In `a` solution the concentration of `CaCl_(2)` is `5M &` that of `MgCl_(2)` is `5 m`. The specific gravity of solution is `1.05`, calculate the concentration of `Cl^(-)` in the solution in terms of Molarity.

Answer

`[Cl^(-)]=13.36M`

Explanation

`CaCl_(2)rarr5M=555g` in `1L` solution or in `1050g` solution
`wt`. Of `("solvant" + MgCl_(2))=1050-555=495g`
`MgCl_(2)rarr5m`
`1000g` solvant `rarr 5` mol of `MgCl_(2)`
`=5xx95=475gMgCl_(2)`
`i.e., 1475 ("solvant" + MgCl_(2))rarr 475g MgCl_(2)`
`495("solvant"+ MgCl_(2)) rarr(475)/(1475)xx495`
`=159.4g MgCl_(2)`
moles of `MgCl_(2)=(159.4)/(95)=1.678`
Total moles of `Cl^(-)`
`=(5+1.678)xx2=13.356`
volume of solution `1L`
Molarity of `Cl^(-) = 13.356 M`