How can 100ml of sodium hydroxide solution with a ph of 13.00 be converted to a sodium hydroxide solution with a ph of 12.00?

Chemical reaction: NaOH(aq) → Na⁺(aq) + OH⁻(aq).
V₁(NaOH) = 100 mL ÷ 1000 mL/L = 0.1 L.
pOH = -log[OH⁻].
pH + pOH = 14.
pOH₁ = 14 - 13 = 1.
pOH₂ = 14 - 12 = 2.
[OH⁻]₁ = 10∧(-pOH₁) = 10⁻¹ M = 0.1 M.
[OH⁻]₂ = 10(-pOH₂) = 10⁻² M = 0.01 M.
[OH⁻]₁ · V₁(NaOH) = [OH⁻]₂ · V₂(NaOH).
0.1 M · 0.1 L = 0.01 M · V₂(NaOH).
V₂(NaOH) = 1 L, we can add 900 mL of water to convert solution of sodium hydroxide from pH=12 to pH=13, or we can add some strong acid.

Ok so co(g) is not pronoused that. the only clue is that youpronouse it the oppisote way. 

Molarity is moles of solute over liters of solution. idk if that is what you are talking about, but you can convert moles of solute to grams of solute using dimensional analyzes by using molar mass found on the periodic table.