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ChemistryClass 122 markseasy

Solutions

Define molarity, molality and mole fraction.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Solutions

Calculate the molality of a solution containing 18 g of glucose (molar mass 180 g/mol) dissolved in 500 g of water.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Solutions

State Henry's law. Give one application.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Solutions

State Raoult's law for a solution of two volatile liquids.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Solutions

What is the Van't Hoff factor? What does its value indicate for association and dissociation?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Solutions

Distinguish between ideal and non-ideal solutions.

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Electrochemistry

What is a galvanic (voltaic) cell? Name its two electrodes and the reaction at each.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Electrochemistry

What is standard electrode potential? What is taken as the reference electrode?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Electrochemistry

Write the relation between the standard Gibbs energy change and the EMF of a cell.

Reveal model answer + marking points

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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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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ChemistryClass 122 markseasy

Electrochemistry

Distinguish between primary and secondary cells with one example each.

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Chemical Kinetics

Define the rate of a chemical reaction. Distinguish between average and instantaneous rate.

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Chemical Kinetics

State the factors that affect the rate of a chemical reaction.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

The rate constant of a first order reaction is 6.93 x 10^-3 s^-1. Calculate its half-life.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

Write the Arrhenius equation and define activation energy.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

How does a catalyst increase the rate of a reaction?

Reveal model answer + marking points

A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. As a result, a larger fraction of the reactant molecules have enough energy to cross the (lowered) energy barrier, so more effective collisions occur and the reaction proceeds faster. The catalyst does not change the enthalpy or the equilibrium position of the reaction, and it is regenerated at the end of the reaction.

Marking-scheme points

  • Provides an alternative path of lower activation energy
  • More molecules can cross the lower energy barrier
  • Does not change the equilibrium; is regenerated
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ChemistryClass 122 marksmedium

Chemical Kinetics

Write the units of the rate constant for a zero order and a first order reaction.

Reveal model answer + marking points

The units of the rate constant depend on the order of the reaction. For a zero order reaction, the rate = k, so the units of k are the same as the rate: mol L^-1 s^-1 (or mol L^-1 time^-1). For a first order reaction, rate = k[R], so k has units of s^-1 (or time^-1), which are independent of concentration. In general, the units of k are (mol L^-1)^(1-n) time^-1 for an nth order reaction.

units of k = (mol L^-1)^(1-n) time^-1

Marking-scheme points

  • Zero order: units of k are mol L^-1 s^-1
  • First order: units of k are s^-1 (time^-1)
  • General: (mol L^-1)^(1-n) time^-1 for order n
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ChemistryClass 122 marksmedium

The p-Block Elements

Name the elements of Group 15 and give their general valence shell electronic configuration and common oxidation states.

Reveal model answer + marking points

The Group 15 elements (the nitrogen family) are nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi). Their general valence shell electronic configuration is ns2 np3 (a half-filled p subshell, which gives extra stability). Their common oxidation states are -3, +3 and +5; the stability of the +5 state decreases and that of the +3 state increases down the group due to the inert pair effect.

ns2 np3

Marking-scheme points

  • Group 15: N, P, As, Sb, Bi
  • General configuration ns2 np3 (half-filled p)
  • Oxidation states -3, +3, +5; +3 more stable down the group
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ChemistryClass 122 marksmedium

The p-Block Elements

Why does ammonia act as a Lewis base and have a higher boiling point than phosphine (PH3)?

Reveal model answer + marking points

Ammonia (NH3) has a lone pair of electrons on the nitrogen atom which it can donate to an electron-deficient species, so it acts as a Lewis base. It has a higher boiling point than phosphine because nitrogen is small and highly electronegative, so NH3 molecules form strong intermolecular hydrogen bonds, whereas PH3 molecules are held only by weak van der Waals forces; more energy is needed to separate the hydrogen-bonded NH3 molecules.

Marking-scheme points

  • NH3 has a lone pair on N -> donates it -> Lewis base
  • N is small and electronegative -> NH3 forms hydrogen bonds
  • PH3 has only weak van der Waals forces -> lower boiling point
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