The volume of 0.25 mol$dm^{-3}$ solution of KOH that would yield 6.5g of of solid KOH on evaporation is (K = 39.0; o = 16.0; H = 1.00)

Step 1: Analyze the question We need the volume of 0.25 mol/dm³ KOH solution that contains 6.5 g of KOH.

Step 2: Concept

• Moles = mass ÷ molar mass
• Volume = moles ÷ concentration

Step 3: Solution

Molar mass of KOH = 39.0 + 16.0 + 1.00 = 56.0 g/mol

Moles of KOH = 6.5 ÷ 56 = 0.11607 mol

Volume = moles ÷ concentration = 0.11607 ÷ 0.25 = 0.4643 dm³

Converting to cm³: 0.4643 × 1000 = 464.3 cm³

Step 4: Answer This matches option A (464.

Tip: The two-step pattern is mass → moles (divide by molar mass), then moles → volume (divide by concentration). Always convert dm³ to cm³ by multiplying by 1000. Also, sanity check: 0.25 mol/dm³ × 1 dm³ would give 14 g, so 6.5 g (less than half of that) must need less than 500 cm³. That quickly points you to A.