0.111g of a metallic chloride requires 0.34g of `AgNO_(3)` for complete precipitation of chlorine. Specific heat of the metal is 0.152 `cal*""^(@)C^(-1)*g^(-1)`. Determine the formula of metallic chloride.

Explanation

0.34g `AgNO_(3)=(0.34)/(170)` [Molecular mass of `AgNO_(3)=170`]
`=2xx10^(-3)` mol of `AgNO_(3)`.
1 mol of `AgNO_(3)` reacts with 1 mol of `Cl^(-)` ions
`therefore 2xx10^(-3)` mol of `AgNO_(3)` reacts with `2xx10^(-3)` mol of `Cl^(-)` ions.
`2xx10^(-3)` mol of `Cl^(-)` ion`=2xx10^(-3)xx35.5`
`=0.071g" of "Cl^(-)` ions.
`therefore`Amount of metal present`=(0.111-0.071)g=0.04g`
`therefore`Equivalent mass of the metal`=(0.04)/(0.071)xx35.5=20`
Approximate atomic mass of the metal`=(6.4)/(0.152)=42.1`
Valency of the metal in the chloride compound
`=(42.1)/(20)=2`
`therefore`Formula of the metallic chloride: `MC l_(2)`.