`A" " 0.2g` sample , which is mixture of `NaCl, NaBr` and `NaI` was dissolved in water and excess of `AgNO_(3)` was added. The precipitate containing `AgCl,AgBr` and `AgI` was filtered, dried and weighed to be `0.412g`. The solid was placed in water and treated with excess of `NaBr`, which converted all `AgCl` into `AgBr`. The precipitate was then weighed to be `0.4881 g`. It was then placed into water and treated with excess of `NaI`, which converted all `AgBr` into `AgI`. The precipitate was then weighed to be `0.5868 g`. What was the percentage of `NaCl, NaBr` and `NaI` in the original mixture :

Answer

`50,20.23,29.77`

Explanation

Working in backward direction In the last step moles of `(AgBr+Agl)`= moles of `AgI`
`implies (0.4881-x)/(188)+(x)/(235)=(0.5868)/(235)impliesx=0.0933g`
Mass `%` of `NaI`
`=(0.0933)/(235)xx150xx(100)/(0.2)=29.77`
Now subtracting mass of `AgI` from `1st` and `2nd` precipitate gives
Mass of `(AgCl+AgBr)=0.3187 g`
and mass of `AgBr=0.3948 g`
Again `(y)/(143.5)+(0.3187-y)/(188)=(0.3948)/(188)impliesy=0.245g`
implies Mass `%` of `NaCl`
`=(0.245)/(143.5)xx58.5xx(100)/(0.2) = 50`
Mass `%` of `NaBr = 20.23`