What is the molarity of the sodium hydroxide sample?

Question

Balanced equation: 2 NaOH + H2SO4 → Na2SO4 + 2H2O 

A neutralization reaction was performed by reacting sodium hydroxide with sulfuric acid (H₂SO₄) and the following data was taken

Volume of sodium hydroxide sample	25.0 mL
Volume of sulfuric acid added	28.2 mL
Concentration of sulfuric acid	0.355 M

Answer 1

2NaOH + H₂SO₄ ==> Na₂SO₄ + 2H₂O  ... balanced equation

moles H₂SO₄ used = 28.2 ml x 1 L / 1000 ml x 0.355 mol/L = 0.0100 moles H₂SO₄

moles NaOH present = 0.0100 mol H₂SO₄ x 2 mol NaOH / 1 mol H₂SO₄ = 0.0200 moles NaOH present

Molarity of NaOH = moles NaOH / liters solution = 0.0200 moles / 0.0250 L = 0.801 M