A 0.2 mL sample of a pure liquid dibasic acid having a density of 1.5 g mL ⁻¹ requires 40 mL of 0.1 M NaOH solution for complete neutralization. The molar mass of the dibasic acid is ________ g mol ⁻¹ .

Step-by-step solution

Mass of the dibasic acid = Volume Density = 0.2 1.5 = 0.3 g Millimoles of NaOH used for neutralization = 40 0.1 = 4 mmol Since the acid is dibasic, 1 mole of acid neutralizes 2 moles of NaOH . Millimoles of acid = 4 2 = 2 mmol = 2 10⁻³ mol Molar mass of the acid = Mass Moles = 0.3 2 10⁻³ = 150 g mol ⁻¹ Answer: 150