1.

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)

Answer»

SOLUTION :`CuSO_(4) +H_(2)Sto 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 equivalentof `CuSO_(4).x H_(2)O` solution = `10/1000`
Weight of `CuSO_(4) . x H_(2)O ` = equivalent `xx ` EQ.wt
`= 10/1000xx (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`


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