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The questions your board exam loves to ask

800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

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ChemistryClass 123 marksmedium

Electrochemistry

State Kohlrausch's law of independent migration of ions and give one application.

Reveal model answer + marking points▾

Kohlrausch's law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the individual contributions of its cations and anions, each migrating independently. Mathematically, the limiting molar conductivity = (number of cations x limiting molar conductivity of cation) + (number of anions x limiting molar conductivity of anion). Applications: it is used to calculate the limiting molar conductivity of weak electrolytes (which cannot be found by extrapolation) and the degree of dissociation of a weak electrolyte.

Marking-scheme points

  • ✓At infinite dilution, molar conductivity = sum of ionic contributions
  • ✓Ions migrate independently
  • ✓Used to find limiting molar conductivity of weak electrolytes and degree of dissociation
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ChemistryClass 122 marksmedium

Electrochemistry

State Faraday's two laws of electrolysis.

Reveal model answer + marking points▾

Faraday's first law states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte (m is proportional to Q = I t). Faraday's second law states that when the same quantity of charge is passed through different electrolytes, the masses of substances deposited or liberated are directly proportional to their chemical equivalent weights.

m proportional to Q = I t

Marking-scheme points

  • ✓First law: mass deposited is proportional to charge passed (m proportional to It)
  • ✓Second law: for same charge, mass is proportional to equivalent weight
  • ✓One Faraday (96500 C) deposits one gram equivalent
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ChemistryClass 122 marksmedium

Electrochemistry

A current of 5 A is passed through a silver nitrate solution for 30 minutes. Calculate the mass of silver deposited. (Atomic mass of Ag = 108, F = 96500 C/mol)

Reveal model answer + marking points▾

Charge passed Q = I t = 5 x (30 x 60) = 5 x 1800 = 9000 C. Silver is deposited by Ag+ + e- -> Ag, so 1 mole of electrons (96500 C) deposits 108 g of silver. Mass of Ag = (108/96500) x 9000 = (108 x 9000)/96500 = 972000/96500 = 10.07 g.

mass = (equivalent mass x Q)/96500

Marking-scheme points

  • ✓Q = I t = 5 x 1800 = 9000 C
  • ✓96500 C deposits 108 g of Ag (Ag+ + e- -> Ag)
  • ✓Mass = (108 x 9000)/96500 = 10.07 g
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