A precipitate of AgCl and AgBr weighs 0.4066 g. On heating in a current of chlroine, the AgBr is converted to AgCl and the mixture loses 0.0725 g in weight. Calculate the % of Cl in the original mixture (Atomic mass of Ag = 108, Cl = 35.5, Br = 80).

A precipitate of AgCl and AgBr weighs 0.4066 g. On heating in a curren

1.	A precipitate of AgCl and AgBr weighs 0.4066 g. On heating in a current of chlroine, the AgBr is converted to AgCl and the mixture loses 0.0725 g in weight. Calculate the % of Cl in the original mixture (Atomic mass of Ag = 108, Cl = 35.5, Br = 80).
Answer» Solution :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%`