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Mole Concept flash cards

Master Mole Concept through 92 NEET-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Mole Concept, question and answer

30 of this chapter's 92 cards, laid out open so you can read straight through. The remaining 62 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.State the Law of Conservation of Mass.

    Mass can neither be created nor destroyed in a chemical reaction. The total mass of reactants equals the total mass of products in a closed system.

    Hint: Given by Lavoisier; nothing is lost or gained.

  2. 2.State the Law of Definite (Constant) Proportions.

    A given chemical compound always contains the same elements combined in a fixed proportion by mass, regardless of its source or method of preparation.

    Hint: e.g. H2O\text{H}_2\text{O} is always 1:81:8 H:O by mass.

  3. 3.State the Law of Multiple Proportions.

    When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers.

    Hint: Dalton; e.g. CO and CO2\text{CO}_2, O masses in 1:21:2.

  4. 4.State Gay-Lussac's Law of Gaseous Volumes.

    When gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the products, provided all volumes are at the same temperature and pressure.

    Hint: Ratios of gas volumes are simple integers.

  5. 5.State Avogadro's Law.

    Equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.

    Hint: Volume \propto number of molecules at fixed T, P.

  6. 6.What is the value of Avogadro's number (NAN_A)?

    NA=6.022×1023N_A = 6.022 \times 10^{23} per mole. It is the number of elementary entities (atoms, molecules, ions) in one mole of a substance.

    Hint: 6.022×1023 mol16.022 \times 10^{23}\ \text{mol}^{-1}.

  7. 7.Define one mole of a substance.

    One mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 g12\ \text{g} of carbon-12, i.e. 6.022×10236.022 \times 10^{23} entities.

    Hint: Amount containing NAN_A particles.

  8. 8.What is molar mass?

    The mass of one mole of a substance, expressed in g mol1\text{g mol}^{-1}. It is numerically equal to the atomic/molecular mass in unified atomic mass units (u).

    Hint: Mass of NAN_A particles in grams.

  9. 9.Define atomic mass unit (u / amu).

    One atomic mass unit is exactly 112\frac{1}{12} of the mass of one carbon-12 atom. 1 u=1.66×1024 g1\ \text{u} = 1.66 \times 10^{-24}\ \text{g}.

    Hint: Reference is 12C^{12}\text{C}.

  10. 10.How is average atomic mass calculated?

    It is the weighted average of the masses of all naturally occurring isotopes: Aˉ=(isotope mass×fractional abundance)\bar{A} = \sum (\text{isotope mass} \times \text{fractional abundance}).

    Hint: Weight each isotope by its abundance.

  11. 11.What is the relation between number of moles nn, mass mm, and molar mass MM?

    n=mMn = \dfrac{m}{M}, where mm is mass in grams and MM is molar mass in g mol1\text{g mol}^{-1}.

    Hint: Moles = given mass / molar mass.

  12. 12.How do you find the number of particles from moles?

    Number of particles =n×NA=mM×6.022×1023= n \times N_A = \dfrac{m}{M} \times 6.022 \times 10^{23}.

    Hint: Multiply moles by Avogadro's number.

  13. 13.What is the molar volume of an ideal gas at STP?

    22.4 L22.4\ \text{L} (or 22400 mL22400\ \text{mL}) per mole at STP (273.15 K273.15\ \text{K}, 1 atm1\ \text{atm}). Modern STP (105 Pa10^5\ \text{Pa}) gives 22.7 L22.7\ \text{L}.

    Hint: 22.4 L mol122.4\ \text{L mol}^{-1} at 0C0^\circ\text{C}, 1 atm1\ \text{atm}.

  14. 14.Define gram atomic mass and gram molecular mass.

    Gram atomic mass is the mass of one mole of atoms (atomic mass in grams); gram molecular mass is the mass of one mole of molecules (molecular mass in grams).

    Hint: Atomic/molecular mass expressed in grams.

  15. 15.What is the difference between atomic mass and molar mass numerically?

    They are numerically equal but have different units: atomic/molecular mass is in u, while molar mass is in g mol1\text{g mol}^{-1}.

    Hint: Same number, different units.

  16. 16.Define empirical formula.

    The empirical formula gives the simplest whole-number ratio of atoms of each element present in a compound.

    Hint: Simplest ratio, e.g. CH2O\text{CH}_2\text{O}.

  17. 17.Define molecular formula.

    The molecular formula gives the actual number of atoms of each element in one molecule of the compound.

    Hint: Actual count, e.g. glucose C6H12O6\text{C}_6\text{H}_{12}\text{O}_6.

  18. 18.What is the relation between molecular and empirical formula?

    Molecular formula=n×Empirical formula\text{Molecular formula} = n \times \text{Empirical formula}, where n=Molecular massEmpirical formula massn = \dfrac{\text{Molecular mass}}{\text{Empirical formula mass}}.

    Hint: nn is a whole number multiplier.

  19. 19.Outline the steps to determine an empirical formula from percentage composition.

    1) Take % as grams. 2) Divide each by its atomic mass to get moles. 3) Divide all by the smallest mole value. 4) Convert to nearest whole numbers (multiply if needed).

    Hint: % → moles → simplest ratio.

  20. 20.How is percentage composition of an element in a compound calculated?

    % element=mass of element in 1 molmolar mass of compound×100\%\ \text{element} = \dfrac{\text{mass of element in 1 mol}}{\text{molar mass of compound}} \times 100.

    Hint: Element mass / total molar mass ×100\times 100.

  21. 21.What is stoichiometry?

    Stoichiometry is the quantitative study of the relationships between amounts (moles, mass, volume) of reactants and products in a balanced chemical reaction.

    Hint: Quantities from balanced equations.

  22. 22.What do the coefficients in a balanced equation represent?

    They give the relative number of moles (and for gases, relative volumes) of reactants and products taking part in the reaction — the mole ratio.

    Hint: Mole ratio of species.

  23. 23.Define limiting reagent.

    The limiting reagent is the reactant that is completely consumed first and therefore limits (determines) the amount of product formed.

    Hint: Runs out first; controls yield.

  24. 24.How do you identify the limiting reagent?

    Divide the moles of each reactant by its stoichiometric coefficient; the reactant giving the smallest quotient is the limiting reagent.

    Hint: Smallest (moles ÷ coefficient) wins.

  25. 25.What is the excess reagent?

    The reactant that remains partly unreacted after the limiting reagent is fully consumed; it is present in more than the stoichiometric amount.

    Hint: Leftover reactant.

  26. 26.Distinguish theoretical yield and actual yield.

    Theoretical yield is the maximum product calculated from stoichiometry; actual yield is the amount actually obtained in the experiment (usually less).

    Hint: Calculated max vs. real obtained.

  27. 27.How is percentage yield calculated?

    % yield=actual yieldtheoretical yield×100\%\ \text{yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100.

    Hint: Actual / theoretical ×100\times 100.

  28. 28.Define molarity (M).

    Molarity is the number of moles of solute dissolved per litre of solution: M=moles of solutevolume of solution (L)M = \dfrac{\text{moles of solute}}{\text{volume of solution (L)}}. Units: mol L1\text{mol L}^{-1}.

    Hint: Moles solute per litre solution.

  29. 29.Define molality (m).

    Molality is the number of moles of solute per kilogram of solvent: m=moles of solutemass of solvent (kg)m = \dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}}. Units: mol kg1\text{mol kg}^{-1}.

    Hint: Moles solute per kg solvent.

  30. 30.Why is molality preferred over molarity for studying temperature-dependent properties?

    Molality depends only on mass, which does not change with temperature, whereas molarity uses volume, which changes with temperature.

    Hint: Mass is temperature-independent.

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