How do you calculate moles of water of crystallization??

Question

In weakly alkaline solution, sulphite is oxidised quantitatively to sulphate with iodine solution:

SO3ˉ I2 H2O = SO42ˉ 2HI

2HI 2HCO3ˉ = 2I ˉ 2H2O 2CO2

20 cm3 of an iodine solution required 35.75 cm3 of a 0.0548 mol dmˉ3 sodium thiosulphate solution for reduction. To a further sample of 20 cm3 of iodine solution was added 20 cm3 of a sodium sulphite solution made by dissolving 1.9400g of Na2SO3.xH2O in 250 cm3 of water. The iodine solution then required 13.25 cm3 of sodium thiosulphate solution for reduction. Calculate the moles of water of crystallisation in the sample of sodium sulphite.

Answer 1

Moles of sodium thiosulphate= 0.0548 * 35.75/1000= 0.0019591.

Mole ratio sodium thiosulphate: I2= 1:1, therefore moles of I2=0.0019591

20 cm^3 of iodine solution was added, therefore again 0.0019591 moles were added.

From 250 cm^3 of the sodium thiosulphate solution 20 cm^3 was taken, therefore, 20/250 * 1.94 g= 0.1552 g was taken.

13.25 cm^3 of sodium thiosulphate still reacted. Therefore 13.25*0.0548/1000= 0.0007261 moles reacted. That means that 0.0019591-0.0007261= 0.001233 moles were reacted during the first mixing.

0.001233 moles= 0.1552 g

1 mole= 126 g

Na2SO3= 126.

Therefore x=0.

I don't think this is correct, but this is what I get, and as far as I know, this is what you should do.