Meritroot Academy website preview · Residential & offline programmes · Secure domain launch coming next
Dehradun, UttarakhandResidential & Offline
Admission guidance+91 7417454936
Meritroot Academy

MERITROOT FREE LEARNING · CLASS 12 / CHEMISTRY

Solutions

Your chapter workspace for understanding concepts, revising key ideas and practising questions. Choose an available resource below.

PYQs4Formulae6Notes2
?

PYQs

Practise questions grouped by chapter and source year.

Practice questions are organised using the years and labels in our source collection. Work through the solutions with your teacher; these are not a complete official examination paper.

Practice question 1

Source year: 20253 marks

Define molarity and molality. [CBSE 2025]

Practice question 2

Source year: 20255 marks

Define molarity, molality, mole fraction. Calculate molality of solution containing 5g NaCl in 100g water. [CBSE 2025]

Practice question 3

Source year: 20243 marks

State and explain Raoult's law for ideal solutions. [CBSE 2024]

Practice question 4

Source year: 20232 marks

What is colligative property? Give examples. [CBSE 2023]

ƒ

Formulae

Revisit important relationships and quick revision formulae.

Molarity

M = moles/L

Molarity

M = moles of solute / volume in L

Molality

m = moles of solute / mass of solvent in kg

Raoult's Law

P = P°×X (P = vapor pressure, X = mole fraction)

Depression in Freezing Point

ΔTf = Kf × m

Nernst Equation

Source year: 2026

The Nernst equation relates the reduction potential of a half-cell to the standard electrode potential, temperature, and activities of the chemical species undergoing reduction and oxidation.

Ecell=Ecell∘−RTnFln⁡Q E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln Q

At 298 K, this simplifies to:

Ecell=Ecell∘−0.0591nlog⁡10Q E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log_{10} Q
≡

Notes

Build understanding with chapter guides and revision notes.

Understand the chapter

Chapter overview

A solution is a homogeneous mixture of two or more components. The component present in larger amount is usually called the solvent, and the component present in smaller amount is called the solute.

1. Types of solutions

SoluteSolventExample
GasGasAir
GasLiquidSoda water
LiquidLiquidAlcohol in water
SolidLiquidSalt in water
SolidSolidAlloys

2. Concentration terms

Mass percentage

Mass percent=mass of solutemass of solution×100\text{Mass percent} = \frac{\text{mass of solute}} {\text{mass of solution}} \times100

Volume percentage

Volume percent=volume of solutevolume of solution×100\text{Volume percent} = \frac{\text{volume of solute}} {\text{volume of solution}} \times100

Mass by volume percentage

mass of solutevolume of solution×100\frac{\text{mass of solute}} {\text{volume of solution}} \times100

Parts per million

ppm=parts of soluteparts of solution×106ppm= \frac{\text{parts of solute}} {\text{parts of solution}} \times10^6

Mole fraction

XA=nAnA+nBX_A=\frac{n_A}{n_A+n_B} XA+XB=1X_A+X_B=1

Molarity

M=moles of solutevolume of solution in litreM= \frac{\text{moles of solute}} {\text{volume of solution in litre}}

Molality

m=moles of solutemass of solvent in kgm= \frac{\text{moles of solute}} {\text{mass of solvent in kg}}

3. Solubility

Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a particular temperature.

Factors affecting solubility:

  • nature of solute,
  • nature of solvent,
  • temperature,
  • pressure for gases.

4. Henry's law

Henry's law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid.

p=KHxp=K_Hx

where:

  • pp = partial pressure of gas,
  • KHK_H = Henry's law constant,
  • xx = mole fraction of gas.

Applications:

  • soda bottles,
  • deep sea diving,
  • oxygen transport,
  • high altitude breathing.

5. Vapour pressure of liquid solutions

Raoult's law for volatile components:

pA=xApA∘p_A=x_Ap_A^\circ pB=xBpB∘p_B=x_Bp_B^\circ

Total pressure:

ptotal=pA+pBp_{\text{total}}=p_A+p_B ptotal=xApA∘+xBpB∘p_{\text{total}} = x_Ap_A^\circ+x_Bp_B^\circ

6. Ideal solutions

An ideal solution obeys Raoult's law over the entire range of concentration.

Conditions:

ΔHmix=0\Delta H_{\text{mix}}=0 ΔVmix=0\Delta V_{\text{mix}}=0

Example:

  • benzene + toluene
  • n-hexane + n-heptane

7. Non-ideal solutions

Non-ideal solutions do not obey Raoult's law.

Positive deviation:

pobserved>pRaoultp_{\text{observed}}>p_{\text{Raoult}}

Negative deviation:

pobserved<pRaoultp_{\text{observed}}<p_{\text{Raoult}}

8. Azeotropes

Azeotropes are mixtures that boil at constant temperature and have same composition in liquid and vapour phase.

Types:

TypeDeviationExample
Minimum boiling azeotropePositive deviationEthanol-water
Maximum boiling azeotropeNegative deviationNitric acid-water

9. Colligative properties

Colligative properties depend on the number of solute particles, not their nature.

Types:

  • relative lowering of vapour pressure,
  • elevation of boiling point,
  • depression of freezing point,
  • osmotic pressure.

10. Relative lowering of vapour pressure

For a non-volatile solute:

pA∘−pApA∘=xB\frac{p_A^\circ-p_A}{p_A^\circ} = x_B

For dilute solution:

pA∘−pApA∘=nBnA\frac{p_A^\circ-p_A}{p_A^\circ} = \frac{n_B}{n_A}

11. Elevation of boiling point

ΔTb=Tb−Tb∘\Delta T_b=T_b-T_b^\circ ΔTb=Kbm\Delta T_b=K_bm

Molar mass of solute:

MB=Kb×wB×1000ΔTb×wAM_B= \frac{K_b\times w_B\times1000} {\Delta T_b\times w_A}

12. Depression of freezing point

ΔTf=Tf∘−Tf\Delta T_f=T_f^\circ-T_f ΔTf=Kfm\Delta T_f=K_fm

Molar mass:

MB=Kf×wB×1000ΔTf×wAM_B= \frac{K_f\times w_B\times1000} {\Delta T_f\times w_A}

13. Osmotic pressure

Osmosis is movement of solvent from dilute solution to concentrated solution through a semipermeable membrane.

Osmotic pressure:

Π=CRT\Pi=CRT

Also:

ΠV=nRT\Pi V=nRT

Molar mass:

MB=wBRTΠVM_B= \frac{w_BRT}{\Pi V}

14. van't Hoff factor

The van't Hoff factor accounts for association or dissociation of solute particles.

i=observed colligative propertynormal colligative propertyi= \frac{\text{observed colligative property}} {\text{normal colligative property}}

For dissociation:

i>1i>1

For association:

i<1i<1

Modified formulae:

ΔTb=iKbm\Delta T_b=iK_bm ΔTf=iKfm\Delta T_f=iK_fm Π=iCRT\Pi=iCRT

15. Solved example

If 0.10.1 mol solute is dissolved in 11 kg water and Kf=1.86K_f=1.86, then:

ΔTf=Kfm\Delta T_f=K_fm ΔTf=1.86×0.1=0.186 K\Delta T_f=1.86\times0.1=0.186\ K

16. Common mistakes

  • Confusing molarity and molality.
  • Forgetting molarity changes with temperature.
  • Using wrong sign for ΔTf\Delta T_f.
  • Forgetting van't Hoff factor for electrolytes.
  • Confusing positive and negative deviation.
  • Treating all solutions as ideal.

17. Quick revision

ConceptFormula
Mole fractionXA=nAnA+nBX_A=\frac{n_A}{n_A+n_B}
MolarityM=nVLM=\frac{n}{V_L}
Molalitym=nkg solventm=\frac{n}{kg\ solvent}
Henry's lawp=KHxp=K_Hx
Raoult's lawpA=xApA∘p_A=x_Ap_A^\circ
RLVPpA∘−pApA∘=xB\frac{p_A^\circ-p_A}{p_A^\circ}=x_B
Boiling point elevationΔTb=Kbm\Delta T_b=K_bm
Freezing point depressionΔTf=Kfm\Delta T_f=K_fm
Osmotic pressureΠ=CRT\Pi=CRT
van't Hoff factori=observednormali=\frac{\text{observed}}{\text{normal}}

Comprehensive Notes: Solutions

Introduction to Solutions

These are the fundamental concepts and key takeaways for this chapter. Make sure to review these before exams.

  • Key Concept 1
  • Key Concept 2
  • Important Definition

More detailed notes will be added here soon.

Ready for a more structured preparation plan?

Explore school-integrated JEE / NEET preparation and residential learning at Meritroot Academy, Dehradun.

Talk to academy admissions ↗

Keep your preparation moving.

WhatsApp ↗