Mole Concept & Molar Masses — IISER Chemistry MCQs with Solutions
Free IISER Chemistry Mole Concept & Molar Masses MCQs with step-by-step solutions (7 questions). Part of Some Basic Concepts of Chemistry. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Mole Concept & Molar Masses · medium · theory
Assertion (A): One mole of any substance always contains 6.022 × 10²³ elementary entities, whatever the substance may be.
Reason (R): The mole is defined as the amount of substance that contains exactly 6.02214076 × 10²³ elementary entities, and an elementary entity may be an atom, molecule, ion or electron.
Choose the correct option.
A. A is true but R is false
B. Both A and R are true and R is the correct explanation of A ✓ Correct
C. Both A and R are true but R is NOT the correct explanation of A
D. A is false but R is true
Solution: By its SI definition, one mole contains a fixed number of entities — the Avogadro number, 6.02214076 × 10²³ ≈ 6.022 × 10²³ — regardless of whether those entities are atoms, molecules, ions or electrons. R states exactly this definition, so it is the correct explanation of A.
Q2 — Mole Concept & Molar Masses · hard · theory
Assertion (A): 16 g of methane (CH₄) and 16 g of oxygen gas (O₂) contain the same number of molecules.
Reason (R): Equal masses of any two gases always contain equal numbers of molecules.
Choose the correct option. (C = 12, H = 1, O = 16)
A. A is true but R is false
B. Both A and R are false ✓ Correct
C. A is false but R is true
D. Both A and R are true and R is the correct explanation of A
Solution: CH₄ has molar mass 16 g mol⁻¹, so 16 g = 1 mol = 6.022 × 10²³ molecules; O₂ has molar mass 32 g mol⁻¹, so 16 g = 0.5 mol = 3.011 × 10²³ molecules. The counts differ, so A is false. Equal masses give equal molecule counts only when the molar masses are equal, so R is false as well.
Q3 — Mole Concept & Molar Masses · medium · numerical
The number of moles of oxygen atoms present in 0.2 mol of sulphuric acid (H₂SO₄) is:
A. 0.2 mol
B. 4 mol
C. 0.05 mol
D. 0.8 mol ✓ Correct
Solution: Each molecule of H₂SO₄ contains 4 oxygen atoms, so moles of O atoms = 4 × 0.2 = 0.8 mol. Reporting the moles of the compound itself gives 0.2 mol; dividing by 4 instead of multiplying gives 0.05 mol; quoting the atoms-per-molecule alone gives 4 mol.
Q4 — Mole Concept & Molar Masses · hard · numerical
Which of the following samples contains the largest number of atoms? (H = 1, C = 12, N = 14, O = 16)
A. 1 g of CH₄
B. 1 g of O₂
C. 1 g of N₂
D. 1 g of H₂ ✓ Correct
Solution: Number of atoms = (mass ÷ molar mass) × (atoms per molecule) × Nₐ. H₂: (1 ÷ 2) × 2 = 1 mol of atoms. CH₄: (1 ÷ 16) × 5 = 0.3125 mol. O₂: (1 ÷ 32) × 2 = 0.0625 mol. N₂: (1 ÷ 28) × 2 = 0.0714 mol. The 1 g of H₂ gives the most atoms (a full 1 mol = 6.022 × 10²³) because hydrogen has the smallest molar mass.
Q5 — Mole Concept & Molar Masses · hard · numerical
The mass of a single atom of an element whose molar mass is 40 g mol⁻¹ (for example calcium) is approximately: (Nₐ = 6.022 × 10²³)
A. 1.51 × 10²² g
B. 6.64 × 10⁻²³ g ✓ Correct
C. 6.64 × 10⁻²² g
D. 2.41 × 10²⁵ g
Solution: Mass of one atom = molar mass ÷ Nₐ = 40 ÷ (6.022 × 10²³) = 6.64 × 10⁻²³ g. A power-of-ten slip gives 6.64 × 10⁻²² g, while multiplying by Nₐ instead of dividing gives the absurdly large 2.41 × 10²⁵ g.
Q6 — Mole Concept & Molar Masses · hard · numerical
For 0.5 mol of a diatomic gas X₂ at STP (molar volume = 22.7 L mol⁻¹), the number of molecules and the volume occupied are respectively: (Nₐ = 6.022 × 10²³)
A. 3.011 × 10²³ molecules and 11.35 L ✓ Correct
B. 6.022 × 10²³ molecules and 22.7 L
C. 3.011 × 10²³ molecules and 22.7 L
D. 1.5055 × 10²³ molecules and 11.35 L
Solution: Molecules = moles × Nₐ = 0.5 × 6.022 × 10²³ = 3.011 × 10²³. Volume at STP = moles × 22.7 = 0.5 × 22.7 = 11.35 L. Treating the sample as 1 mol gives 6.022 × 10²³ molecules and 22.7 L; mixing a correct molecule count with the 1 mol volume gives the third option.
Q7 — Mole Concept & Molar Masses · hard · numerical
At STP (molar volume = 22.7 L mol⁻¹), the volume occupied by 11.2 g of nitrogen gas (N₂, molar mass = 28 g mol⁻¹) is:
A. 9.08 L ✓ Correct
B. 22.7 L
C. 11.35 L
D. 8.96 L
Solution: n = 11.2 ÷ 28 = 0.4 mol, so volume = 0.4 × 22.7 = 9.08 L. Assuming 1 mol gives 22.7 L; using 0.5 mol gives 11.35 L; using the older 22.4 L mol⁻¹ value (0.4 × 22.4) gives 8.96 L.