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).

1.020 g of metallic oxide contains 0.540g of the metal . Calculate the

1.	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).
Answer» SOLUTION :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)`.