Question

a metal forms two oxides the higher oxide contains 80% metal. 0.72 gram of the lower oxide gave 0.8 gram of higher oxide when oxidized .Then the ratio of the weight of oxygen that combines with the fixed weight of the metal in the two oxides will be

Answer 1

80% of higher oxide 0.80 gms : 0.64 gm of metal in the higher oxide.

Thus before oxidation, there is 0.64 gm of metal in the lower oxide.

So there is 0.72 gms - 0.64 gms = 0.08 gms of oxygen in lower oxide.

and 0.80 - 0.64 = 0.16 gms of oxygen in lower oxide. The proportions of metal : oxygen in both oxides are:
0.64 : 0.08 = 8 : 1
0.64 : 0.16 = 8 : 2

So in the higher oxide there is double the amount of oxygen as compared to the lower oxide. The content of oxygen that reacts with metal are in the integer ratio : 1 : 2.