5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL) of this solution would be required to prepare 500 mL of 0.1 M solution? _____________
Given : Molar Mass of Na, O and H is 23, 16 and 1 g mol⁻¹ respectively

Solution

Chemistry ma’am · 3 steps

Find the stock's molarity first, then dilution is just M₁V₁ = M₂V₂: moles of NaOH stay the same.

1. Moles of NaOH. Molar mass of NaOH is 23 + 16 + 1 = 40 g mol⁻¹. So 5 g is 5/40 = 0.125 mol.
2. Stock molarity. These 0.125 mol sit in 450 mL, that is 0.45 L. So M₁ = 0.125/0.45 = 5/18 ≈ 0.278 M.

   V₁ = (M₂ V₂)/M₁ = (0.1 × 500)/(5/18) = (50 × 18)/5 mL

3. So. V₁ = 180 mL. Take 180 mL of the stock and dilute it with water to 500 mL.

The common slip: 200. Dividing the 0.125 mol by the final 500 mL instead of the 450 mL stock gives 0.25 M and a volume of 200 mL.

The idea to keep. Dilution adds only water, so the moles of solute do not change. Get the stock molarity from its own volume, then use M₁V₁ = M₂V₂.

Answer from the NTA final answer key. Solution by PrepWiser.