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CBSE Class 12 Chemistry

Class 12 · CBSEChemistry50 則筆記，位於 11 個資料夾，352 KB

cookie 製作的考試準備筆記，共 50 篇，分為 11 個部分，依照課程主題編排。

Notes for CBSE Class 12 Chemistry (043), in a folder for each of the syllabus's ten units in its order, with one note for each chapter or group of related topics and the practicals gathered in a final folder. They follow the 2026-27 syllabus, so topics it removed or assesses only formatively are not included. Delete any folder your course leaves out once the notes are yours.

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內容

  • Solutions
    • Types of solutions and concentration6 KB
    • Solubility and Henry's law6 KB
    • Vapour pressure, Raoult's law and non-ideal solutions7 KB
    • Colligative properties, molar mass and the van't Hoff factor12 KB
  • Electrochemistry
    • Galvanic cells and electrode potentials8 KB
    • The Nernst equation, Gibbs energy and concentration cells6 KB
    • Conductance of electrolytic solutions and Kohlrausch's law8 KB
    • Electrolysis and Faraday's laws7 KB
    • Batteries, fuel cells and corrosion8 KB
  • Chemical kinetics
    • Rate of reaction and the factors affecting it6 KB
    • Rate law, order and molecularity7 KB
    • Integrated rate equations and half-life6 KB
    • Temperature dependence, Arrhenius equation and collision theory7 KB
  • d- and f-block elements
    • Position, electronic configuration and trends of the d-block7 KB
    • Oxidation states, colour, magnetism, catalysis, interstitial compounds and alloys8 KB
    • Potassium dichromate and potassium permanganate6 KB
    • Lanthanoids and actinoids7 KB
  • Coordination compounds
    • Werner's theory and the terms used7 KB
    • Nomenclature of coordination compounds5 KB
    • Isomerism in coordination compounds5 KB
    • Bonding in coordination compounds9 KB
    • Importance and applications of coordination compounds4 KB
  • Haloalkanes and haloarenes
    • Classification, nomenclature and the carbon-halogen bond6 KB
    • Preparation and physical properties of haloalkanes and haloarenes7 KB
    • Nucleophilic substitution, stereochemistry and other reactions of haloalkanes9 KB
    • Haloarenes and polyhalogen compounds8 KB
  • Alcohols, phenols and ethers
    • Classification, nomenclature, structure and preparation of alcohols8 KB
    • Properties and reactions of alcohols9 KB
    • Preparation, properties and reactions of phenols7 KB
    • Preparation, properties and reactions of ethers6 KB
  • Aldehydes, ketones and carboxylic acids
    • Nomenclature, structure and preparation of aldehydes and ketones6 KB
    • Physical properties and nucleophilic addition of aldehydes and ketones6 KB
    • Oxidation, reduction, alpha-hydrogen reactions and tests for aldehydes and ketones7 KB
    • Nomenclature, structure and preparation of carboxylic acids5 KB
    • Properties, reactions and uses of carboxylic acids8 KB
  • Amines
    • Structure, classification, nomenclature and preparation of amines6 KB
    • Physical properties and basicity of amines6 KB
    • Chemical reactions and identification of amines8 KB
    • Diazonium salts6 KB
  • Biomolecules
    • Classification of carbohydrates, glucose and fructose8 KB
    • Disaccharides, polysaccharides and the role of carbohydrates6 KB
    • Amino acids and proteins7 KB
    • Enzymes, vitamins, nucleic acids and hormones8 KB
  • Practicals
    • Preparing sols, dialysis and emulsions7 KB
    • Experiments on reaction rate, enthalpy change and cell potential9 KB
    • Paper chromatography5 KB
    • Preparing inorganic and organic compounds7 KB
    • Finding the molarity of potassium permanganate by titration6 KB
    • Tests for functional groups, carbohydrates, fats and proteins8 KB
    • Qualitative analysis of salts10 KB

第一則筆記

Solutions / Types of solutions and concentration

Solutions and their types

A solution is a homogeneous mixture of two or more substances, meaning that its composition and properties are the same throughout. The component present in the largest amount is the solvent and determines the physical state of the solution; the others are solutes. A solution of two components is a binary solution.

Any of the three states of matter can act as solvent or solute, so there are nine combinations. The ones a student meets most often are these.

SoluteSolventExample
gasgasair (oxygen in nitrogen)
gasliquidoxygen dissolved in water, carbon dioxide in a soft drink
gassolidhydrogen absorbed in palladium
liquidliquidethanol in water
liquidsolidmercury in sodium (an amalgam)
solidliquidsugar or salt in water
solidsolidbrass (zinc in copper), a substitutional alloy

The notes that follow deal mostly with liquid solutions, in which the solvent is a liquid, because that is where concentration, vapour pressure and colligative properties are measured.

Expressing concentration

The concentration of a solution is the amount of solute present in a given amount of solution or solvent. It can be stated in several ways, and the choice depends on whether temperature will change.

Mass percentage
mass %=mass of solutetotal mass of solution×100 \text{mass \%} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100

A solution that is 10% glucose by mass holds 10 g of glucose in every 100 g of solution, which means 90 g of water.

Volume percentage
volume %=volume of solutetotal volume of solution×100 \text{volume \%} = \frac{\text{volume of solute}}{\text{total volume of solution}} \times 100

This suits liquid in liquid solutions. A 35% (v/v) ethylene glycol antifreeze has 35 mL of ethylene glycol in 100 mL of solution.

Mass by volume percentage

This is the mass of solute in grams per 100 mL of solution, and it is the form used in medicine and pharmacy. A 0.9% (w/v) sodium chloride solution has 0.9 g of salt in 100 mL.

Parts per million

For a very dilute solution, where the solute is a trace, the concentration is given in parts per million (ppm):

ppm=number of parts of the solutetotal number of parts of all components×106 \text{ppm} = \frac{\text{number of parts of the solute}}{\text{total number of parts of all components}} \times 10^{6}

The parts may be by mass, by volume or by number of moles. Dissolved oxygen in a lake or a pollutant in drinking water is usually quoted in mass ppm.

Mole fraction

The mole fraction of a component is its amount of substance divided by the total amount of substance in the solution. For components A and B,

xA=nAnA+nB,xB=nBnA+nB,xA+xB=1 x_\mathrm{A} = \frac{n_\mathrm{A}}{n_\mathrm{A} + n_\mathrm{B}}, \qquad x_\mathrm{B} = \frac{n_\mathrm{B}}{n_\mathrm{A} + n_\mathrm{B}}, \qquad x_\mathrm{A} + x_\mathrm{B} = 1

For a solution with many components the mole fractions still sum to 1. Mole fraction has no unit and does not change with temperature, which is why it appears in Raoult's law and Henry's law.

Molarity

Molarity (MM) is the amount of solute in moles in one litre (one dm3\pu{dm3}) of solution:

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

Its unit is mol L−1\pu{mol L-1}. Molarity changes with temperature, because the volume of the solution expands or contracts while the amount of solute stays the same.

Molality

Molality (mm) is the amount of solute in moles per kilogram of solvent:

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

Its unit is mol kg−1\pu{mol kg-1}. Mass does not vary with temperature, so molality, like mole fraction, is independent of temperature. That makes it the natural choice for boiling point and freezing point calculations.

Worked examples

Molarity from a mass and a volume

A solution is made by dissolving 4.0 g of sodium hydroxide (molar mass 40 g mol−1\pu{40 g mol-1}) in water and making it up to 250 mL.

n=4.040=0.10 molM=0.100.250 L=0.40 mol L−1 \begin{aligned} n &= \frac{4.0}{40} = 0.10\ \text{mol} \\ M &= \frac{0.10}{0.250\ \text{L}} = 0.40\ \text{mol L}^{-1} \end{aligned}
Molality and mole fraction

12.0 g of glucose (molar mass 180 g mol−1\pu{180 g mol-1}) is dissolved in 250 g of water.

nglucose=12.0180=0.0667 molm=0.06670.250 kg=0.267 mol kg−1 \begin{aligned} n_\text{glucose} &= \frac{12.0}{180} = 0.0667\ \text{mol} \\ m &= \frac{0.0667}{0.250\ \text{kg}} = 0.267\ \text{mol kg}^{-1} \end{aligned}

For the mole fraction, 250 g of water is 250/18=13.9250/18 = 13.9 mol, so

xglucose=0.06670.0667+13.9=4.78×10−3 x_\text{glucose} = \frac{0.0667}{0.0667 + 13.9} = 4.78 \times 10^{-3}

A second case uses equal amounts. 46 g of ethanol (1.0 mol) and 36 g of water (2.0 mol) together give xethanol=1/(1+2)=0.333x_\text{ethanol} = 1/(1+2) = 0.333 and xwater=0.667x_\text{water} = 0.667.

Converting between mass percentage and molarity

A bottle of concentrated hydrochloric acid is labelled 36.5% HCl by mass with a density of 1.18 g mL−1\pu{1.18 g mL-1}. Take 1000 mL of the acid.

mass of solution=1000×1.18=1180 gmass of HCl=0.365×1180=430.7 gnHCl=430.736.5=11.8 mol \begin{aligned} \text{mass of solution} &= 1000 \times 1.18 = 1180\ \text{g} \\ \text{mass of HCl} &= 0.365 \times 1180 = 430.7\ \text{g} \\ n_\text{HCl} &= \frac{430.7}{36.5} = 11.8\ \text{mol} \end{aligned}

The molarity is therefore 11.8 mol L−1\pu{11.8 mol L-1}. The density is the link: it turns the mass of solution into a volume.

Parts per million

Sea water of density 1.03 g mL−1\pu{1.03 g mL-1} contains 6.0⋅10−3 g\pu{6.0e-3 g} of dissolved oxygen in 1 L. One litre has mass 1030 g, so

ppm=6.0×10−31030×106=5.8 ppm \text{ppm} = \frac{6.0 \times 10^{-3}}{1030} \times 10^{6} = 5.8\ \text{ppm}

Relating the units

For a solution in which the solute has molar mass M2M_2, the molarity and molality are related through the density dd (in g mL−1\pu{g mL-1}) of the solution by

m=1000 M1000 d−M M2 m = \frac{1000\,M}{1000\,d - M\,M_2}

For a dilute aqueous solution the density is close to 1 and the mass of solute is small, so molarity and molality have nearly the same numerical value. For a concentrated solution the two differ noticeably, and the derivation above, which works from a fixed mass of solution, is safer than remembering the formula.

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