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


Result

x =  546.2 g

Solution:

x=546.2=27315=546.2  g x=546.2 = \dfrac{ 2731 }{ 5 } = 546.2 \ \text{ g }



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