1.020 g of metallic oxide contains 0.540 g of the metal. If the specific heat of the metal, M is 0.216 $\text{ cal }{deg}^{-1}{ \mathrm{g}}^{-1}$. The molecular formula of its oxide is

Mass of oxygen in the oxide = (l .O2O- 0.540)= 0.480g

$\text{ Equivalent mass of the metal }=\frac{0.540}{0.480}\times 8=9.0$

According to Dulong and Petit's law,

$\text{ Approx. atomic mass }=\frac{6.4}{\text{ Sp. heat }}=\frac{6.4}{0.216}=29.63$

$\text{ Valency of the metal }=\frac{\text{ At. mass }}{\text{ Eq. mass }}=\frac{29.63}{9.0}=3$

$\text{ Hence, the formula of the oxide is }{\mathrm{M}}_2{\mathrm{O}}_3\text{. }$