What is the molar mass of a monoprotic acid if 33.30 mL of 0.0823 M NaOH is required to neutralize a 0.254-g sample?

MW = 94.07 g/mol

Explanation:

We are here given a neutralization reaction which is a 1: 1 molar relation. So calculating the # mol NaOH reacted we will know the #mol of the monoprotic acid, and since we are given the mass of sample we can calculate the molar mas of the compund because # mol = mass / MW.

# mol NaOH = 0.0823 mol/L x 33 mL/1000 mL/L = 0.0027 mol

# mol = m/MW ∴ MW: m/# mol

MW : 0.254 g / 0.0027 mol = 94.07 g/mol