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]

Correct answer is D

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

See all Chemistry questions & answers

See more JAMB questions & answers

Similar Questions

Aptitude Tests

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]

Similar Questions

Aptitude Tests