Mass and volume data for titration of primary standard acid and unknown acid with sodium hydroxide?

part A Run 1 Mass of weighing paper 0.4452 Mass of weighing paper and H2C2O4*2H2O 0.6812 Initial reading of buret (mL) 0 Final reading of buret (mL) 32.00 QuestionsPart A questions Run1 Mass of H2C2O4*2H2O used (g) 0.6812 – 0.4452 = 0.2360 grams Moles of H2C2O4*2H2O (mol) 0.2360 grams @ 126.07 g/mol = 0.001872 moles acid Number of protons available for reaction OH- 2 H+ / molecule acidMoles of OH- which reacted *Mol) @ 2 NaOH & 1 H2C2O4 –>0.001872 moles acid => twice => 0.003744 moles NaOHVolume of NaOH solution used (mL) 32.00 mlMolarity of NaOH solution (M) 0.003744 moles NaOH / 0.03200 litres = 0.1170 Molar NaOH Average molarity of NaOH (M) you will need to plug the other two trialls through & use the average molarity in the next procedure’s questions=================================================Part B Run 1 Mass of unknown acid suggested for titration (0.25 g) 3 reactive protons per molecule of unknown acid Mass of weighing pape(0.4452g) Mass of weighing paper and unknown (0.6952 g) Initial reading of buret(0mL) Final reading of buret(38.90 mL) QuestionsPart B questions Run1 Mass of unknown acid used for titration (g) (0.6952 g) – (0.4452g) = 0.2500 g Volume of NaOH solution used (38.90mL) Moles of NaOH which reacted (mol) 0.03890 litres @ 0.1170 Molar NaOH = 0.0045513 moles NaOHMoles of unknown acid 1 H3X & 3 NaOH –> …. 0.0045513 moles NaOH @ 1 mole acidH3X / 3 moles NaOH = 0.001517 moles of acidMolecular weight of unknown acid (g/mol) 0.2500 g acid / 0.0045513 moles NaOH = 54.93 g/molAverage molecular weight of unknown acid (g/mol) remember to do part A for each trial, average the molarity of NaOH,… then use that average NaOH molarity for part B... Show More

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