Question

A 0.317 g sample of a monoprotic acid is dissolved in water and titrated with 0.100 M KOH. What is the molar mass of the acid if 16.5 mL of the KOH solution is required to o neutralize the sample? molar mass: g/mol

Answer 1

molar mass acid = 192 g/mol

Explanation

Concentration KOH = 0.100 M

volume KOH = 16.5 mL = 0.0165 L

moles KOH = (concentration KOH) * (volume KOH in Liter)

moles KOH = (0.100 M) * (0.0165 L)

moles KOH = 0.00165 mol

Since acid is monoprotic, moles acid = moles KOH

moles acid = 0.00165 mol

molar mass acid = (mass acid) / (moles acid)

molar mass acid = (0.317 g) / (0.00165 mol)

molar mass acid = 192.12 g/mol