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Moles and Avogadro's Numbers

150 questions found

Practice Questions

During the reaction 2H₂ + O₂ → 2H₂O, complete reaction of 44.8 L of oxygen at STP produces

A. 22.4 L of H₂O(g)
B. 44.8 L of H₂O(g)
C. 89.6 L of H₂O(g)
D. 67.2 L of H₂O(g)

At the same conditions of temperature and pressure, gas volumes follow mole ratios. One volume of O₂ produces two volumes of gaseous H₂O. Thus, 44.8 L O₂ forms 89.6 L H₂O(g). Concept tested: Gas volume ratios at STP.

nmdcat.online Chemistry NMDCAT MCQs
Jul 13, 2026

The coefficients in a balanced equation represent mole ratios. Here, 2 moles of H₂ react with 1 mole of O₂. The other ratios do not match the balanced equation. Concept tested: Mole ratio from a balanced equation.

nmdcat.online Chemistry NMDCAT MCQs
Jul 13, 2026

The mass of carbon present in 22 g of carbon dioxide (CO₂) is

A. 12 g
B. 3 g
C. 6 g
D. 8 g

22 g CO₂ = 0.5 mole. One mole of CO₂ contains one mole of carbon atoms. Mass of carbon = 0.5 × 12 = 6 g.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

Moles of O₂ = 9.6 ÷ 32 = 0.3 mole. Required KClO₃ = 0.3 × 2/3 = 0.2 mole. Mass = 0.2 × 122.5 = 24.5 g.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

The volume ratio H₂:Cl₂ is 1:1. Therefore, 11.2 dm³ of hydrogen requires 11.2 dm³ of chlorine.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

Number of moles = 6.022 × 10²¹ ÷ 6.022 × 10²³ = 0.01 mole. Atomic mass = 0.64 ÷ 0.01 = 64 g/mol (64 u).

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

The number of molecules in 1 dm³ of an ideal gas at STP is approximately

A. 6.022 × 10²³
B. 2.69 × 10²²
C. 1.81 × 10²⁴
D. 3.01 × 10²²

One mole occupies 22.4 dm³. Therefore, molecules in 1 dm³ = (6.022 × 10²³) ÷ 22.4 ≈ 2.69 × 10²².

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

49 g H₂SO₄ = 0.5 mole. NaOH:H₂SO₄ = 2:1. Therefore, NaOH required = 1 mole = 40 g.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

The total number of protons in 9 g of water (H₂O) is

A. 6.022 × 10²³
B. 3.011 × 10²⁴
C. 6.022 × 10²⁴
D. 1.5055 × 10²⁴

9 g H₂O = 0.5 mole. Each molecule contains 10 protons. Total protons = 0.5 × 10 × Nₐ = 3.011 × 10²⁴.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

Fe:H₂ = 3:4. For 3 moles H₂, Fe = 2.25 moles. Mass = 2.25 × 56 = 126 g.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

143.5 g = 0.5 mole washing soda. Water = 0.5 × 10 = 5 moles.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

32.5 g Zn = 0.5 mole. It gives 0.5 mole H₂. Volume = 0.5 × 22.4 = 11.2 dm³.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

2 moles C₂H₆ require 7 moles O₂. Mass = 7 × 32 = 224 g.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

0.5 × 6 = 3 moles of oxygen atoms. Therefore, the oxygen subscript is 6.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

Moles of O₂ = 16 ÷ 32 = 0.5 mole. By Avogadro's law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Therefore, the same volume of CO₂ also contains 0.5 mole.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

Mole ratio S:H₂S = 3:2. Moles of S = 0.5 × 3/2 = 0.75 mole. Mass = 0.75 × 32 = 24 g.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

The total number of electrons in 1.7 g of ammonia gas (NH₃) is (Atomic numbers: N = 7, H = 1)

A. 6.022 × 10²³
B. 3.011 × 10²³
C. 6.022 × 10²⁴
D. 3.011 × 10²⁴

1.7 g NH₃ = 0.1 mole. Each NH₃ molecule has 10 electrons. Therefore, total electrons = 0.1 × 10 × Nₐ = 6.022 × 10²³.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

A gas has a density of 1.964 g/dm³ at STP. The molar mass of this gas is approximately

A. 22 g/mol
B. 32 g/mol
C. 44 g/mol
D. 28 g/mol

Molar mass = Density × 22.4 = 1.964 × 22.4 ≈ 44 g/mol.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026

The mole ratio of O₂ to CH₄ is 2:1. For 2.5 moles CH₄, moles of O₂ = 2.5 × 2 = 5.0 moles. At STP, 1 mole of gas occupies 22.4 dm³, so volume = 5.0 × 22.4 = 112.0 dm³.

nmdcat.online Chemistry NMDCAT MCQs
Jul 2, 2026
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