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General Chemistry flash cards

Master General Chemistry through 97 JEE Advanced-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.

General Chemistry, question and answer

30 of this chapter's 97 cards, laid out open so you can read straight through. The remaining 67 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.What is matter?

    Anything that has mass and occupies space (volume). It exists in three physical states: solid, liquid and gas.

    Hint: Mass + volume.

  2. 2.Distinguish the three physical states of matter by shape and volume.

    Solid: fixed shape and fixed volume. Liquid: no fixed shape (takes container shape) but fixed volume. Gas: neither fixed shape nor fixed volume.

    Hint: Which are 'fixed'?

  3. 3.On what basis is matter classified physically vs chemically?

    Physical classification: solid, liquid, gas. Chemical classification: pure substances (elements, compounds) and mixtures (homogeneous, heterogeneous).

    Hint: Two different classification schemes.

  4. 4.Define a pure substance.

    A substance with a fixed (constant) composition throughout and characteristic properties. Its constituents cannot be separated by physical methods. Examples: elements and compounds.

    Hint: Constant composition.

  5. 5.Difference between an element and a compound.

    Element: made of only one kind of atom (e.g. FeFe, OO). Compound: two or more elements chemically combined in a fixed mass ratio; components separable only by chemical methods (e.g. H2OH_2O).

    Hint: One kind of atom vs chemically combined.

  6. 6.Homogeneous vs heterogeneous mixture — key difference.

    Homogeneous: uniform composition throughout, one phase (e.g. salt solution, air). Heterogeneous: non-uniform composition, more than one phase visible (e.g. sand + iron filings).

    Hint: Uniform or not?

  7. 7.How does a mixture differ from a compound?

    Mixture: variable composition, components keep their own properties, separable by physical means, no energy change on formation. Compound: fixed composition, new properties, separable only chemically, energy change on formation.

    Hint: Variable vs fixed composition.

  8. 8.State the Law of Conservation of Mass.

    In any physical or chemical change, mass can neither be created nor destroyed. Total mass of reactants equals total mass of products. (Lavoisier)

    Hint: Mass before = mass after.

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

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

    Hint: Same compound, same mass ratio.

  10. 10.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. (Dalton)

    Hint: Fixed mass of A, small whole-number masses of B.

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

    When gases react, they do so in volumes bearing a simple whole-number ratio to one another and to the volumes of gaseous products, measured at the same temperature and pressure.

    Hint: Volumes in simple whole-number ratios.

  12. 12.State Avogadro's Law.

    Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

    Hint: Equal V, equal T, equal P → equal number of molecules.

  13. 13.List the main postulates of Dalton's atomic theory.

    1) Matter consists of indivisible atoms. 2) Atoms of a given element are identical in mass and properties. 3) Compounds form when atoms combine in fixed whole-number ratios. 4) Atoms are neither created nor destroyed in a chemical reaction.

    Hint: Indivisible atoms, fixed ratios, conserved.

  14. 14.Which law explains why COCO and CO2CO_2 exist as distinct compounds?

    The Law of Multiple Proportions: for a fixed 12 g of carbon, oxygen masses are 16 g and 32 g, a simple 1:21:2 ratio.

    Hint: Same two elements, two compounds.

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

    One atomic mass unit is exactly 112\frac{1}{12} of the mass of one atom of carbon-12. 1u=1.66056×1024g1\,u = 1.66056\times10^{-24}\,g.

    Hint: 1/12 of a C-12 atom.

  16. 16.What is average atomic mass and why is it usually non-integer?

    It is the weighted average of the isotopic masses of an element, weighted by their fractional natural abundances. Non-integer because elements are mixtures of isotopes.

    Hint: Weighted by isotopic abundance.

  17. 17.Define molecular mass.

    The sum of the atomic masses of all atoms present in one molecule of a substance. It is the relative mass of a molecule compared to 112\frac{1}{12} of a C-12 atom.

    Hint: Add atomic masses of all atoms.

  18. 18.Define formula mass and when it is used instead of molecular mass.

    Formula mass is the sum of atomic masses of atoms in one formula unit. It is used for ionic compounds (e.g. NaClNaCl), which do not exist as discrete molecules.

    Hint: For ionic / non-molecular substances.

  19. 19.What is the mole (SI definition, amount of substance)?

    A mole is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12 g of carbon-12, i.e. 6.022×10236.022\times10^{23} entities.

    Hint: Amount containing Avogadro number of entities.

  20. 20.State the value and meaning of Avogadro's number NAN_A.

    NA=6.022×1023 mol1N_A = 6.022\times10^{23}\ \text{mol}^{-1}. It is the number of elementary entities (atoms, molecules, ions) in one mole of a substance.

    Hint: 6.022×10236.022\times10^{23} per mole.

  21. 21.Define molar mass.

    The mass of one mole of a substance, numerically equal to its atomic/molecular/formula mass expressed in gmol1g\,mol^{-1}.

    Hint: Mass of 1 mol, in g/mol.

  22. 22.Give the three key relations connecting moles (nn) to mass, number, and gas volume.

    n=given massmolar mass=number of entitiesNA=V at STP22.4 Ln = \dfrac{\text{given mass}}{\text{molar mass}} = \dfrac{\text{number of entities}}{N_A} = \dfrac{V\ \text{at STP}}{22.4\ \text{L}}.

    Hint: Mass, particle count, STP volume.

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

    22.4 L (22400 mL)22.4\ \text{L}\ (22400\ \text{mL}) per mole at STP (0 °C, 1 atm). At the newer STP (0 °C, 1 bar) it is 22.7 L22.7\ \text{L}.

    Hint: 22.422.4 L at 273 K, 1 atm.

  24. 24.Define gram atomic mass and gram molecular mass.

    Gram atomic mass = atomic mass expressed in grams = mass of 1 mole of atoms. Gram molecular mass = molecular mass in grams = mass of 1 mole of molecules.

    Hint: Atomic/molecular mass in grams = 1 mole.

  25. 25.What is percentage composition of an element in a compound?

    % 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}}\times100.

    Hint: Mass of element / molar mass × 100.

  26. 26.Define empirical formula.

    The formula giving the simplest whole-number ratio of the atoms of each element in a compound (e.g. CH2OCH_2O for glucose).

    Hint: Simplest whole-number ratio.

  27. 27.Define molecular formula and its relation to empirical formula.

    The molecular formula gives the actual number of atoms of each element in one molecule. 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: Actual atoms = n × empirical.

  28. 28.List 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 mole values by the smallest. 4) Round to nearest whole numbers (multiply if fractional) → subscripts.

    Hint: % → moles → ÷ smallest → whole numbers.

  29. 29.For empirical formula CH2OCH_2O (mass 30) and molecular mass 180, find the molecular formula.

    n=18030=6n = \dfrac{180}{30} = 6, so molecular formula =C6H12O6= C_6H_{12}O_6 (glucose).

    Hint: n = molecular mass / empirical mass.

  30. 30.What is the mole ratio in a balanced chemical equation used for?

    The stoichiometric coefficients give the mole ratio in which reactants combine and products form; this ratio is the basis for all stoichiometric mass/volume/particle calculations.

    Hint: Coefficients = mole ratio.

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