The specific heat at constant volume of a certain metal is 420 J kg^(-1). Write the chemical formula of its chloride, if contains 0.345 fraction of the metal (NCERT).

The specific heat at constant volume of a certain metal is 420 J kg^(-

1.	The specific heat at constant volume of a certain metal is 420 J kg^(-1). Write the chemical formula of its chloride, if contains 0.345 fraction of the metal (NCERT).
Answer» Solution :Given S = 420 `Jkg^(-1)K^(-1)`. From Dulong - Petit's law (Atomic mass) (specific heat) = 25200 `JkgK^(-1)`. Hence ATOMICMASS of the metal `=(25200)/(420) = 60` 60 g combine with (35.5) n g of CHLORINE To produce (60 + n(35.5))g of metal chloride, FRACTION of metal in its chloride `= (60)/(60 + 35.5n) = 0.34.5)`. Fraction of metal that combines with 35.5n of chlorine `= (X)/(x + 35.5n) = 0.345` Solving x = 56.1 = 56 which is the atomic mass of iron. `therefore` Chemical formula of metal chloride is `FeCI_(3)`.