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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
Calculate the number of atoms present per unit cell in simple cubic, body-centred cubic (bcc) and face-centred cubic (fcc) structures.
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A corner atom is shared by 8 unit cells (contributes 1/8), a face atom by 2 (contributes 1/2) and a body-centre atom belongs entirely to one cell (contributes 1). Simple cubic: 8 corners x 1/8 = 1 atom. Body-centred cubic (bcc): (8 x 1/8) + 1 (centre) = 2 atoms. Face-centred cubic (fcc): (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 atoms.
Marking-scheme points
- ✓Corner atom contributes 1/8, face atom 1/2, body-centre 1
- ✓Simple cubic: 1 atom; bcc: 2 atoms
- ✓fcc: 1 + 3 = 4 atoms
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The Solid State
An element with atomic mass 56 g/mol has a bcc structure with edge length 288 pm. Calculate its density. (NA = 6.022 x 10^23)
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For bcc, Z = 2. Edge a = 288 pm = 2.88 x 10^-8 cm, so a^3 = (2.88 x 10^-8)^3 = 2.39 x 10^-23 cm^3. Density d = Z M/(a^3 NA) = (2 x 56)/(2.39 x 10^-23 x 6.022 x 10^23) = 112/14.39 = 7.79 g/cm^3.
d = Z M/(a^3 NA)
Marking-scheme points
- ✓d = Z M/(a^3 NA); bcc Z = 2
- ✓a^3 = (2.88e-8)^3 = 2.39 x 10^-23 cm^3
- ✓d = 112/14.39 = 7.79 g/cm^3
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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 are colligative properties? Name the four colligative properties.
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Colligative properties are those properties of a solution that depend only on the number of solute particles present and not on their chemical nature. The four colligative properties are: (1) relative lowering of vapour pressure; (2) elevation of boiling point; (3) depression of freezing point; and (4) osmotic pressure. They are used to determine the molar mass of a solute.
Marking-scheme points
- ✓Depend only on the number of solute particles, not their nature
- ✓Relative lowering of vapour pressure and elevation of boiling point
- ✓Depression of freezing point and osmotic pressure
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Solutions
When 3 g of a non-volatile solute is dissolved in 100 g of water, the freezing point is depressed by 0.93 K. Calculate the molar mass of the solute. (Kf for water = 1.86 K kg/mol)
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The depression of freezing point is given by delta Tf = Kf x molality = Kf x (w2 x 1000)/(M2 x w1), where w2 = mass of solute, w1 = mass of solvent, M2 = molar mass of solute. So 0.93 = 1.86 x (3 x 1000)/(M2 x 100) = 55.8/M2. Therefore M2 = 55.8/0.93 = 60 g/mol.
delta Tf = Kf x (w2 x 1000)/(M2 x w1)
Marking-scheme points
- ✓delta Tf = Kf x (w2 x 1000)/(M2 x w1)
- ✓0.93 = 1.86 x (3 x 1000)/(M2 x 100) = 55.8/M2
- ✓M2 = 60 g/mol
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Solutions
Calculate the osmotic pressure of a 0.1 M glucose solution at 300 K. (R = 0.0821 L atm K^-1 mol^-1)
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Osmotic pressure is given by pi = C R T, where C is the molar concentration. Here C = 0.1 mol/L, R = 0.0821 L atm K^-1 mol^-1, T = 300 K. So pi = 0.1 x 0.0821 x 300 = 2.463 atm. Osmotic pressure is a colligative property used to find the molar mass of macromolecules.
pi = C R T
Marking-scheme points
- ✓pi = C R T
- ✓= 0.1 x 0.0821 x 300
- ✓pi = 2.463 atm
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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
Calculate the standard EMF of a Daniell cell given the standard electrode potentials E(Cu2+/Cu) = +0.34 V and E(Zn2+/Zn) = -0.76 V.
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In the Daniell cell, zinc (lower/more negative potential) is the anode and copper is the cathode. The standard EMF = E(cathode) - E(anode) = E(Cu2+/Cu) - E(Zn2+/Zn) = (+0.34) - (-0.76) = 0.34 + 0.76 = 1.10 V. Since the EMF is positive, the cell reaction is spontaneous.
E(cell) = E(cathode) - E(anode)
Marking-scheme points
- ✓Zn = anode (more negative), Cu = cathode
- ✓EMF = E(cathode) - E(anode)
- ✓= 0.34 - (-0.76) = 1.10 V (spontaneous)
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Electrochemistry
Write the Nernst equation for a general electrode and explain the terms.
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The Nernst equation gives the electrode (or cell) potential under non-standard conditions. For the reaction M(n+) + n e- -> M, the electrode potential E = E(standard) - (2.303 RT/nF) log(1/[M(n+)]). At 298 K, substituting the constants, E = E(standard) - (0.0591/n) log(1/[M(n+)]), where E(standard) is the standard electrode potential, n is the number of electrons transferred, F is the Faraday constant, and the term in brackets is the reaction quotient. It shows how potential varies with concentration.
E = E(standard) - (0.0591/n) log Q
Marking-scheme points
- ✓Gives potential under non-standard conditions
- ✓E = E(standard) - (0.0591/n) log Q at 298 K
- ✓n = electrons transferred; depends on ion concentration
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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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