WAEC 2011 Chemistry · Question 28

What is the amount of magnesium that would contain 1.20 x 1024 particles? [Mg = 24, Avogadro's constant = 6.02 x 1023]

Options

A.0.5 moles
B.2.0 moles
C.12.0 moles
D.24.0 moles

Answer: B. 2.0 moles

Explanation written with AI and checked against the answer key

Analyze the question You are asked for the amount (in moles) of magnesium that contains 1.20 × 10²⁴ particles (atoms). Concept One mole of any substance contains 6.02 × 10²³ particles (Avogadro's constant, L or N_A). So: Number of moles = number of particles ÷ Avogadro's constant Solution n = (1.20 × 10²⁴) ÷ (6.02 × 10²³) Rewrite 1.20 × 10²⁴ as 12.0 × 10²³: n = (12.0 × 10²³) ÷ (6.02 × 10²³) n ≈ 1.99 ≈ 2.0 moles The molar mass of Mg (24 g/mol) is not needed here. It would only be used if the question asked for the mass (2 × 24 = 48 g). Answer: B. 2.0 moles. Tip: When the question gives a number of particles and asks for moles, divide by Avogadro's constant. Watch the powers of 10: 10²⁴ is ten times 10²³, so 1.20 × 10²⁴ is about 2 times 6.02 × 10²³. Also, extra data like Mg = 24 is often a distractor.

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