Finding % purity of the acid solution?

Question

6g sample of a solution of monoprotic acid (F.W= 60g/mole) require 40ml of 0.1 M NaOH solution for neutralization

the ans are

A 1%

B 2%

C 3%

D 4%

Ty alot

Answer 1

H+ + OH- = H2O

mol NaOH used for neutralization = 0.040 * 0.1 = 0.004mol

mol H+ : mol OH- = 1:1

mol H+ = 0.004 mol

since the acid is monoprotic 1 mol H+ is released from 1 mol of the acid.

therefore mass of acid = 0.004 * 60 = 0.24 g

Hence % purity = .24/6 * 100 = 4%