻¹, i.e. about 1.67% (1). (b) NaCl (i = 2): C = 0.278 ÷ 2 = 0.139 mol L⁻¹; 0.139 × 58.5 = 8.1 g L⁻¹, i.e. about 0.81% (1).

9. Q9 · 1 mark · MCQHarder board level

   The van't Hoff factor for K₄[Fe(CN)₆], assuming complete dissociation, and for a solute that completely dimerises are respectively:
   (a) 4 and 2 (b) 5 and 0.5 (c) 5 and 2 (d) 11 and 0.5

   Hint

   The complex ion stays together.

   Answer

   (b) K₄[Fe(CN)₆] → 4K⁺ + [Fe(CN)₆]⁴⁻ gives 5 ions; complete dimerisation halves the particles, i = 0.5.

10. Q10 · 3 marks · Short answerHOTS / competency

    Equal masses of glucose (180 g mol⁻¹) and sucrose (342 g mol⁻¹) are dissolved in the same volume of water at the same temperature. Which solution has the higher osmotic pressure, and by what factor? Which would show the larger depression of freezing point? (3 marks)

    Hint

    Equal mass, different molar mass: compare moles.

    Answer

    Glucose: it has more moles because its molar mass is smaller (1). π is proportional to moles, so π(glucose) ÷ π(sucrose) = 342 ÷ 180 = 1.9 (1). Glucose also shows the larger depression of freezing point, as it has more particles (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    Suppose a student mixes acetone and chloroform and finds that the mixture warms up. (a) Sketch, in words, how the total vapour pressure–composition curve differs from Raoult's law. (b) Give the signs of ΔmixH and ΔmixV with reasons. (c) Name the type of azeotrope that can form and state whether its boiling point is above or below those of both components. (d) Give one other pair of liquids that behaves the same way. (5 marks)

    Hint

    Warming means new attractions are stronger.

    Answer

    (a) The total vapour pressure curve lies below the straight Raoult's law line, possibly with a minimum (1). (b) ΔmixH < 0 because hydrogen bonds form between chloroform's H and acetone's O, releasing heat; ΔmixV < 0 because the stronger attractions pull molecules closer (2). (c) Maximum boiling azeotrope; it boils above both pure components (1). (d) Phenol + aniline, or nitric acid + water (1).

12. Q12 · 3 marks · Short answerHOTS / competency

    A salt AB₂ ionises in water as AB₂ → A²⁺ + 2B⁻. A 0.020 m aqueous solution of it freezes at −0.093 °C. (a) Calculate the van't Hoff factor and the degree of dissociation. (b) Estimate the osmotic pressure of this solution at 300 K, taking molarity ≈ molality. (c) Suggest why i is less than 3 even though AB₂ is ionic. ($K_f$ = 1.86 K kg mol⁻¹, R = 0.083 L bar mol⁻¹ K⁻¹) (3 marks)

    Hint

    Three ions per formula unit: i = 1 + 2α.

    Answer

    (a) Expected ΔT_f = 1.86 × 0.020 = 0.0372 K; i = 0.093 ÷ 0.0372 = 2.5; 1 + 2α = 2.5, so α = 0.75 (75%) (1½). (b) π = iCRT = 2.5 × 0.020 × 0.083 × 300 ≈ 1.2 bar (1). (c) Some oppositely charged ions stay attracted to each other as ion pairs, so fewer independent particles act than complete dissociation predicts (½).

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. 1.Which concentration terms are independent of temperature?Show answer

   Mass percentage, ppm, mole fraction and molality, because they depend only on masses.

2. 2.State Henry's law.Show answer

   The partial pressure of a gas in the vapour phase is proportional to its mole fraction in the solution: p = K_H x.

3. 3.What happens to K_H when temperature rises, and what does that mean for solubility?Show answer

   K_H increases, so the solubility of the gas decreases.

4. 4.What is anoxia and why does it happen?Show answer

   Weakness and inability to think clearly at high altitude, caused by the low partial pressure of oxygen, so less oxygen dissolves in blood and tissues.

5. 5.Give the conditions for an ideal solution.Show answer

   It obeys Raoult's law at all compositions, with ΔmixH = 0 and ΔmixV = 0.

6. 6.Name one minimum boiling and one maximum boiling azeotrope.Show answer

   Minimum: ethanol–water (about 95% ethanol by volume). Maximum: nitric acid–water (about 68% HNO₃ by mass, boiling at 393.5 K).

7. 7.What are the values of K_b and K_f for water?Show answer

   K_b = 0.52 K kg mol⁻¹ and K_f = 1.86 K kg mol⁻¹.

8. 8.What concentration of NaCl is isotonic with blood cells?Show answer

   About 0.9% (mass/volume), called normal saline.

9. 9.Why is osmotic pressure preferred for finding the molar mass of proteins?Show answer

   It is measured at room temperature and gives a large, measurable value even for very dilute solutions.

10. 10.What is the van't Hoff factor for ethanoic acid in benzene, and why?Show answer

    Close to 0.5, because the acid forms dimers through hydrogen bonding.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why molarity changes with temperature but molality does not?
• ☐Can you use p = K_H x to find a mole fraction and a mass of dissolved gas?
• ☐Can you state two effects of Henry's law on divers and climbers?
• ☐Can you find the total vapour pressure and the vapour composition of an ideal mixture?
• ☐Can you give one example each of positive and negative deviation with the signs of ΔH and ΔV?
• ☐Can you derive the molar mass formula from ΔT_b = K_b m?
• ☐Can you explain isotonic, hypertonic and hypotonic solutions using red blood cells?
• ☐Can you write i for NaCl, Na₂SO₄, K₄[Fe(CN)₆] and a dimerising acid, and find α from i?

Solutions: frequently asked questions