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)

Step-by-step solution

millimole of NaOH = 0 . 24 × 25 For nuetralisation, equal moles of monoacidic base and monobasic acid must be present: ∴ millimole of acid = 0 . 24 × 25 ⇒ mass of acid = 0 . 24 × 25 × 24 . 2   g for pure acid, V = w d ; ( d = 1 . 21   kg / L = 1 . 21   g / ml ) ∴ V = 0 . 24 × 25 × 24 . 2 1 . 12 × 10 - 3 = 120 × 10 - 3 ml = 12 × 10 - 2 ml