How to solve this calculation (about chemical)

Question

 It is time for the titration. You measure 15.0 mL of the H₂C₂O₄ solution with a 100-mL graduated cylinder and add it to a 125-mL Erlenmeyer flask. Then you add two drops of the indicator, phenolphthalein, to the oxalic acid solution in the flask. You dispense the NaOH solution from the buret into the flask (while swirling the flask) until you reach the endpoint. At the endpoint, the solution in the flask turns light pink. You repeat the titration two additional times for a total of three trials. The initial and final volumes of NaOH used in each trial are given to you below. Calculate the moles of H₂C₂O₄ used in the titration, the moles of NaOH used in the titration, the volume of NaOH used, the molarity of NaOH in each trial, the average molarity of the NaOH solution, and the percent error. To calculate the average molarity of the NaOH solution, average the molarities of the three trials. Show ALL calculations on a separate sheet of paper. 

H₂C₂O₄ (aq) + 2 NaOH (aq) --> Na₂C₂O₄ (aq) + 2 H₂O (l)

(I)       (II)        (III)

Volume of H₂C₂O₄ used  15.0 0 mL    15.0 0   mL   15.00____mL

1) Moles H₂C₂O₄ used    _______mol     ______mol      ______mol

2) Moles NaOH            _____mol    _____mol    _____mol

Initial Buret Reading    _____0.00_mL  __0.05_mL     _0.10__mL

Final Buret Reading      ____14.45mL     14.38_mL   15.40_mL

Volume NaOH used      ______mL     ______mL    ____mL

3) M NaOH                    ____M   ____M  ____M                  

4) Average Molarity of NaOH      ____________M 

True Molarity of NaOH          0.1000 M

5) Percent Error                _________

How to solve these 1) to 5) question?

Answer 1

Hello, Ooyeon,

I have difficulty understanding the numbers given. What is the difference of "True" and "average" NaOH molarity? Was "average molarity that of NaOH or H₂C₂O_4?

I'm not sure which number is the established value to calculate % error.

Bob