Some copper (ll) sulphate pentahydrate (CuSO4 5H2O), was heated at 120 oC with the following results. Wt of crucible = 10.00 g; Wt of crucible + CuSO4 5H2O =14.98g; Wt of crucible + residue = 13.54 g. How many molecules of water of crystalization were lost?
[H = 1, Cu = 63.5, O = 16, S = 32]
[H = 1, Cu = 63.5, O = 16, S = 32]
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Correct Answer: Option D
Explanation:
CuSO4 + 5H 2O
Wt CuSO4 + 5H2O = 14.98 - 10.00
= 4.98 gm
Wt of residue or CuSO4 = 13.54 - 10.00 =3.54
Wt of water lost = 4.98 - 3.54 = 1.44
Molecules of H2O lost = 5 - 1 = 4
CuSO4 + 5H 2O
Wt CuSO4 + 5H2O = 14.98 - 10.00
= 4.98 gm
Wt of residue or CuSO4 = 13.54 - 10.00 =3.54
Wt of water lost = 4.98 - 3.54 = 1.44
Molecules of H2O lost = 5 - 1 = 4