1.245 g of `CuSO_(4). xH_(2)O` was dissolved in water and `H_(2)S` gas was passed through it will till CuS was completely precipitated . The `H_(2)SO_(4)` produced in the filtrate required 100 ml of 0.1 M NaOH solution . Calculate x (approximately)

Explanation

`CuSO_(4) +H_(2)S to CuS +H_(2)SO_(4)`
`{:(" m.e of "CuSO_(4).x H_(2)O " solution = m.e of "H_(2)SO_(4)),( " = m.e of 10 mL of N NaOH"),(" "1xx10 =10 ):}`
` :. ` number of equivalent of `CuSO_(4).x H_(2)O` solution = `10/1000`
Weight of `CuSO_(4) . x H_(2)O ` = equivalent `xx ` eq.wt
`= 10/1000 xx (159.5 +18x)/2 `
`{ " eq . wt of "CuSO_(4) .x H_(2)O = (159.5 +18x)/2 }`
Thus , `10/1000 xx (159.5 +18x)/2 = 1.245 ` (given)
18 x = 89.5
`x approx 5`