TiTrAtIoN CaCuLaTiOn can u please heeeeeeelp me?

Question

A 1.575g of a sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250 cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.16M of NaOH. Calculate the vaue of n

Answer 1

The molecular weight of the compound is
MW = 2+2*12+4*16 +n(2+16) =90+18n

The concentration of the acid is Ca=mole/V and since mole=mass/MW you have
Ca=mass/(MW*V)= 1.575/(MW*0.250)

The stoichiometry of the titration is 1:2 = mole acid:mole base =>
2*mole acid= mole base =>
2*Ca*Va= Cb*Vb =>
2*1.575/(MW*0.250) *25=0.16*15.6 =>
MW =2*1.575*25 / (0.16*15.6*0.250) =126

But MW=90+18n, so 126=90+18n=> 18n=36 => n=2