1.245 g of a sample of `CuSO_4.xH_2O` was dissolved in water and `H_2S` passed till CuS was complete precipitated. The filtrate contained liberated `H_2SO_4`, which required 20 " mL of " `(N)/(2)` NaOH for complete neutralisation. Calculate x, the number of molecules of water associated with `CuSO_4(Cu=63.6)`

Explanation

`CuSO_4xxH_2O` will liberate equivalent amount of `H_2SO_4`
`therefore20" mL of " (N)/(20)NaOH=10m" Eq of "H_2SO_4`
`=10m" Eq of "CuSO_4.xH_2O`
`=10xx10^(-3)xx" Eq of "CuSO_4.xH_2O`
`therefore10xx10^(-3)xxE=1.245g`
`therefore` Equivalent weight `124.5 g`
`therefore` Molecular weight `=124.5xx2=249`
`CuSO_4.xH_2O=249`
`159.6+18x=249`
`thereforex=4.9approx5`