5. Assuming complete decomposition, the volume of Co, released at S.T.P. on heating 9.55 g Baco, will be (Atomic weight of Ba = 137) (1) 1.12 L (2) 0.84 L (3) 2.24 L no and 32 g HCI? (Molar mass (4) 4.06 L

5. Assuming complete decomposition, the volume of Co, released at S.T.

1.	5. Assuming complete decomposition, the volume of Co, released at S.T.P. on heating 9.55 g Baco, will be (Atomic weight of Ba = 137) (1) 1.12 L (2) 0.84 L (3) 2.24 L no and 32 g HCI? (Molar mass (4) 4.06 L
Answer» n:-Assuming complete decomposition, the VOLUME of Co, released at S.T.P. on HEATING 9.55 g Baco, will be (Atomic weight of Ba = 137).Given:-when heated BaCo = 9.55g.Atomic weight of Barium CARBONATE = 137.Solution:-In question mole of weight of Ba is given but not a Carbon and oxygen is given because they are common.So,137 of Ba + 12 of Carbon + 16 × 3 of oxygen. molecular weight of BaCo3 = molecular weight of Ba + molecular weight of C + 3 × molecular weight of O.Answer:- = 1.085l