MolarMass_Acid_223

Madison Maison Molar Mass of an Acid Lab Chem 105 section 9 Calculations 1. Concentration of NaOH 107.501 g − 105.619 g = 1.882 1.882 g× 1 moloxalic acid 126.07 g = 0.0149 mols oxalicacid 0.0149 mols 0.05 L = 0.298 M oxalic acid 0.298 M × 0.01 L = 0.00298 mols 0.00298 mols× 2 mol NaOH 1 mol H 2 C 2 O 4 = 0.00596 mol NaOH Flask 1 20.1 mL − 2.6 mL = 17.5 mL = 0.0175 L 0.00596 mol 0.0175 L = 0.341 M NaOH Flask 2 17.5 mL − 1.9 mL = 15.6 mL = 0.0156 L 0.00596 mol 0.0156 L = 0.382 M NaOH 2. Average Concentration of NaOH 0.341 M + 0.382 M 2 = 0.361 M NaOH 3. Molar mass of Unknown 1 Unknown 1 = 15.643 g − 15.215 g = 0.428 g Titration 1 13.6 mL − 1.00 mL = 12.6 mL = 0.0126 L 0.0126 × 0.382 mol 1 L = 0.004813 mols NaOH 0.004813 mols NaOH × 1 moloxalicacid 1 mol NaOH = 0.004813 molsoxalic acid 0.004813 mol 0.01 L = 0.4813 M NaOH