Mole Concept & Molar Masses — JEE Main Chemistry MCQs with Solutions
Free JEE Main 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 · numerical
The number of molecules present in 8.8 g of carbon dioxide (CO₂; C = 12, O = 16) is: (Nₐ = 6.022 × 10²³)
A. 2.6497 × 10²³
B. 6.022 × 10²³
C. 1.2044 × 10²³ ✓ Correct
D. 1.2044 × 10²²
Solution: Molar mass of CO₂ = 12 + 2 × 16 = 44 g mol⁻¹, so n = 8.8 ÷ 44 = 0.2 mol. Molecules = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³. Taking 8.8 g as 1 mol gives 6.022 × 10²³; a power-of-ten slip gives 1.2044 × 10²²; using a wrong molar mass of 20 (8.8 ÷ 20 = 0.44) gives 2.6497 × 10²³.
Q2 — Mole Concept & Molar Masses · hard · numerical
The total number of atoms present in 3.6 g of glucose (C₆H₁₂O₆, molar mass = 180 g mol⁻¹) is: (Nₐ = 6.022 × 10²³)
A. 2.891 × 10²³ ✓ Correct
B. 2.891 × 10²²
C. 1.2044 × 10²²
D. 2.891 × 10²⁴
Solution: n = 3.6 ÷ 180 = 0.02 mol, giving 0.02 × 6.022 × 10²³ = 1.2044 × 10²² molecules. Each glucose molecule contains 6 + 12 + 6 = 24 atoms, so total atoms = 24 × 1.2044 × 10²² = 2.891 × 10²³. Forgetting to multiply by 24 atoms per molecule leaves 1.2044 × 10²² (the molecule count); the others are power-of-ten slips.
Q3 — Mole Concept & Molar Masses · hard · numerical
What mass of magnesium (Mg = 24) contains the same number of atoms as are present in 12 g of carbon (C = 12)?
A. 48 g
B. 6 g
C. 24 g ✓ Correct
D. 12 g
Solution: 12 g of carbon = 12 ÷ 12 = 1 mol = 6.022 × 10²³ atoms. To hold the same number of atoms we need 1 mol of Mg = 1 × 24 = 24 g. Assuming equal masses give equal atom counts (12 g) ignores the different atomic masses; 48 g and 6 g come from doubling or halving in error.
Q4 — Mole Concept & Molar Masses · hard · numerical
At STP (molar volume = 22.7 L mol⁻¹), the number of molecules present in 5.675 L of nitrogen gas (N₂) is: (Nₐ = 6.022 × 10²³)
A. 1.5055 × 10²²
B. 1.5055 × 10²³ ✓ Correct
C. 6.022 × 10²³
D. 3.011 × 10²³
Solution: n = volume ÷ molar volume = 5.675 ÷ 22.7 = 0.25 mol. Molecules = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³. Treating the sample as 1 mol gives 6.022 × 10²³; using 0.5 mol gives 3.011 × 10²³; a power-of-ten slip gives 1.5055 × 10²².
Q5 — Mole Concept & Molar Masses · hard · numerical
3.4 g of a gas occupies 2.27 L at STP (molar volume = 22.7 L mol⁻¹). The molar mass of the gas is:
A. 68 g mol⁻¹
B. 34 g mol⁻¹ ✓ Correct
C. 17 g mol⁻¹
D. 3.4 g mol⁻¹
Solution: n = 2.27 ÷ 22.7 = 0.1 mol, so molar mass = mass ÷ moles = 3.4 ÷ 0.1 = 34 g mol⁻¹ (the gas is H₂S). Dividing by 0.2 instead gives 17 g mol⁻¹; taking the raw mass as the molar mass gives 3.4 g mol⁻¹; doubling gives 68 g mol⁻¹.
Q6 — Mole Concept & Molar Masses · hard · numerical
The number of oxygen atoms present in 9.8 g of sulphuric acid (H₂SO₄, molar mass = 98 g mol⁻¹) is: (Nₐ = 6.022 × 10²³)
A. 6.022 × 10²²
B. 2.4088 × 10²²
C. 6.022 × 10²³
D. 2.4088 × 10²³ ✓ Correct
Solution: n = 9.8 ÷ 98 = 0.1 mol, giving 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules. Each H₂SO₄ molecule has 4 oxygen atoms, so oxygen atoms = 4 × 6.022 × 10²² = 2.4088 × 10²³. Forgetting the factor of 4 leaves 6.022 × 10²² (the molecule count); the others are scaling slips.
Q7 — Mole Concept & Molar Masses · hard · numerical
A container holds 1.7 g of ammonia (NH₃, molar mass = 17 g mol⁻¹). The total number of hydrogen atoms in the sample is: (Nₐ = 6.022 × 10²³)
A. 1.8066 × 10²²
B. 6.022 × 10²³
C. 6.022 × 10²²
D. 1.8066 × 10²³ ✓ Correct
Solution: n = 1.7 ÷ 17 = 0.1 mol, giving 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules. Each NH₃ molecule has 3 hydrogen atoms, so H atoms = 3 × 6.022 × 10²² = 1.8066 × 10²³. Forgetting to multiply by 3 leaves 6.022 × 10²² (the molecule count).