`1.878 g` of `MBr_(X)` when heated in a stream of `HCl` gas was comletely converted to chloride `MCl_(X)` which weighted `1.0 g` The specific heat of metal is `0.14 cal g^(-1)`. Calculate the molemular weight of the metal bromide.

Explanation

`MBr_(x) + xHCl rarr MCl_(x) + xHBr`
weight of `MBr_(x) = 1.878 g`
weight of `MCl_(x) = 1.0 g`
For the reaction
Eqivalent of `MBr_(x) =` Equivalent of `MCl_(x)`
`(1.878)/(E + 80) = (1.0)/(E + 35.5)`
`E = 15.18`
`:'` Atomic weight `xx` Specific heat `~~ 6.4`
`:.` Atomic weight of metal `M = (6.4)/(0.14) = 45.71`
`:.` Valency of metal `= ("Atomic Weight")/(Ew) = (45.71)/(15.18)`
`~~ 3.01 = 3` (integer)
`:.` Exact atomic weight of metal `= 15.18 xx 3 = 45.54`
`:.` Molecular weight of `MBr_(x) = 45.54 + 80 xx 3 = 285.54`