9.7g of hydrated copper sulphate on heating loses 3.5 g of water. What is the percentage of water of crystallization?

Explanation

Let the formula of hydrated copper sulphate be `CuSO_(4)*xH_(2)O`
`CuSO_(4)*xH_(2)O=63.5+32+64+18x=(159.5+18x)g`
Amount of water present in `CuSO_(4)*xH_(2)O=18x`g
`therefore`Amount of water produced from `(159.5+18x)g` of
Hydrated copper sulphate on heating=18x g
`therefore`Amount of water produced from 9.7g of hydrated copper
sulphate on heating `=(18x xx9.7)/(159.5+18x)g`
As given in the question, `(18x xx9.7)/(159.5+18x )=3.5` or, x=5
`therefore`Molecular formula of hydrated copper sulphate:
`CuSO_(4)*5H_(2)O`
`CuSO_(4)*5H_(2)O=(63.5+32+64+5xx18)g=249.5`
`therefore`Amount of water present`=(5xx18)/(249.5)xx100=36.07%`.