Crystal water

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

How much crystal water, instead of 24 molecules, contains 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

Final Answer:

x =  546.2 g

Step-by-step explanation:

$x=546.2=546.2\text{g}$

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