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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.
ChemistryClass 122 markseasy
The Solid State
Distinguish between crystalline and amorphous solids with one example each.
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Crystalline solids have a regular, long-range ordered arrangement of particles, sharp melting points, definite geometrical shapes and are anisotropic (properties differ with direction); e.g. sodium chloride and diamond. Amorphous solids have only a short-range order (irregular arrangement), no sharp melting point (they soften over a range), no definite shape and are isotropic (same properties in all directions); e.g. glass and rubber.
Marking-scheme points
- ✓Crystalline: long-range order, sharp melting point, anisotropic (NaCl)
- ✓Amorphous: short-range order, no sharp melting point, isotropic (glass)
- ✓Amorphous solids soften over a range of temperature
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The Solid State
What is a unit cell? Distinguish between a primitive and a centred unit cell.
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A unit cell is the smallest repeating three-dimensional portion of a crystal lattice which, when repeated in different directions, generates the entire crystal. In a primitive (simple) unit cell, the constituent particles are present only at the corners of the unit cell. In a centred unit cell, particles are present at positions other than the corners as well, such as body-centred (one at the centre of the body), face-centred (one at the centre of each face) or end-centred.
Marking-scheme points
- ✓Unit cell = smallest repeating unit of a crystal lattice
- ✓Primitive: particles only at the corners
- ✓Centred: extra particles (body-, face- or end-centred)
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The Solid State
Distinguish between Schottky and Frenkel defects.
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Schottky defect is a vacancy defect in which an equal number of cations and anions are missing from their lattice sites, so the density of the solid decreases; it occurs in ionic solids with high coordination number and similar cation and anion sizes (e.g. NaCl, KCl). Frenkel defect is a dislocation defect in which an ion (usually the smaller cation) leaves its lattice site and occupies an interstitial site, so the density remains unchanged; it occurs in solids with a large difference in the sizes of the ions (e.g. AgCl, ZnS).
Marking-scheme points
- ✓Schottky: equal cations and anions missing; density decreases (NaCl)
- ✓Frenkel: smaller ion shifts to an interstitial site; density unchanged (AgCl)
- ✓Both are point defects in ionic solids
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The Solid State
State the packing efficiency and coordination number of simple cubic, bcc and fcc structures.
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Packing efficiency is the percentage of the total space occupied by the particles. Simple cubic: packing efficiency 52.4 percent, coordination number 6. Body-centred cubic (bcc): packing efficiency 68 percent, coordination number 8. Face-centred cubic (fcc, also ccp/hcp): packing efficiency 74 percent (the highest), coordination number 12.
Marking-scheme points
- ✓Simple cubic: 52.4 percent, coordination number 6
- ✓bcc: 68 percent, coordination number 8
- ✓fcc/ccp/hcp: 74 percent (highest), coordination number 12
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Solutions
Define molarity, molality and mole fraction.
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Molarity (M) is the number of moles of solute per litre of solution (mol/L). Molality (m) is the number of moles of solute per kilogram of solvent (mol/kg); it is independent of temperature. Mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all components in the solution; it is dimensionless and the sum of the mole fractions equals 1.
M = n/V(L); m = n/mass of solvent(kg)
Marking-scheme points
- ✓Molarity = moles of solute/litre of solution (mol/L)
- ✓Molality = moles of solute/kg of solvent (mol/kg), temperature independent
- ✓Mole fraction = moles of component/total moles (dimensionless)
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Solutions
Calculate the molality of a solution containing 18 g of glucose (molar mass 180 g/mol) dissolved in 500 g of water.
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Moles of glucose = mass/molar mass = 18/180 = 0.1 mol. Mass of solvent (water) = 500 g = 0.5 kg. Molality = moles of solute/mass of solvent in kg = 0.1/0.5 = 0.2 mol/kg (0.2 m).
molality = moles of solute/mass of solvent (kg)
Marking-scheme points
- ✓Moles of glucose = 18/180 = 0.1 mol
- ✓Mass of water = 0.5 kg
- ✓Molality = 0.1/0.5 = 0.2 m
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Solutions
State Henry's law. Give one application.
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Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the liquid. Mathematically, p = KH x (mole fraction of gas), where KH is Henry's law constant. Applications: it explains why soft drinks (aerated under high pressure) fizz when opened, and why deep-sea divers can suffer from bends (nitrogen dissolving in blood under high pressure).
p = KH x (mole fraction)
Marking-scheme points
- ✓Solubility of a gas is proportional to its partial pressure
- ✓p = KH x (mole fraction of gas)
- ✓Application: aerated drinks, deep-sea diving (bends)
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Solutions
State Raoult's law for a solution of two volatile liquids.
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Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution. For components A and B, pA = pA(pure) x xA and pB = pB(pure) x xB, and the total vapour pressure = pA + pB. For a solution of a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute.
pA = pA(pure) x xA
Marking-scheme points
- ✓Partial pressure of each component is proportional to its mole fraction
- ✓pA = pA(pure) x xA
- ✓Total pressure = pA + pB
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Solutions
What is the Van't Hoff factor? What does its value indicate for association and dissociation?
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The Van't Hoff factor (i) is the ratio of the observed (experimental) value of a colligative property to the calculated (theoretical) value assuming no association or dissociation; equivalently, i = (actual number of particles after association/dissociation)/(number of particles before). For a solute that dissociates (like NaCl), i is greater than 1; for a solute that associates (like acetic acid in benzene), i is less than 1; and for a solute that neither associates nor dissociates, i = 1.
i = observed value/calculated value
Marking-scheme points
- ✓i = observed colligative property/calculated value
- ✓Dissociation: i greater than 1 (e.g. NaCl)
- ✓Association: i less than 1 (e.g. acetic acid in benzene)
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Solutions
Distinguish between ideal and non-ideal solutions.
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An ideal solution is one that obeys Raoult's law over the entire range of concentration; the enthalpy of mixing and the volume change on mixing are zero (e.g. benzene and toluene). A non-ideal solution does not obey Raoult's law; it shows either positive deviation (when A-B interactions are weaker than A-A and B-B, giving higher vapour pressure, e.g. ethanol and water) or negative deviation (when A-B interactions are stronger, giving lower vapour pressure, e.g. chloroform and acetone).
Marking-scheme points
- ✓Ideal: obeys Raoult's law; delta H(mix) = 0 and delta V(mix) = 0 (benzene-toluene)
- ✓Non-ideal positive deviation: higher vapour pressure (ethanol-water)
- ✓Non-ideal negative deviation: lower vapour pressure (chloroform-acetone)
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Electrochemistry
What is a galvanic (voltaic) cell? Name its two electrodes and the reaction at each.
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A galvanic cell is an electrochemical device that converts chemical energy into electrical energy through a spontaneous redox reaction (e.g. the Daniell cell). It has two electrodes: the anode, where oxidation (loss of electrons) takes place and which is the negative terminal; and the cathode, where reduction (gain of electrons) takes place and which is the positive terminal. A salt bridge connects the two half-cells and maintains electrical neutrality.
Marking-scheme points
- ✓Converts chemical energy into electrical energy (spontaneous redox)
- ✓Anode: oxidation, negative terminal
- ✓Cathode: reduction, positive terminal; salt bridge maintains neutrality
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Electrochemistry
What is standard electrode potential? What is taken as the reference electrode?
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The standard electrode potential is the potential difference developed between an electrode and its solution of unit concentration (1 M) at 298 K and 1 bar pressure, measured with respect to a reference electrode. The standard hydrogen electrode (SHE) is taken as the reference electrode and is assigned a potential of exactly zero volt. Electrodes are compared with the SHE to obtain their standard potentials.
Marking-scheme points
- ✓Electrode potential under standard conditions (1 M, 298 K, 1 bar)
- ✓Reference: standard hydrogen electrode (SHE)
- ✓SHE assigned a potential of zero volt
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Electrochemistry
Write the relation between the standard Gibbs energy change and the EMF of a cell.
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The standard Gibbs energy change of a cell reaction is related to the standard EMF by delta G(standard) = -n F E(standard), where n is the number of electrons transferred, F is the Faraday constant (96500 C/mol) and E(standard) is the standard cell EMF. A positive EMF gives a negative delta G, meaning the cell reaction is spontaneous. This relation also links electrochemistry with thermodynamics.
delta G(standard) = -n F E(standard)
Marking-scheme points
- ✓delta G(standard) = -n F E(standard)
- ✓n = electrons transferred; F = 96500 C/mol
- ✓Positive EMF gives negative delta G (spontaneous)
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Electrochemistry
State Faraday's two laws of electrolysis.
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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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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)
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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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Electrochemistry
Distinguish between primary and secondary cells with one example each.
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A primary cell is one in which the redox reaction occurs only once and cannot be reversed, so it cannot be recharged and is discarded after use; e.g. the dry cell (Leclanche cell) and the mercury cell. A secondary cell is one that can be recharged by passing current through it in the opposite direction, so it can be used again and again; e.g. the lead storage battery and the nickel-cadmium cell.
Marking-scheme points
- ✓Primary cell: cannot be recharged, used once (dry cell)
- ✓Secondary cell: rechargeable, reusable (lead storage battery)
- ✓Secondary cells are recharged by passing current in reverse
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Chemical Kinetics
Define the rate of a chemical reaction. Distinguish between average and instantaneous rate.
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The rate of a chemical reaction is the change in the concentration of a reactant or product per unit time. The average rate is the change in concentration over a measurable time interval (delta concentration/delta time). The instantaneous rate is the rate of the reaction at a particular instant of time, obtained by making the time interval very small (the derivative d[concentration]/dt). Its units are usually mol L^-1 s^-1.
rate = -d[R]/dt = +d[P]/dt
Marking-scheme points
- ✓Rate = change in concentration per unit time
- ✓Average rate = delta concentration/delta time over an interval
- ✓Instantaneous rate = rate at a particular instant (d[c]/dt)
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Chemical Kinetics
State the factors that affect the rate of a chemical reaction.
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The rate of a chemical reaction is affected by: (1) the nature and concentration of the reactants (rate usually increases with concentration); (2) temperature (rate generally increases with a rise in temperature); (3) the presence of a catalyst (which increases the rate by providing an alternative path of lower activation energy); (4) the surface area of solid reactants (greater surface area gives a faster rate); and (5) for photochemical reactions, the intensity of light.
Marking-scheme points
- ✓Concentration of reactants and their nature
- ✓Temperature (rate increases with temperature)
- ✓Catalyst, surface area and (for photochemical reactions) light
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Chemical Kinetics
The rate constant of a first order reaction is 6.93 x 10^-3 s^-1. Calculate its half-life.
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For a first order reaction, the half-life is independent of the initial concentration and is given by t(1/2) = 0.693/k. Substituting k = 6.93 x 10^-3 s^-1: t(1/2) = 0.693/(6.93 x 10^-3) = 100 s. Thus the half-life of the reaction is 100 seconds.
t(1/2) = 0.693/k
Marking-scheme points
- ✓First order half-life t(1/2) = 0.693/k (independent of concentration)
- ✓= 0.693/(6.93 x 10^-3)
- ✓t(1/2) = 100 s
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Chemical Kinetics
Write the Arrhenius equation and define activation energy.
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The Arrhenius equation relates the rate constant to temperature: k = A e^(-Ea/RT), where k is the rate constant, A is the frequency (pre-exponential) factor, Ea is the activation energy, R is the gas constant and T is the absolute temperature. Activation energy (Ea) is the minimum extra energy that the reactant molecules must possess (above their average energy) for a collision to be effective and lead to a reaction. A higher Ea means a slower reaction.
k = A e^(-Ea/RT)
Marking-scheme points
- ✓k = A e^(-Ea/RT)
- ✓Ea = minimum extra energy needed for an effective collision
- ✓Higher Ea -> slower reaction
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