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Answer 1

From the question it is clear that Initial weight of hydrated CuSO4 is 30 grams.Final weight after complete water molecules loss is 19.1773.hence weight of water molecules lost= 30-19.1773 =10.8223.

now CuSO4.xH2O was the initial compound now it is CuSO4 finally.

For 1 mole of CuSO4.H20 where only one molecule of water is there the ratio of= (Gram molecular weight of CuSO4)/Gram molecular weight of H2O)

= 159.61/18.016 = 8.8593

Now in problem the ratio of =( weight of CuSO4 )/Gram molecular weight of H2O)

= 19.1773/10.8223 = 1.77

Hence no of moles of H20 dehydrated=x= 8.8593/1.77 = 5.005 which is 5.

so the formula of the compound is CuSO4.5H20.