Question

A 0.173 g sample of a monoprotic acid is dissolved in water and titrated with 0.120 M NaOH. What is the molar mass of the acid if 10.5 mL of the NaOH solution is required to neutralize the sample? molar mass: 274.6 molar mass: 274.6 g/mol g/mol

Answer 1

SInce acid is monoprotic

mols of NaOH consumed = moles of acid present

mols of NaOH consumed = molarity * volume
= 0.120 M * 0.0105 Liters = number of mols of acid

mass of acid used = 0.173 g

molar mass of acid = mass of acid/ number of mols of acid
= 0.173g/0.120 M * 0.0105 Liters
= 137.3 g/mol

round to 3 sig.fig.

Answer : molar mass = 137.g/mol