0.5 g `CaBr_2` was dissolved in water and the solution is acidified with nitric acid, 50 " mL of " standard `0.1N` `AgNO_3` is added and the solution is shaken thoroughly, the remaining `Ag^(o+)` ions required 15 " mL of " 0.1 N `NH_4CNS` solution using ferric alum as the indicator. Calculate the percentage of `CaBr_2` in the sample.

Explanation

`[Mw(CaBr_2)=40+2xx80=200g]`
`[Ew(CaBr_2)=(200)/(2)=100g(n=2)]`
`CaBr_2-=AgNO_3-=NH_4CNS`
`mEq-=mEq-=mEq`
`Br^(ɵ)-=Ag^(o+)-=CNS^(ɵ)`
`mEq-=mEq-=mEq`
m" Eq of "`CNS^(ɵ)=15xx0.1=1.5mEq`
Excess of `Ag^(o+)` ions reacted with `CNS^(ɵ)-=1.5mEq`
Total `Ag^(o+)=50xx0.1N=5mEq`
`Ag^(o+)` ions reacted with `Br^(ɵ)=5-1.5=3.5mEq`
`=3.5mEq` of `CaBr_2`
Weight of `CaBr_2=3.5xx10^(-3)EqxxEw(CaBr_2)`
`=3.5xx10^(-3)xx100=0.35g`
`%` of `CaBr_2=(0.35xx100)/(0.5)=70%`