JEE Advanced levelSingle correct optionDifficulty 3 of 3mt038
A green mineral is heated strongly in the absence of air. It loses mass, giving off a gas that turns lime water milky together with water vapour, and leaves a black solid that dissolves in dilute HX2SOX4 to give a blue solution. The mineral and its percentage loss in mass on heating (with Cu=63.5, C=12, O=16, H=1) are
Amalachite, CuCOX3⋅Cu(OH)X2; 28.1%
Bcopper glance, CuX2S; 28.1%
Cmalachite, CuCOX3⋅Cu(OH)X2; 71.9%
Dmalachite, CuCOX3⋅Cu(OH)X2; 19.9%
Answer: A
A green basic carbonate that gives COX2, water and black CuO (soluble in acid to give blue CuX2+) is malachite: CuCOX3⋅Cu(OH)X22CuO+COX2+HX2O. Its molar mass is 221 and the residue 2CuO weighs 159, so the loss is (221−159)/221×100=28.1%. Counting only COX2 would give 19.9%.