1.020 g of metallic oxide contains 0.540 g of the metal. If the specific heat of the metal, M is 0.216  cal deg-1 g-1. The molecular formula of its oxide is

Mass of oxygen in the oxide = (l .O2O- 0.540)= 0.480g Equivalent mass of the metal =0.5400.480×8=9.0According to Dulong and Petit's law, Approx. atomic mass =6.4 Sp. heat =6.40.216=29.63 Valency of the metal = At. mass  Eq. mass =29.639.0=3 Hence, the formula of the oxide is M2O3.