Crystal water

The chemist wanted to check the content of water of crystallization of chromic potassium alum K₂SO₄ * Cr₂ (SO₄) 3 * 24 H₂O, which was a long time in the laboratory. From 96.8 g of K₂SO₄ * Cr₂ (SO₄) 3 * 24 H₂O prepared 979 cm³ solution of base. Subsequently, the 293 cm³ of the stock solution, a solution of barium nitrate, and a precipitate precipitated weighing 19.6 g.

How many crystal water instead of 24 molecules contain salt?
             
Additional data to compute:
M (K₂SO₄ * Cr₂ (SO₄) 3 * 24 H₂O) = 998.2 g / mol, M(BaSO₄) = 233.4 g / mol
M(K₂SO₄ * Cr₂ (SO₄) 3) = 566.2 g / mol, M (H₂O) = 18 g / mol

Correct answer:

x =  546.2 g

Step-by-step explanation:

$x=546.2={\displaystyle \frac{2731}{5}}\text{ g}=546.2\text{ g}$

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