17
A 0.5222 −g sample of an unknown monoprotic acid was titrated with 9.98×10−2 M NaOH. The equivalence point of the titration occurs at 23.86 mL
Part A
Determine the molar mass of the unknown acid.
M= g/mol
Ans 17 :
Mol NaOH = molarity x volume (L)
= 9.98 x 10-2 M x 0.02386 L
= 0.00238 mol
since the acid is monoprotic , so the number of mol of acid will also be 0.00238 mol
Molar mass of acid = mass / mol
putting values :
= 0.5222 g / 0.00238 mol
= 219.3 g/mol
17 A 0.5222 −g sample of an unknown monoprotic acid was titrated with 9.98×10−2 M NaOH....