What is the mass of solute in 500$cm^{3}$ of 0.005$moldm^{-3}$ $H_{2}SO_{4}$? ( H = 1, S = 32.0, O=16.0)

Analyze the question: We need the mass of H₂SO₄ dissolved in 500 cm³ of a 0.005 mol/dm³ solution.

Concept:

• Moles = concentration × volume (in dm³)
• Mass = moles × molar mass

Solution:

1. Molar mass of H₂SO₄ = 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 g/mol

2. Convert the volume: 500 cm³ = 0.5 dm³

3. Moles = 0.005 × 0.5 = 0.0025 mol

4. Mass = 0.0025 × 98 = 0.245 g

Answer: The calculation gives 0.245 g, which is option C.

Tip: Always convert cm³ to dm³ by dividing by 1000 before using concentration in mol/dm³. A quick shortcut is mass = concentration × molar mass × volume (dm³), so 0.005 × 98 = 0.49 g per dm³, and half a litre gives 0.245 g. Option A (0.490 g) is the trap for students who forget to convert 500 cm³ to 0.5 dm³ and use 1 dm³ instead.