The density of a monobasic strong acid (Molar mass 24.2 g/mol) is 1.21 kg/L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is __________ × 10⁻² mL (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Use the density to find how many moles of acid sit in one litre, then match moles of acid to NaOH.

1. Molarity from density. The density is 1.21 kg L⁻¹ = 1210 g L⁻¹, so one litre of the acid holds 1210 g of it. Divide by the molar mass to get the moles in that litre.

   M_(acid) = (1210 g L⁻¹)/(24.2 g mol⁻¹) = 50 mol L⁻¹

2. Moles of NaOH. 25 mL × 0.24 M = 6 mmol of NaOH. A monobasic acid gives one H⁺ per molecule, so you need exactly 6 mmol of acid.
3. Volume of acid. V = (6 mmol)/(50 mmol mL⁻¹) = 0.12 mL.
4. So. 0.12 mL = 12 × 10⁻² mL, so the answer is 12.

The common slip: 12000. Reading 1.21 kg/L as 1.21 g/L makes the acid 0.05 M and the volume 120 mL, which gives 12000 in units of 10⁻² mL.

The idea to keep. For a pure liquid, molarity is density in g/L divided by molar mass. For neutralisation, match the moles of H⁺ to the moles of OH⁻, and remember that mL times M gives mmol.

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