Crystal water

The chemist wanted to check the content of water of 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 prepared 979 cm3 solution of base. Subsequently, the 293 cm3 of the stock solution, barium nitrate solution, and a 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

Correct answer:

x =  546.2 g

Step-by-step explanation:

x=546.2=546.2 g



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