Some Basic Concepts of Chemistry MCQs

In a chemical reaction atoms are only rearranged; they are not created or annihilated.

The total mass of all reactants therefore equals the total mass of all products.

Thus the law of conservation of mass states that mass is neither created nor destroyed.

One mole contains Avogadro’s number of entities, \(N_A = 6.022\times10^{23}\).

For 2 moles, multiply: \(2 \times 6.022\times10^{23}=1.2044\times10^{24}\).

Thus the number of CO₂ molecules in 2 moles is \(1.2044\times10^{24}\).

Add atomic masses: N = 14, H = 1, S = 32, O = 16.

Molar mass = 2(N + 4H) + S + 4O = 2(14 + 4×1) + 32 + 4×16.

Calculate: 2(18) = 36; 4×16 = 64; sum = 36 + 32 + 64 = 132 g mol⁻¹.

Thus the correct answer is 132 g/mol.

Mole ratio C:H:O = 1:2:1 means one carbon atom, two hydrogen atoms and one oxygen atom combine.

The simplest whole‑number ratio directly gives the empirical formula.

Write the symbols with their respective subscripts: C₁H₂O₁ → CH₂O.

Thus the empirical formula is CH2O.

The balanced equation shows 1 mol N₂ reacts with 3 mol H₂.

For 3 mol N₂ the reaction would need 3 × 3 = 9 mol H₂, but only 3 mol H₂ are present.

Since the available H₂ is far less than required, H₂ is exhausted first.

Thus H₂ is the limiting reagent.

One mole of any substance contains Avogadro’s number of entities, NA = 6.022 × 10²³.

Carbon‑12 is defined so that 12 g of it corresponds exactly to one mole.

Therefore 12 g of carbon‑12 contains NA atoms, which is 6.022 × 10²³.

Mass = moles × molar mass

0.2 mol × 40 g mol⁻¹ = 8 g

Thus the mass of 0.2 mole of NaOH is 8 g. 8 g.

Empirical formula mass = 12 + 2×1 = 14 g mol⁻¹.

Molar mass / empirical mass = 42 / 14 = 3, so the molecular formula contains three empirical units.

Multiply each subscript in CH₂ by 3 → C₃H₆.

Thus the molecular formula is C3H6.

Molar mass of CaCO₃ = Ca 40 + C 12 + 3·O 16 = 100 g mol⁻¹, so 100 g of CaCO₃ corresponds to 1 mol.

The decomposition reaction is 1 mol CaCO₃ → 1 mol CaO + 1 mol CO₂, so 1 mol of CaO is produced.

Molar mass of CaO = Ca 40 + O 16 = 56 g mol⁻¹, therefore the mass of CaO obtained is 56 g.

56 g.

Mole fraction of a component = moles of that component / total moles of all components.

Summing over all components gives Σ (ni / Σ ni) = (Σ ni) / (Σ ni).

The ratio simplifies to 1, independent of the number of species.

Hence the sum of mole fractions of all components in a solution is 1.

Carbon combines with a fixed mass of carbon, first with 1 g O (forming CO) and then with 2 g O (forming CO₂).

The masses of oxygen that combine with the same mass of carbon are in a simple whole‑number ratio (1:2).

This is exactly the statement of the law that when elements 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.

Hence the law illustrated is the Law of multiple proportions.

At STP (0 °C, 1 atm) one mole of an ideal gas occupies 22.4 L.

Divide the given volume by the molar volume: 11.2 L ÷ 22.4 L = 0.5.

Thus 11.2 L of any ideal gas corresponds to 0.5 mol. 0.5 mol.

Molar mass of O₂ = 2 × 16 g mol⁻¹ = 32 g mol⁻¹.

Number of moles = mass / molar mass = 8 g ÷ 32 g mol⁻¹.

8 ÷ 32 = 0.25.

Thus the amount of O₂ is 0.25 mol.

Find the simplest whole‑number ratio of C to H in C₆H₆.

Divide both subscripts by the greatest common divisor, which is 6.

C₆ ÷ 6 = 1 and H₆ ÷ 6 = 1, giving the ratio C₁H₁ → CH.

Hence the empirical formula is CH.

The balanced equation shows 2 mol Mg react with 1 mol O₂, so the mole ratio Mg : O₂ is 2 : 1.

For 0.5 mol Mg, divide by 2 to get the required O₂: 0.5 mol ÷ 2 = 0.25 mol.

Thus 0.25 mol of O₂ is needed, matching the given answer.

The law of definite proportions says that in a pure compound the masses of the constituent elements are always in the same proportion, regardless of the sample size.

Thus, if you take any amount of the compound, the mass of element A to the mass of element B remains constant.

This constant relationship is expressed as a fixed ratio by mass, which is the statement of the law.

Hence the correct answer is Ratio by mass.

NH₃ contains 1 N atom and 3 H atoms, total 4 atoms per molecule.

One mole of any substance contains Avogadro’s number (6.022 × 10²³) of entities.

Thus 1 mole of NH₃ has 4 × 6.022 × 10²³ atoms.

4 × 6.022 × 10²³.

One mole of any element contains Avogadro’s number of atoms and its mass in grams equals the element’s atomic mass (relative atomic mass).

The relative atomic mass of sodium (Na) is 23 (from the periodic table).

Thus 1 mol of Na atoms has a mass of 23 g.

23 g.

Empirical formula HO has a molar mass of (1 + 16)=17 g mol⁻¹.

Given molecular mass = 34 g mol⁻¹, the molecular mass is 34 ÷ 17 = 2 times the empirical mass.

Therefore the molecular formula contains twice the atoms of the empirical unit: (HO)₂ → H₂O₂.

Hence the correct answer is H2O2.

Moles of water formed = 36 g ÷ 18 g mol⁻¹ = 2 mol.

From the balanced equation, 2 mol H₂O require 1 mol O₂, so O₂ needed = 2 mol ÷ 2 = 1 mol.

At STP 1 mol gas occupies 22.4 L, therefore volume of O₂ required = 22.4 L.

22.4 L.