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]
A. 1 B. 2 C. 3 D. 4 E. 5
Correct Answer: 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
