The correct option is C X2O3
Let the mass of each oxide be 100 g.
For I oxide:
Oxygen = 27.6 % (given)
Metal = 72.4 % (given)
Molecular formula =X3O4 (given)
Let atomic mass of X be ‘a′ g.
So, 3a3a+64×100=72.4;
solving, we get ′a′=56 g
For II oxide:
Oxygen is 30 g, so metal will be 70 g.
So, the ratio of moles of X:O will be 7056:3016=1.25:1.875 or 2:3
Hence, the compound is X2O3.