J.R. S. answered 06/29/22
Ph.D. University Professor with 10+ years Tutoring Experience
Look at the reaction between oxalic acid (H2C2O4) and potassium hydroxide (KOH)
H2C2O4 + 2KOH = K2C2O4 + 2H2O ... balanced equation
molar mass H2C2O4* 2H2O = 126.1 g / mol
mols H2C2O4* 2H2O present = 0.851 g x 1 mol / 126.1 g = 0.006749 mols
mols KOH needed for neutralization = 0.006749 mols H2C2O4 x 2 mol KOH / mol H2C2O4 = 0.0135 mols KOH
Molarity of KOH = 0.0135 mols / 39.1 ml x 1000 ml / L = 0.345 mol/L = 0.345 M KOH (3 sig. figs.)