InterviewSolution
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InterviewSolution
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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. |
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Answer» `[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%` |
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