Titration of 0.5083 g of an unknown monoprotic acid dissolved in 25.00 mL of water requires 36.95 mL of 0.1067 M NaOH to reach the endpoint.What is the molar mass of the acid?
A) 128.9 g/mol
B) 0.007756 g/mol
C) 190.6 g/mol
D) 1.468 g/mol
E) 2.718 g/mol

Question

Quizplus · Accepted Answer

The moles of NaOH used are calculated by multiplying the volume (in liters) by the molarity: 0.03695 L * 0.1067 mol/L = 0.003944 mol. Since the acid is monoprotic, it reacts in a 1:1 ratio with NaOH, so there are 0.003944 mol of acid. The molar mass is the mass of the acid divided by the moles: 0.5083 g / 0.003944 mol = 128.9 g/mol.