1.020 g of metallic oxide contains 0.540g of the metal . Calculate the equivalent mass of the metal and hence its atomic mass with the help of Dulong and Petit's law. Taking the symbol for the metal s M. find the molecular formula of the oxide. The specific heat of the metal is 0.216 cal `"deg"^(-1) g^(-1)`.

Explanation

Mass of oxygen in the oxide
`=(1.020-0.540)=0.480g`
Equivalent mass of the metal `=(0.540)/(0.480)xx8=9.0`
According to Dulong and Petit's law,
Approx, atomic mass `=(6.4)/("Sp.heat")=(6.4)/(0.216)=29.63`
Valency of the metal =`("At.mass)/("Eq.mass")=(29.63)/(9.0)=3`
Hence, the formula of the oxide is `M_(2)O_(3)`.