What Volume of 0.100M NaOH is needed to neutralized 3.00 grams of benzoic acid?

Answer 1

C6H5COOH (s) + NaOH (aq) --> C6H5COONa (aq) + H2O (l)

One mole of NaOH is required for each mole of Benzoic acid.

3.00 g / 122.12 g/mol = moles of Benzoic acid

moles of Benzoic acid / 0.100 moles/Liter = Liters of NaOH required.

Liters required * 1000 = mL required