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ChemistryClass 112 markseasy

Some Basic Concepts of Chemistry

State the law of conservation of mass and the law of definite proportions.

Reveal model answer + marking points

Law of conservation of mass: matter can neither be created nor destroyed in a chemical reaction; the total mass of reactants equals the total mass of products. Law of definite proportions: a given chemical compound always contains the same elements combined in the same fixed proportion by mass, irrespective of its source or method of preparation.

Marking-scheme points

  • Conservation of mass: mass of reactants = mass of products
  • Definite proportions: fixed ratio of elements by mass
  • Example: water always 1:8 H:O by mass
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ChemistryClass 112 markseasy

Some Basic Concepts of Chemistry

Calculate the number of moles present in 11 g of carbon dioxide (CO2).

Reveal model answer + marking points

Molar mass of CO2 = 12 + 2(16) = 44 g/mol. Number of moles = given mass / molar mass = 11 / 44 = 0.25 mol.

n = m / M

Marking-scheme points

  • Molar mass of CO2 = 44 g/mol
  • moles = mass / molar mass
  • n = 11/44 = 0.25 mol
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ChemistryClass 113 marksmedium

Some Basic Concepts of Chemistry

How many molecules and how many atoms are present in 0.25 mol of CO2? (Avogadro number = 6.022 x 10^23)

Reveal model answer + marking points

Number of molecules = 0.25 x 6.022 x 10^23 = 1.506 x 10^23 molecules. Each CO2 molecule has 3 atoms (1 C + 2 O), so number of atoms = 3 x 1.506 x 10^23 = 4.52 x 10^23 atoms.

N = n x N_A

Marking-scheme points

  • molecules = n x N_A = 0.25 x 6.022e23 = 1.506e23
  • Atoms per CO2 molecule = 3
  • atoms = 3 x 1.506e23 = 4.52e23
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ChemistryClass 112 markseasy

Some Basic Concepts of Chemistry

Define one mole and state the value of Avogadro's number.

Reveal model answer + marking points

One mole is the amount of a substance that contains as many elementary entities (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12. This number of entities is Avogadro's number, N_A = 6.022 x 10^23 mol^-1.

1 mol = 6.022 x 10^23 particles

Marking-scheme points

  • Mole = amount containing N_A entities
  • Reference: atoms in 12 g of C-12
  • N_A = 6.022 x 10^23 per mole
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ChemistryClass 113 markshard

Some Basic Concepts of Chemistry

A compound contains 24.27% carbon, 4.07% hydrogen and 71.65% chlorine by mass. Its molar mass is 98.96 g/mol. Find its empirical and molecular formula.

Reveal model answer + marking points

Divide each percentage by atomic mass: C = 24.27/12 = 2.02, H = 4.07/1 = 4.07, Cl = 71.65/35.5 = 2.02. Divide by the smallest (2.02): C = 1, H = 2, Cl = 1. Empirical formula = CH2Cl (empirical mass = 12 + 2 + 35.5 = 49.5). n = molar mass / empirical mass = 98.96/49.5 = 2. Molecular formula = C2H4Cl2.

n = molar mass / empirical formula mass

Marking-scheme points

  • Moles of atoms: C 2.02, H 4.07, Cl 2.02
  • Simplest ratio 1 : 2 : 1 -> empirical CH2Cl (mass 49.5)
  • n = 98.96/49.5 = 2 -> molecular C2H4Cl2
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ChemistryClass 112 markseasy

Some Basic Concepts of Chemistry

Define molarity and molality. State their units.

Reveal model answer + marking points

Molarity (M) is the number of moles of solute dissolved per litre of solution; unit mol/L (or M). Molality (m) is the number of moles of solute dissolved per kilogram of solvent; unit mol/kg (or m). Molality is independent of temperature since it uses mass, whereas molarity changes with temperature because volume changes.

M = n_solute / V(L); m = n_solute / mass_solvent(kg)

Marking-scheme points

  • Molarity = moles of solute / litre of solution (mol/L)
  • Molality = moles of solute / kg of solvent (mol/kg)
  • Molality is temperature independent
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ChemistryClass 113 marksmedium

Some Basic Concepts of Chemistry

Calculate the molarity of a solution prepared by dissolving 5 g of NaOH in enough water to make 450 mL of solution.

Reveal model answer + marking points

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles of NaOH = 5/40 = 0.125 mol. Volume = 450 mL = 0.450 L. Molarity = 0.125 / 0.450 = 0.278 M (approximately 0.28 M).

M = n / V(L)

Marking-scheme points

  • Molar mass NaOH = 40 g/mol
  • moles = 5/40 = 0.125 mol
  • M = 0.125/0.450 = 0.28 M
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ChemistryClass 113 markshard

Some Basic Concepts of Chemistry

3.0 g of H2 reacts with 29 g of O2 to form water (2H2 + O2 -> 2H2O). Identify the limiting reagent and calculate the mass of water formed.

Reveal model answer + marking points

Moles of H2 = 3/2 = 1.5 mol; moles of O2 = 29/32 = 0.906 mol. From the equation, 2 mol H2 need 1 mol O2. For 1.5 mol H2, O2 required = 0.75 mol, but 0.906 mol O2 is available, so O2 is in excess and H2 is the limiting reagent. Water formed = moles of H2 (since 2H2 -> 2H2O) = 1.5 mol = 1.5 x 18 = 27 g.

mass = moles x molar mass

Marking-scheme points

  • moles H2 = 1.5, moles O2 = 0.906
  • H2 needs O2 in 2:1 ratio -> only 0.75 mol O2 needed
  • H2 is limiting; water = 1.5 mol = 27 g
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ChemistryClass 112 marksmedium

Some Basic Concepts of Chemistry

Define mole fraction and mass percent of a component in a solution.

Reveal model answer + marking points

Mole fraction of a component = number of moles of that component / total number of moles of all components in the solution; it is dimensionless and the sum of mole fractions equals 1. Mass percent of a component = (mass of the component / total mass of solution) x 100.

x_A = n_A / n_total; mass% = (mass component / total mass) x 100

Marking-scheme points

  • Mole fraction x_A = n_A / (n_A + n_B); sum = 1
  • Dimensionless quantity
  • Mass percent = (mass of component / total mass) x 100
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ChemistryClass 113 marksmedium

Structure of Atom

State the main postulates of Bohr's model of the hydrogen atom.

Reveal model answer + marking points

1) The electron revolves around the nucleus only in certain fixed circular orbits of definite energy called stationary states, without radiating energy. 2) Angular momentum of the electron is quantised: mvr = nh/2pi, where n = 1, 2, 3... 3) Energy is emitted or absorbed only when an electron jumps from one orbit to another, and the energy difference equals hv (delta E = E2 - E1 = h v).

mvr = nh/2pi ; delta E = h v

Marking-scheme points

  • Fixed stationary orbits with no energy loss
  • Quantised angular momentum mvr = nh/2pi
  • Energy change on jump: delta E = h v
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ChemistryClass 112 marksmedium

Structure of Atom

Calculate the energy of a photon of light of wavelength 4000 Angstrom. (h = 6.626 x 10^-34 J s, c = 3 x 10^8 m/s)

Reveal model answer + marking points

Wavelength = 4000 Angstrom = 4000 x 10^-10 m = 4 x 10^-7 m. Energy E = hc/lambda = (6.626 x 10^-34 x 3 x 10^8) / (4 x 10^-7) = 1.988 x 10^-25 / 4 x 10^-7 = 4.97 x 10^-19 J.

E = hc / lambda

Marking-scheme points

  • Convert 4000 Angstrom = 4 x 10^-7 m
  • E = hc/lambda
  • E = 4.97 x 10^-19 J
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ChemistryClass 113 marksmedium

Structure of Atom

The radius of the first Bohr orbit of hydrogen is 0.529 Angstrom. Calculate the radius of the third orbit.

Reveal model answer + marking points

For a hydrogen atom, r_n = n^2 x r_1. For n = 3: r_3 = 3^2 x 0.529 = 9 x 0.529 = 4.761 Angstrom.

r_n = n^2 x r_1

Marking-scheme points

  • r_n is proportional to n^2 for hydrogen
  • r_3 = 9 x r_1
  • r_3 = 9 x 0.529 = 4.761 Angstrom
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ChemistryClass 112 markseasy

Structure of Atom

State Heisenberg's uncertainty principle and give its mathematical form.

Reveal model answer + marking points

It is impossible to determine simultaneously and with absolute accuracy both the position and the momentum (or velocity) of a microscopic particle such as an electron. The product of the uncertainties in position (delta x) and momentum (delta p) is at least of the order of h/4pi.

delta x . delta p >= h / 4pi

Marking-scheme points

  • Cannot measure position and momentum exactly at once
  • Applies to microscopic particles like electrons
  • delta x . delta p >= h/4pi
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ChemistryClass 113 marksmedium

Structure of Atom

Calculate the de Broglie wavelength of an electron moving with a velocity of 2.05 x 10^7 m/s. (mass of electron = 9.1 x 10^-31 kg, h = 6.626 x 10^-34 J s)

Reveal model answer + marking points

de Broglie wavelength lambda = h / (m v) = (6.626 x 10^-34) / (9.1 x 10^-31 x 2.05 x 10^7). Denominator = 1.866 x 10^-23. lambda = 3.55 x 10^-11 m = 0.355 Angstrom.

lambda = h / (m v)

Marking-scheme points

  • lambda = h / mv
  • mv = 9.1e-31 x 2.05e7 = 1.866e-23
  • lambda = 3.55 x 10^-11 m
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ChemistryClass 113 marksmedium

Structure of Atom

What are the four quantum numbers? State what each one describes.

Reveal model answer + marking points

1) Principal quantum number (n): gives the main energy level/shell and size of the orbital (n = 1, 2, 3...). 2) Azimuthal (angular momentum) quantum number (l): gives the subshell and shape of the orbital (l = 0 to n-1, i.e. s, p, d, f). 3) Magnetic quantum number (m_l): gives the orientation of the orbital in space (m_l = -l to +l). 4) Spin quantum number (m_s): gives the direction of electron spin (+1/2 or -1/2).

l = 0 to (n-1); m_l = -l ... +l

Marking-scheme points

  • n: shell / size and energy
  • l: subshell / shape (0 to n-1)
  • m_l: orientation (-l to +l); m_s: spin (+/-1/2)
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ChemistryClass 113 marksmedium

Structure of Atom

Write the electronic configurations of chromium (Z = 24) and copper (Z = 29). Explain why they are exceptions to the expected order.

Reveal model answer + marking points

Cr (Z = 24): [Ar] 3d5 4s1 (not 3d4 4s2). Cu (Z = 29): [Ar] 3d10 4s1 (not 3d9 4s2). One 4s electron shifts to 3d because exactly half-filled (3d5) and completely filled (3d10) subshells have extra stability due to symmetrical distribution of electrons and greater exchange energy.

Marking-scheme points

  • Cr = [Ar] 3d5 4s1; Cu = [Ar] 3d10 4s1
  • Half-filled and fully-filled d subshells are extra stable
  • Cause: symmetry + maximum exchange energy
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ChemistryClass 112 marksmedium

Structure of Atom

Calculate the wave number of the spectral line when an electron in a hydrogen atom jumps from n = 3 to n = 2. (Rydberg constant R = 1.097 x 10^7 m^-1)

Reveal model answer + marking points

Wave number (nu bar) = R (1/n1^2 - 1/n2^2) = 1.097 x 10^7 (1/2^2 - 1/3^2) = 1.097 x 10^7 (1/4 - 1/9) = 1.097 x 10^7 x (5/36) = 1.523 x 10^6 m^-1. This is the H-alpha line of the Balmer series.

nu bar = R (1/n1^2 - 1/n2^2)

Marking-scheme points

  • nu bar = R(1/n1^2 - 1/n2^2), n1=2, n2=3
  • 1/4 - 1/9 = 5/36
  • nu bar = 1.523 x 10^6 m^-1 (Balmer series)
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ChemistryClass 112 markseasy

Structure of Atom

State Pauli's exclusion principle and Hund's rule of maximum multiplicity.

Reveal model answer + marking points

Pauli's exclusion principle: no two electrons in an atom can have the same set of all four quantum numbers; an orbital can hold at most two electrons with opposite spins. Hund's rule of maximum multiplicity: electron pairing in orbitals of the same subshell (degenerate orbitals) does not occur until each orbital is singly occupied, and all singly filled orbitals have parallel spin.

Marking-scheme points

  • Pauli: no two electrons share all four quantum numbers
  • Max 2 electrons per orbital, opposite spins
  • Hund: singly fill degenerate orbitals first, parallel spins
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ChemistryClass 112 markseasy

Structure of Atom

State the Aufbau principle and the (n + l) rule for filling of orbitals.

Reveal model answer + marking points

Aufbau principle: in the ground state of an atom, electrons are filled into orbitals in order of increasing energy, i.e. the lowest energy orbital is filled first. (n + l) rule: the orbital with the lower (n + l) value has lower energy and is filled first; if two orbitals have the same (n + l) value, the one with the lower n is filled first (e.g. 4s (n+l=4) is filled before 3d (n+l=5)).

energy order by increasing (n + l)

Marking-scheme points

  • Fill lowest energy orbitals first
  • Lower (n + l) = lower energy = filled first
  • Equal (n+l): lower n filled first (4s before 3d)
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ChemistryClass 112 markseasy

Classification of Elements and Periodicity

State the modern periodic law. How does it differ from Mendeleev's periodic law?

Reveal model answer + marking points

Modern periodic law: the physical and chemical properties of elements are a periodic function of their atomic numbers. Mendeleev's law was based on atomic mass, whereas the modern law is based on atomic number (number of protons), which removed anomalies such as the position of argon and potassium.

Marking-scheme points

  • Properties are periodic function of atomic number
  • Mendeleev: based on atomic mass
  • Atomic number basis removes mass-order anomalies
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