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