Chapter 1 of 10

10%

CBSE Class 12 Chemistry★ Unit I · Solutions, CBSE states no chapter-wise split

Solutions

Types of solutions and the four ways of expressing concentration, solubility and Henry's law, Raoult's law for volatile and non-volatile solutes, ideal and non-ideal solutions and azeotropes, the four colligative properties and their use in finding molar mass, and abnormal molar masses through the van't Hoff factor. CBSE Class 12 Chemistry Chapter 1.

⏱ 13 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

Podcast

Listen as a conversation

A teacher & a student talk through the chapter — in your language.

Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

…

Study Guide

Night-before notes

Key points, glossary and the questions most likely to show up.

…

VocabularyBuild your vocabularyLexiGenome mines the key words and schedules them with spaced repetition.Mine words →

By the end of this chapter you'll be able to…

1Classify solutions by the states of solute and solvent
2Convert between mass percentage, mole fraction, molality and molarity
3Explain why gas solubility falls with temperature and apply Henry's law in both its forms
4Apply Raoult's law to volatile and non-volatile solutes and find the vapour composition
5Distinguish ideal from non-ideal solutions and relate deviations to the enthalpy of mixing
6Use the four colligative properties to determine molar mass
7Apply the van't Hoff factor to solutes that dissociate or associate

💡

Why this chapter matters

Solutions is the numerical backbone of Class 12 Chemistry. Concentration conversions, Raoult's law and the four colligative properties recur throughout physical chemistry, and this chapter carries the largest exercise set in the book at 41 questions plus 12 intext.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

🔗

Some Basic Concepts of Chemistry (Class 11 Ch 1): the mole concept, molar mass and concentration units are used in almost every question here

🔗

States of Matter (Class 11 Ch 5): vapour pressure, intermolecular forces and the behaviour of gases underpin Henry's law and Raoult's law

🔗

Equilibrium (Class 11 Ch 7): Le Chatelier's principle explains the temperature dependence of gas solubility, and Ka is needed for the weak-acid van't Hoff questions

🔗

Thermodynamics (Class 11 Ch 6): the sign of the enthalpy of mixing is what distinguishes positive from negative deviation

Solutions

1. Check this before you revise anything

This was Chapter 2 in the previous edition. Six whole chapters have been removed from the Class 12 Chemistry book — The Solid State, Surface Chemistry, General Principles and Processes of Isolation of Elements, The p-Block Elements, Polymers, and Chemistry in Everyday Life. The surviving ten are renumbered contiguously, so Solutions has moved from 2 to 1.

If you are working from an older guide or a past paper, every chapter number in Class 12 Chemistry will be one or more higher than it is now.

This chapter has two separate question sets, unlike Physics:

Intext Questions, 12 of them, appearing in boxes as each topic is introduced.
Exercises, 41 of them, at the end of the chapter.

Both sets are numbered 1.1, 1.2, 1.3 …, so an intext question and an exercise share the same label. When a question is quoted without saying which set it belongs to, check the content rather than the number.

The book answers only the intext questions, and only some of them. A short key at the end of the chapter gives answers to intext 1.1 to 1.5 and 1.9 to 1.12. Intext 1.6, 1.7 and 1.8 have no printed answer, and none of the 41 exercises do. Every answer on this site has been worked out independently, and where the book does supply one, ours agrees with it.

At 41 questions, this is the largest exercise set in the Chemistry book, larger than any single Physics chapter.

Textbook section	Topic
1.1	Types of solutions
1.2	Expressing concentration of solutions
1.3	Solubility of a solid in a liquid, and of a gas in a liquid
1.4	Vapour pressure of liquid solutions; Raoult's law
1.5	Ideal and non-ideal solutions
1.6	Colligative properties and determination of molar mass
1.7	Abnormal molar masses and the van't Hoff factor

2. Types of Solutions and Concentration (Textbook 1.1 to 1.2)

A solution is a homogeneous mixture of two or more non-reacting substances. The component in excess is the solvent and the rest are solutes. Classifying by the state of the solution and of the solute gives nine types, from gas in gas (air) through solid in liquid (glucose in water) to solid in solid (copper in gold).

Four ways of expressing concentration, and the distinctions between them earn marks constantly:

Measure	Definition	Unit	Temperature dependent?
Mass percentage	mass of component / total mass of solution × 100	%	No
Mole fraction $x$	moles of component / total moles	none	No
Molality $m$	moles of solute / kg of solvent	mol kg $^{- 1}$	No
Molarity $M$	moles of solute / litre of solution	mol L $^{- 1}$	Yes

The one distinction to fix firmly: molality is per kilogram of solvent, molarity per litre of solution. Because molality involves only masses, and mass does not change with temperature, molality is temperature-independent. Molarity is not, since volume expands on heating.

Where the density comes in. A density is needed only to convert a mass into a volume, so it is used for molarity and nothing else. Exercises 1.5 and 1.8 both give a density and both use it only at that one step.

The 100 g trick. Whenever a concentration is given as a percentage by mass, take 100 g of solution. The percentage then becomes a mass in grams directly, and everything else follows. This works for Exercises 1.4, 1.5 and 1.9.

Parts per million is used for trace quantities: 15 ppm means 15 parts in $1 0^{6}$ by mass.

3. Solubility and Henry's Law (Textbook 1.3)

Solids in liquids (1.3.1) follow like dissolves like: polar solutes dissolve in polar solvents and non-polar in non-polar. A saturated solution is in dynamic equilibrium with undissolved solute.

For most solids, dissolution is endothermic, so solubility rises with temperature. Where dissolution is exothermic, it falls.

Gases in liquids (1.3.2) behave the opposite way. Dissolving a gas is exothermic, so by Le Chatelier's principle warming drives dissolved gas out and solubility falls with temperature — which is why boiling expels dissolved air and why warm rivers hold less oxygen.

Henry's law relates solubility to pressure:

$p = K_{H} x$

where $p$ is the partial pressure of the gas and $x$ its mole fraction in solution.

Reading $K_{H}$ the right way round: a larger $K_{H}$ means a lower solubility. $K_{H}$ also rises with temperature, which is the same statement as gases becoming less soluble on heating.

Check the units before substituting. Exercise 1.35 quotes $K_{H}$ for the molality form, so the answer comes out directly as a molality with no mole-fraction conversion at all. Missing that turns a one-line problem into a wrong one.

Four applications worth knowing: soda bottles sealed under CO $_{2}$ pressure; the bends suffered by divers surfacing too fast as dissolved nitrogen bubbles out; anoxia at high altitude from the low partial pressure of oxygen; and oxygen-enriched air for patients.

4. Raoult's Law, and Deviations (Textbook 1.4 to 1.5)

For two volatile liquids (1.4.1 to 1.4.2), each component's partial pressure is proportional to its mole fraction in the liquid:

$p_{A} = p_{A}^{\circ} x_{A}, p_{total} = p_{A}^{\circ} x_{A} + p_{B}^{\circ} x_{B}$

For a non-volatile solute (1.4.3) only the solvent contributes, and the law takes the form of a relative lowering:

$\frac{p ^{\circ} - p}{p ^{\circ}} = x_{solute}$

The liquid and the vapour have different compositions. By Dalton's law the mole fraction in the vapour is $y_{A} = p_{A} / p_{total}$ , and the vapour is always richer in the more volatile component. That difference is what makes fractional distillation possible, and it is the point of Intext 1.8 and Exercise 1.38.

Ideal solutions (1.5.1) obey Raoult's law at every composition, because the A-B interaction is the same strength as A-A and B-B. Consequently:

$Δ_{mix} H = 0, Δ_{mix} V = 0$

Benzene with toluene, and n-hexane with n-heptane, are the standard examples.

Non-ideal solutions (1.5.2) arise when the unlike interaction differs in strength:

	Positive deviation	Negative deviation
A-B interaction	Weaker than A-A and B-B	Stronger than A-A and B-B
Vapour pressure	Higher than predicted	Lower than predicted
$Δ_{mix} H$	Positive, endothermic	Negative, exothermic
$Δ_{mix} V$	Positive	Negative
Example	Ethanol and water	Chloroform and acetone
Azeotrope formed	Minimum boiling	Maximum boiling

Chloroform and acetone deviate negatively because a hydrogen bond forms between chloroform's hydrogen and acetone's carbonyl oxygen — a new interaction stronger than either pure liquid had. That is the whole explanation required by Exercise 1.37.

Azeotropes are constant-boiling mixtures that distil unchanged, so they cannot be separated by fractional distillation. Ethanol and water form a minimum-boiling azeotrope at about 95% ethanol, which is why absolute alcohol cannot be obtained by distillation alone.

5. The Four Colligative Properties (Textbook 1.6)

A colligative property depends only on the number of solute particles, not on their chemical identity. There are four:

$(i) \frac{p ^{\circ} - p}{p ^{\circ}} = x_{solute} (ii) Δ T_{b} = K_{b} m$

$(iii) Δ T_{f} = K_{f} m (iv) Π = C R T$

Boiling point rises and freezing point falls. A non-volatile solute lowers the vapour pressure, so a higher temperature is needed to reach atmospheric pressure, while the solid-liquid equilibrium is reached at a lower temperature.

$K_{b}$ and $K_{f}$ belong to the solvent, not the solute. For water $K_{b} = 0.52$ and $K_{f} = 1.86$ K kg mol $^{- 1}$ . Intext 1.11 uses acetic acid as solvent, so its own $K_{f} = 3.9$ applies instead.

Take the elevation from the right baseline. Intext 1.10 gives the boiling point of water at 750 mm Hg as $99.6 3^{\circ}$ C, so the elevation to reach $10 0^{\circ}$ C is $0.37$ K, not zero.

Osmosis and osmotic pressure (1.6.4). Solvent flows through a semipermeable membrane from the dilute side to the concentrated side. The osmotic pressure is the pressure that must be applied to stop that flow:

$Π = \frac{n}{V} R T = C R T$

Osmotic pressure is the method of choice for large molecules. Exercise 1.12 finds a polymer of molar mass 185,000 giving a perfectly measurable 31 Pa, where the other three properties would produce changes far too small to detect.

Reverse osmosis (1.6.5). Applying a pressure greater than the osmotic pressure drives solvent backwards, from concentrated to dilute. This is how seawater is desalinated.

Isotonic, hypertonic and hypotonic describe solutions of equal, greater and smaller osmotic pressure relative to a reference — which is why a 0.9% saline drip matches blood and leaves red cells intact.

6. Abnormal Molar Masses and the van't Hoff Factor (Textbook 1.7)

When a solute dissociates or associates in solution, the number of particles differs from the number of formula units dissolved, and the molar mass calculated from a colligative property comes out wrong.

The van't Hoff factor measures the discrepancy:

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

Behaviour	$i$	Observed molar mass	Example
Dissociation	$> 1$	Lower than actual	KCl, CaCl $_{2}$ , weak acids
Association	$< 1$	Higher than actual	Benzoic acid in benzene
Neither	$= 1$	Correct	Glucose, urea

Every colligative expression then carries $i$ :

$Δ T_{f} = i K_{f} m, Π = i C R T$

Relating $i$ to the degree of dissociation. For a solute giving $n$ particles:

$i = 1 + (n - 1) α$

so for a weak acid giving two particles, $i = 1 + α$ — used in Exercises 1.32 and 1.33.

For a weak acid the degree of dissociation follows from $K_{a}$ :

$α = \frac{K _{a}}{c} when α is small, K_{a} = \frac{c α ^{2}}{1 - α}$

Why the substituted acetic acids differ. Exercise 1.31 asks why depression increases from acetic to trichloroacetic to trifluoroacetic acid. Electron-withdrawing halogens stabilise the carboxylate anion, so dissociation increases, $i$ increases, and the depression increases with it. Fluorine, being more electronegative than chlorine, has the strongest effect.

Omitting $i$ is the classic error. In Exercise 1.40 the CaCl $_{2}$ factor of 2.47 changes the answer from 8.45 g to 3.42 g.

Summary

A solution is a homogeneous mixture; nine types follow from the states of solution and solute.
Molality is per kg of solvent; molarity per litre of solution. Molality is temperature-independent, molarity is not.
Density is needed only to find a volume, so it is used for molarity alone.
Take 100 g of solution whenever a mass percentage is given.
Solids usually dissolve better on heating; gases always dissolve worse, since dissolution is exothermic.
Henry's law $p = K_{H} x$ ; a larger $K_{H}$ means lower solubility, and $K_{H}$ rises with temperature.
Check whether $K_{H}$ is quoted for mole fraction or for molality before substituting.
Raoult's law: $p_{A} = p_{A}^{\circ} x_{A}$ for volatile components; relative lowering $= x_{solute}$ for a non-volatile one.
The vapour is always richer in the more volatile component, which is what fractional distillation exploits.
Ideal solutions obey Raoult's law throughout, with $Δ_{mix} H = 0$ and $Δ_{mix} V = 0$ .
Positive deviation: weaker A-B forces, higher vapour pressure, endothermic, minimum-boiling azeotrope.
Negative deviation: stronger A-B forces, lower vapour pressure, exothermic, maximum-boiling azeotrope.
Chloroform with acetone deviates negatively because a hydrogen bond forms between them.
Azeotropes distil unchanged and cannot be separated by fractional distillation.
The four colligative properties depend on the number of particles alone.
$Δ T_{b} = K_{b} m$ , $Δ T_{f} = K_{f} m$ , $Π = C R T$ ; $K_{b}$ and $K_{f}$ belong to the solvent.
Osmotic pressure is preferred for large molecules, since it stays measurable where the others do not.
Reverse osmosis drives solvent backwards under a pressure above the osmotic pressure, and desalinates seawater.
$i$ is greater than 1 for dissociation and less than 1 for association, with $i = 1 + (n - 1) α$ .
Every colligative formula carries $i$ when the solute dissociates; omitting it is the standard error.
This chapter was Chapter 2 in the previous edition; six chapters have been removed from the book.
It carries two question sets sharing the same numbering, and the book prints answers for only part of the intext set.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Mass percentage

mass of component divided by total mass of SOLUTION, times 100

The denominator is the whole solution, not the solvent alone

Mole fraction

x_A = n_A divided by the total number of moles

Dimensionless, and all mole fractions sum to one

Molality

m = moles of solute per kilogram of SOLVENT

Unit mol/kg; independent of temperature because it involves only masses

Molarity

M = moles of solute per litre of SOLUTION

Unit mol/L; depends on temperature because volume expands on heating

Parts per million

ppm = mass of component divided by total mass, times 10^6

Used for trace contamination, as in Exercise 1.9

Henry's law, mole fraction form

p = K_H x

A larger K_H means LOWER solubility; K_H rises with temperature

Henry's law, molality form

p = K_H m

Exercise 1.35 quotes K_H this way, so no mole-fraction conversion is needed

Raoult's law for volatile components

p_A = p_A(pure) x_A, and p_total = p_A(pure) x_A + p_B(pure) x_B

Applies to each component of an ideal binary liquid solution

Raoult's law for a non-volatile solute

(p(pure) - p)/p(pure) = x_solute

The relative lowering of vapour pressure equals the solute mole fraction

Vapour phase composition

y_A = p_A / p_total, by Dalton's law

The vapour is always richer in the more volatile component

Ideal solution

Obeys Raoult's law at all compositions, with enthalpy and volume of mixing both zero

A-B interactions equal in strength to A-A and B-B

Positive deviation

Vapour pressure higher than predicted; enthalpy of mixing positive; minimum boiling azeotrope

A-B interactions weaker than A-A and B-B, as in ethanol and water

Negative deviation

Vapour pressure lower than predicted; enthalpy of mixing negative; maximum boiling azeotrope

A-B interactions stronger, as in chloroform and acetone which hydrogen bond

Elevation of boiling point

Delta T_b = K_b m

K_b belongs to the SOLVENT; for water it is 0.52 K kg/mol

Depression of freezing point

Delta T_f = K_f m

K_f belongs to the SOLVENT; for water it is 1.86 K kg/mol

Osmotic pressure

Pi = (n/V) R T = C R T

Preferred for large molecules, since it stays measurable where the other properties do not

Molar mass from freezing point depression

M = (K_f x w x 1000) divided by (Delta T_f x W)

w is the solute mass in g and W the solvent mass in g

van't Hoff factor

i = observed colligative property divided by calculated value = normal molar mass divided by observed molar mass

Greater than 1 for dissociation, less than 1 for association

van't Hoff factor and degree of dissociation

i = 1 + (n - 1) alpha

For a solute giving n particles; for a weak acid giving two, i = 1 + alpha

Colligative properties with dissociation

Delta T_f = i K_f m, and Pi = i C R T

Omitting i is the classic error, as in Exercise 1.40

Dissociation constant of a weak acid

K_a = c alpha squared divided by (1 - alpha)

When alpha is small, alpha is approximately the square root of K_a over c

⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Confusing molality with molarity

✓ Molality is per kilogram of SOLVENT and molarity per litre of SOLUTION. Only molality is temperature-independent, since it involves masses alone.

WATCH OUT

✗ Treating the total mass given as the mass of solvent in a molality problem

✓ In Intext 1.4 the 2.5 kg is the whole solution. Split it as w kg of water plus the solute mass, then solve. Taking it as solvent gives 37.5 g instead of 36.95 g.

WATCH OUT

✗ Using density when it is not needed

✓ Density converts a mass into a volume, so it is required for molarity and for nothing else. Molality and mole fraction never need it.

WATCH OUT

✗ Omitting the water of crystallisation from a molar mass

✓ Co(NO3)2 6H2O has molar mass 291, not 183. Write the full formula out before computing, as Intext 1.3(a) requires.

WATCH OUT

✗ Assuming gas solubility rises with temperature as solid solubility usually does

✓ Dissolving a gas is exothermic, so by Le Chatelier's principle heating drives it out. Gas solubility always falls with temperature.

WATCH OUT

✗ Substituting into the wrong form of Henry's law

✓ Check the units of K_H first. Exercise 1.35 quotes it for molality, so the answer is a molality directly and no mole-fraction step is needed.

WATCH OUT

✗ Reporting the liquid composition when the vapour composition is asked for

✓ Use Dalton's law: the mole fraction in the vapour is the partial pressure divided by the total. The vapour is always richer in the more volatile component.

WATCH OUT

✗ Taking K_b or K_f as a property of the solute

✓ They are properties of the SOLVENT. Intext 1.11 uses acetic acid as solvent, so its K_f of 3.9 applies rather than water's 1.86.

WATCH OUT

✗ Measuring the boiling point elevation from 100 degrees Celsius regardless of pressure

✓ Intext 1.10 gives the boiling point of water at 750 mm Hg as 99.63 degrees, so the elevation is 0.37 K. Using 100 degrees as the baseline gives zero and no answer.

WATCH OUT

✗ Forgetting the van't Hoff factor for an ionic solute

✓ In Exercise 1.40 including i = 2.47 changes the answer from 8.45 g to 3.42 g. Any dissociating or associating solute needs i in every colligative formula.

WATCH OUT

✗ Confusing the two question sets, which share the same numbering

✓ Intext questions and end-of-chapter exercises are both numbered 1.1, 1.2 and so on. Identify a question by its content, not by its number alone.

WATCH OUT

✗ Assuming the book supplies answers to the exercises

✓ It answers only intext 1.1 to 1.5 and 1.9 to 1.12. Intext 1.6, 1.7 and 1.8 and all 41 exercises have no printed answer.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Intext Questions

Mass percentage and mole fraction, molarity and molality conversions, Henry's law constants for H2S and CO2, liquid and vapour composition from Raoult's law, relative lowering of vapour pressure, boiling point elevation, freezing point depression and osmotic pressure of a polymer

Questions

Ex s

Exercises

Types of solutions and concentration definitions, concentration conversions using density, Henry's law, positive and negative deviations, molar mass from vapour pressure lowering and from freezing point depression, osmotic pressure, intermolecular interactions and solubility trends, and van't Hoff factor problems for weak acids and ionic solutes

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Solutions?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•A solution is a homogeneous mixture; nine types follow from the states of solution and solute
•Molality is per kg of solvent and is temperature-independent
•Molarity is per litre of solution and does depend on temperature
•Density is needed only for molarity
•Take 100 g of solution whenever a mass percentage is given
•Solid solubility usually rises with temperature; gas solubility always falls
•Henry's law p = K_H x; larger K_H means lower solubility
•Check whether K_H is quoted for mole fraction or molality
•Raoult's law for volatile components: p_A = p_A(pure) x_A
•Relative lowering of vapour pressure equals the solute mole fraction
•The vapour is always richer in the more volatile component, by Dalton's law
•Ideal solutions have zero enthalpy and zero volume of mixing
•Positive deviation: weaker A-B forces, endothermic, minimum boiling azeotrope
•Negative deviation: stronger A-B forces, exothermic, maximum boiling azeotrope
•Chloroform and acetone deviate negatively because they hydrogen bond
•Azeotropes cannot be separated by fractional distillation
•The four colligative properties depend on particle number alone
•Delta T_b = K_b m, Delta T_f = K_f m, Pi = C R T
•K_b and K_f are properties of the solvent, not the solute
•Osmotic pressure is preferred for polymers and proteins
•Reverse osmosis drives solvent from concentrated to dilute and desalinates seawater
•i greater than 1 means dissociation, less than 1 means association
•i = 1 + (n - 1) alpha, so a weak acid giving two particles has i = 1 + alpha
•Omitting i for an ionic solute is the classic error
•This chapter was Chapter 2 in the previous edition; six chapters were removed from the book
•It has two question sets sharing the same numbering, and the book answers only part of the intext set

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit I: Solutions, no chapter-wise split published by CBSE

Question type	Marks each	Typical count	What it tests
Concentration Units and Conversions, Henry's Law and Gas Solubility	2-3	1	Molality against molarity, use of density, ppm, and Henry's law in both forms
Raoult's Law and Vapour Pressure, Ideal and Non-Ideal Solutions	3-4	1	Partial and total pressures, vapour composition, deviations and azeotropes
Colligative Properties and Molar Mass, van't Hoff Factor and Abnormal Molar Mass	4-5	1	The four colligative properties, molar mass determination, and dissociation or association

Prep strategy

Take 100 g of solution whenever a mass percentage appears
Write down whether the denominator is solvent or solution before computing any concentration
Use density only when a volume is needed, which means molarity alone
Check the units of K_H before choosing which form of Henry's law to use
Ask whether the solute dissociates or associates before writing any colligative formula
Note that K_b and K_f belong to the solvent, and read which solvent the question uses

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Desalination of seawater

Reverse osmosis pushes water backwards through a semipermeable membrane under a pressure greater than the osmotic pressure, leaving the salt behind.

Antifreeze in car radiators

Ethylene glycol depresses the freezing point and raises the boiling point of the coolant, keeping it liquid across a much wider temperature range.

Intravenous fluids

Saline drips are made isotonic with blood at about 0.9% so that red blood cells neither swell and burst nor shrink.

Carbonated drinks and diving

Henry's law explains both the fizz when a bottle is opened and the bends suffered by divers who surface too quickly.

De-icing roads

Salt spread on roads dissolves in the surface water and depresses its freezing point, so ice melts at temperatures below zero.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Take 100 g of solution the moment a mass percentage appears

State whether your denominator is solvent or solution in every concentration calculation

Reach for the density only when you need a volume

Read the units of K_H before deciding which form of Henry's law applies

Decide whether the solute dissociates or associates before writing any colligative equation

Check which liquid is the solvent, since K_b and K_f belong to it

For vapour composition questions, finish with Dalton's law rather than stopping at the liquid mole fractions

Show the molality as an explicit intermediate step in molar mass determinations

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Activity and activity coefficients, which replace mole fraction when a solution is far from ideal

STRETCH

The Gibbs-Duhem equation relating the chemical potentials of the two components of a binary solution

STRETCH

Phase diagrams of binary liquid mixtures, including the derivation of azeotropic composition

STRETCH

Osmotic pressure of polyelectrolyte solutions and the Donnan membrane equilibrium

🚀

JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainMolar mass from a two-stage vapour pressure measurementRaoult's law with an unknown solvent pressure

A solution of 30 g of a non-volatile solute in 90 g of water has vapour pressure 2.8 kPa at 298 K. Adding 18 g more water raises it to 2.9 kPa. Find the molar mass of the solute and the vapour pressure of pure water.

Stuck? Show the approach

Write Raoult's law for both states with two unknowns, then divide one equation by the other to eliminate the pure solvent pressure.

Show the full solution

Let $a = 30/ M$ be the moles of solute. Before adding water, $n_{H_{2} O} = 5$ , so $2.8 = p^{\circ} \frac{5}{5 + a}$ . After adding, $n_{H_{2} O} = 6$ , so $2.9 = p^{\circ} \frac{6}{6 + a}$ . Dividing eliminates $p^{\circ}$ : $\frac{2.8}{2.9} = \frac{5 ( 6 + a )}{6 ( 5 + a )}$ , which gives $84 + 16.8 a = 87 + 14.5 a$ , so $2.3 a = 3$ and $a = 1.304$ . Hence $M = \frac{30}{1.304} = 23$ g mol $^{- 1}$ . Substituting back, $p^{\circ} = 2.8 \times \frac{6.304}{5} = 3.53$ kPa.

Answer: Molar mass = 23 g/mol; vapour pressure of pure water = 3.53 kPa

The trap

Trying to solve either equation alone. Two unknowns need both equations, and dividing is what removes the awkward one.

JEE MainWeak acid dissociation and freezing pointvan't Hoff factor from Ka

10 g of CH $_{3}$ CH $_{2}$ CHClCOOH is added to 250 g of water. Given $K_{a} = 1.4 \times 1 0^{- 3}$ and $K_{f} = 1.86$ K kg mol $^{- 1}$ , find the depression in freezing point.

Stuck? Show the approach

Find the molality, then the degree of dissociation from Ka, then build i before applying the depression formula.

Show the full solution

The molar mass of C $_{4}$ H $_{7}$ ClO $_{2}$ is $122.5$ g mol $^{- 1}$ , so $n = 10/122.5 = 0.0816$ mol and $m = 0.0816/0.250 = 0.3265$ mol kg $^{- 1}$ . For a weak acid with small $α$ , $α = K_{a} / c = 1.4 \times 1 0^{- 3} /0.3265 = 0.0655$ . Since one molecule gives two particles, $i = 1 + α = 1.0655$ . Therefore $Δ T_{f} = i K_{f} m = 1.0655 \times 1.86 \times 0.3265 = 0.65$ K. Ignoring the dissociation would give 0.61 K, so the partial ionisation matters.

Answer: Depression in freezing point = 0.65 K

The trap

Treating the acid as undissociated. Any weak acid needs i, and i is built from the degree of dissociation.

JEE AdvancedNegative deviation and hydrogen bondingInterpreting experimental vapour pressure data

Acetone and chloroform have pure vapour pressures of 741.8 and 632.8 mm Hg at 328 K. The measured total pressures for the mixture lie consistently below the Raoult's law line. Identify the deviation and explain it.

Stuck? Show the approach

Compare the measured totals with the ideal straight line, then reason from the intermolecular interactions involved.

Show the full solution

For an ideal mixture $p_{total} = 632.8 + 109 x_{acetone}$ , a straight line between the two pure values. At $x_{acetone} = 0.234$ the measured total is $110.1 + 469.4 = 579.5$ mm Hg against an ideal $658.3$ mm Hg, and the same shortfall appears at every composition. This is negative deviation. The cause is that acetone's carbonyl oxygen forms a hydrogen bond with the hydrogen of chloroform, an interaction stronger than the acetone-acetone and chloroform-chloroform interactions it replaces. Molecules therefore escape less readily, so the vapour pressure falls below the ideal value, mixing is exothermic and the volume contracts.

Answer: Negative deviation, caused by hydrogen bonding between acetone and chloroform, with negative enthalpy and volume of mixing

The trap

Reading a lower vapour pressure as weaker interactions. It is the opposite: stronger unlike interactions hold molecules in the liquid.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 BoardHigh

JEE MainHigh

NEETHigh

JEE AdvancedMedium

AI helpers for this chapter

Generate more practice on the fly, or have the chapter explained at a different level.

⚡

Generate 5 more questions

Fresh MCQs + short-answer items based on this chapter, plus answer keys.

🧠

Explain at a different level

Re-explain the chapter at the level you actually want.

📝 My notes

Saved on this device — sign in to sync · how this works

0/600

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Molality is the number of moles of solute per kilogram of solvent, while molarity is the number of moles per litre of solution. Two things follow. First, molality counts only the solvent in its denominator whereas molarity counts the whole solution, so the two differ noticeably in concentrated solutions. In Exercise 1.8 the same antifreeze is 17.95 molal but only 9.11 molar. Second, molality involves only masses and is therefore independent of temperature, while molarity depends on a volume that expands on heating. This is why molality is used whenever temperature changes, as in all the colligative property work.

Because the two processes differ in sign of enthalpy. Dissolving a gas in a liquid is exothermic, since gas molecules are brought closer together and attractive forces with the solvent are established. By Le Chatelier's principle, adding heat to a process that already releases heat pushes the equilibrium backwards, so dissolved gas is expelled. Most solids dissolve endothermically, so heating favours dissolution instead. This is why boiling drives dissolved air out of water and why warm rivers hold less oxygen for fish.

Only when you need a volume, which in practice means only for molarity. Mass percentage, mole fraction and molality all involve masses and moles, so no density is required. The standard method is to take 100 g of solution, split it into solute and solvent using the percentage, convert to moles, and then use the density once at the end to turn that 100 g into a volume in litres. Exercises 1.5 and 1.8 both supply a density and both use it at exactly that single step.

Compare the strength of the interaction between unlike molecules with that between like molecules. If A-B attraction is weaker than A-A and B-B, molecules escape more easily, the vapour pressure is higher than Raoult's law predicts, and the deviation is positive with endothermic mixing. If A-B attraction is stronger, molecules are held more tightly, the vapour pressure is lower, and the deviation is negative with exothermic mixing. The sign of the enthalpy of mixing therefore tells you the direction immediately. Ethanol with water is the standard positive case and chloroform with acetone, which hydrogen bond together, the standard negative one.

Because it remains measurable when the other colligative properties do not. A polymer has an enormous molar mass, so even a gram of it supplies very few moles and the solution is extremely dilute. The resulting elevation of boiling point or depression of freezing point would be a tiny fraction of a degree, far below what a thermometer can resolve. Osmotic pressure, however, is measured at room temperature and stays appreciable. Intext 1.12 gives 31 pascals for a polymer of molar mass 185,000, which is small but perfectly measurable with a standard osmometer.

Whenever the solute does not stay as the same number of particles it was dissolved as. Ionic solutes dissociate, so one formula unit of CaCl2 gives close to three ions and i is above 1. Weak acids dissociate partially, so i is a little above 1 and is found from the degree of dissociation as i = 1 + alpha. Some solutes associate instead, such as benzoic acid dimerising in benzene, giving i below 1. Only solutes such as glucose and urea, which neither dissociate nor associate, have i equal to 1 and can be treated without it. In Exercise 1.40 leaving i out changes the answer from 3.42 g to 8.45 g.

Because NCERT numbers both sets by chapter. The intext questions appear in boxes as each topic is introduced and run 1.1 to 1.12, while the end-of-chapter exercises run 1.1 to 1.41, so the first twelve labels are used twice over. When a question is quoted without saying which set it belongs to, identify it by its content instead. On this site the two sets are kept as separate groups so there is no ambiguity.

Only partly. A short key at the end of the chapter answers intext questions 1.1 to 1.5 and 1.9 to 1.12. Intext 1.6, 1.7 and 1.8 have no printed answer, and none of the 41 end-of-chapter exercises do. Every solution on this site has been worked out independently from the chapter's own methods, and for the nine intext questions where the book does print an answer, ours agrees with it.

Practice more, ask doubts, find a tutor

CBSE Class 12 mock tests

CBSE Class 12 PYQs

Ask in Q&A

Chemistry flashcards

Related chapters

Students who read Solutions usually read these next.

Electrochemistry

Galvanic and electrolytic cells, standard electrode potentials measured against the hydrogen electrode, the Nernst equation and its link to Gibbs energy and the equilibrium constant, conductivity and molar conductivity with Kohlrausch's law, Faraday's laws of electrolysis, and batteries, fuel cells and corrosion. CBSE Class 12 Chemistry Chapter 2.

Chemical Kinetics

Average and instantaneous rate, rate laws and the difference between order and molecularity, integrated rate equations for zero and first order reactions with half-life, the Arrhenius equation and activation energy, the effect of a catalyst, and collision theory. CBSE Class 12 Chemistry Chapter 3.

d- and f-Block Elements

The d-block elements (TRANSITION METALS) sit in the middle of the periodic table — and they are the MOST chemically diverse group. This chapter covers their properties, compounds, and the f-block (lanthanoids and actinoids). CBSE Class 12 Chemistry Chapter 4.

Coordination Compounds

A COORDINATION COMPOUND is formed when a central metal ion is SURROUNDED by a group of molecules or ions (ligands). These compounds are VITALLY important in biology, medicine, and industry. CBSE Class 12 Chemistry Chapter 5.

🧑‍🏫