Given:

Mass of hydrated salt (Na\(_3\)AsO\(_4\)(H\(_2\)O)\(_{12}\)) = 38.9g

Mass of hydrated salt (Na\(_3\)AsO\(_4\)) =  X

Molar mass of hydrated salt (Na\(_3\)AsO\(_4\)(H\(_2\)O)\(_{12}\)) = (23 x 3) + 75 + (16 x 4) + (18 x 12) = 424g/mol

Molar mass of hydrated salt (Na\(_3\)AsO\(_4\)) = (23 x 3) + 75 + (16 x 4) = 208g/mol

Using the formula,

  \(\frac{mass of hydrated salt}{mass of anhydrous salt}\) =  \(\frac{Molar mass of hydrated salt}{Molar mass of anhydrous salt}\)

 \(\frac{38.9}{X}\) =  \(\frac{424}{208}\)

 X = 19.083 = 19.1%

The mass of anhydrous salt is obtained to be 19.1%

However, we need to get the total mass of the solution, and this we can get by first solving for the total mass of water that make up the solution.

Mass of water in hydrated salt = mass of hydrated salt - mass of anhydrous salt

 =  38.9 - 19.1 = 19.8g

Also, recall that 38.9g salt was dissolved in 100g of water ⇒ total mass of water = 100 + 19.8 = 119.8g

Total mass of solution = mass of Na\(_3\)AsO\(_4\) + mass of H\(_2\)O

  = 19.1 + 119.8 = 138.9g

Therefore, %Mass of Na\(_3\)AsO\(_4\) in the saturation solution = \(\frac{mass of anhydrous salt}{Total mass of solution}\times 100\)

 = \(\frac{19.1}{138.9}\times 100\)

 = 13.75%