A 0.125-g sample of a monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH. The equivalence point is reached after adding 20.77 mL of base. What is the molar mass of the unknown acid?
Find: we have to calculate the molar mass of unknown acid.
Mass of monoprotic acid = 0.125 g
Concentration of NaOH = 0.1003 M
Volume = 20.77mL = 0.02077 L
(0.1003 M NaOH)(.02077 L)= .002083
Then I take the .125 g sample of unknown acid and divide that by .002083
.125/.002083 = 60.0 g/mol of unknown acid
Thus the molar mass of unknown acid is 60.0 g/mol