J.R. S. answered 11/16/22
Ph.D. University Professor with 10+ years Tutoring Experience
I do believe that you asked the same question (with different amounts of material), just the other day. I answered it then, and will do so again, but you really should be able to do these yourself after reviewing my previous answer. Unless you didn't understand any or part of that answer. If so, you should state so.
H2C2O4 + 2NaOH ==> Na2C2O4 + 2H2O ... balanced equation for the reaction taking place
Molar mass oxalic acid = 90.03 g / mol
moles H2C2O4 present = 68 mg x 1 g /1000 mg x 1 mol / 90.03 g = 7.55x10-4 moles
moles NaOH needed = 7.55x10-4 mol H2C2O4 x 2 mol NaOH / mol H2C2O4 = 1.51x10-3 mol NaOH
Concentration of NaOH = 1.51x10-3 mol / 64.7 ml x 1000 ml / L = 0.0233 mols / L = 0.0233 M