Chemistry Notes for class 12 Chapter 2 Solutions
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Chemistry. Calculate the mole fraction of ethylene glycol (C. 2 2. = 12 × 2 + 1 × 6 + 16 × 2 = 62 g mol–1. Moles of C2H6O2 =.
HYDROGEN
2 Commercial Production of. Dihydrogen. The commonly used processes are outlined below: (i) Electrolysis of acidified water using platinum electrodes gives
Class XII Chemistry Ch. 2: Solutions Important formulae & Concepts
Number of gram equivalent of solute. Normality. Volumeof solution in litres. = 8. o. 1. 1. 2 o. 1 p p.
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CHEMISTRY ACIDS BASES & SALTS www.topperlearning.com. 2. Acids
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classes of polymers and different Chemistry. 2. Semi-synthetic polymers. Cellulose derivatives as cellulose acetate (rayon) and cellulose.
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Darzen procedure is the best method for preparing alkyl halides from alcohols since both the by products (SO2 and HCl) are gaseous and escape easily. 2. Free
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Chapter Two - ELECTROSTATIC POTENTIAL AND CAPACITANCE
In Chapters 6 and 8 (Class XI) the notion of potential energy was 12 is the distance between points 1 and 2. Since electrostatic force is conservative
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Chemistry
Ch. 2: Solutions
Important formulae & Concepts
1. Mass percentage of a component (w/w)
Mass of component in solution100Toal mass of solution× 2.Volume percentage of a component (v/v)
Volume of the component = 100Total volume of solution×3. Mole fraction of a component (x) =
Number of moles of the component
Total number of moles of all components
4.6Number of parts of componentParts per million10Total number of parts of all components of solution=×
5. Number of moles of soluteMolarityVolumeof solution in litres= 6. Number of moles of soluteMolalityMass of solvent in kilograms= 7. Number of gram equivalent of soluteNormalityVolumeof solution in litres= 8. o 1 1 2o 1 p pxp-= 9.Δ = -0
b bT T T b 2b2 1K 1000 wTM ×w× ×Δ =
10. 0 f fT T TΔ = - f 2f2 1K 1000 wTM ×w× ×Δ =
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12.22w RTM =Vπ
13.Normal molar massiAbnormal molar mass
Observedcolligative property
Calculatedcolligative property
Totalnumberofmolesofparticlesafterassoci
ation/dissociation on/dissociation14. Inclusion of van"t Hoff factor modified the equations for colligative
properties as: -=o 1 1 2 o11 n.n p pip× ×Δ =b 2b
2 1K 1000 wT .M ×wi
× ×Δ =f 2f
2 1K 1000 wT .M ×wi
=π2n RT.Vi15. According to Raoult"s law for a solution of volatile liquids the partial
vapour pressure of each component in the solution is directly proportional to its mole fraction. p1 = po1 x1 ; p2 = po2 x2
Using Dalton"s law of partial pressures the total pressure of solution is calculated. o o o total 1 2 1 2p =p +(p -p )xquotesdbs_dbs17.pdfusesText_23[PDF] class 12 chemistry chapter 2 pdf
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