Is this calculation correct up here? (about chemistry)

Question

A student prepared a standard oxalic acid (H2C2O4·2H2O) solution and performed a titration to determine the molarity of an unknown sodium hydroxide (NaOH) solution.

Use the data given below to find the molarity of the NaOH solution.

Given: Atomic masses: H: 1.008; C: 12.01; O: 16.00; 1 L = 1000 mL.

Mass of H2C2O4·2H2O 1.3245 g 

a) Molar mass of H2C2O4·2H2O ?

2x1.008 + 2x12.01 + 4x16 + 2 x 18 = 126.04 g/mol

What is this question's formula exactly?

b) Moles of H2C2O4·2H2O ?

1.3245 x 1mol/126.04 g mol = 0.010509 mol

Volume of the H2C2O4·2H2O solution 250.0 mL 

c) Molarity of the oxalic acid solution ?

250.0 ml x 1L/1000ml = 0.25 L

M = 0.010509 / 0.25 L = 0.04204 M

d) Moles of oxalic acid in 20.00 mL of the above prepared solution ?

0.04204 x 0.02 L = 0.0008408 moles

Is my answer/calculation correct up to here?

and How to fill these answer below?

e) Moles of NaOH at the equivalence point ?

f) Volume of NaOH solution used in titration 19.70 mL 

g) Molarity of the NaOH solution ?

Answer 1

Your answers appear to be correct up to this point.

e) moles NaOH at equivalence. Equivalence point is when stoichiometric mol ratios are equal. Oxalic acid is a dicarboxylic acid of the form H2A so it takes 2 moles NaOH for each mole of oxalic acid,

HOOCCOOH + 2NaOH ==> NaOOCCOONa

If you are looking for moles NaOH at equivalence in titration of 20 ml (not in 250 ml), then it is 2 x 0.0008408 moles = 0.001681 moles NaOH

f) molarity of NaOH = 0.001681 moles/0.01970 L = 0.08536 M