A precipitate of `AgCl` and `AgBr` weighs `0.4066g`. On heating in a current of chlorine, the `AgBr` is converted to `AgCl` and the mixutre loses `0.0725 g` in weight. Find the `%` of `Cl` in original mixture.

Explanation

Suppose the precipitate contains 'a' g of AgCl and 'b' g of AgBr. Then
`a+b=0.4066g`
On heating the mixture in a current of `Cl_(2),AgBr` changes to `AgCl` according to the reaction
`underset(188g)(AgBr)overset(Cl_(2))rarr underset(143.5g)(AgCl)`
Thus, 188g AgBr gives AgCl = 143.5g
`therefore" b g AgBr will give AgCl"=(143.5)/(188)xxbg`
As AgCl remians as such (= a g), therefore total mass of the mixture now `=a+(143.5)/(188)xxb`
`therefore" "(a+b)-(a+(143.5)/(188)b)=0.0725" (Given value of loss in mass)"`
`"or "b(1-(143.5)/(188))=0.0725" or "bxx(44.5)/(188)=0.0725`
`"or "b=0.0725 xx(188)/(44.5)=0.3063"g (mass of AgBr)"`
`therefore" Mass of agCl"=0.4066-0.3063=0.1003g`
Mass of Cl in 0.1003 g of `AgCl =(35.5)/(143.5)xx0.1003=0.025g`
`therefore %" of Cl in the mixture "=(0.025)/(0.4066)xx100=6.15%`