The mass of 800cm3 of gas Q at s.t.p. is 1.0g. What is the molar mass of Q? [Avogadros constant = 22.4dm3 $mol^{-1}$`]

Analyze the question: We are given the mass of a gas (1.0 g) and its volume at s.t.p. (800 cm³). We need its molar mass. The "Avogadro constant" in the question is really the molar volume of a gas at s.t.p. (22.4 dm³/mol).

Concept: At s.t.p., 1 mole of any gas occupies 22.4 dm³ (22,400 cm³). So:

Molar mass = (mass of gas ÷ volume of gas) × molar volume

Solution:

1. Convert the volume to dm³: 800 cm³ = 0.8 dm³
2. Find the number of moles: n = 0.8 ÷ 22.4 = 0.0357 mol
3. Molar mass = mass ÷ moles = 1.0 ÷ 0.0357 ≈ 28 g/mol

Alternatively, if 800 cm³ weighs 1.0 g, then 22,400 cm³ weighs: (22,400 × 1.0) ÷ 800 = 28.0 g

Checking the options: The calculation gives 28.0 g/mol, which matches option C.

Tip: For these questions, remember the shortcut: Molar mass = (mass × 22.4) ÷ volume in dm³. Always convert cm³ to dm³ by dividing by 1000 first. Also, 28 g/mol is the molar mass of N₂ or CO, which is a common value in WASSCE questions.