A mixture of pure AgCl and pure AgBr is foundto contain 60.94%Ag by mass. Whatare mass percentages of Cland Br in themixture ?(Ag = 108 , Cl = 35 . 5 Br = 80)

A mixture of pure AgCl and pure AgBr is foundto contain 60.94%Ag by ma

1.	A mixture of pure AgCl and pure AgBr is foundto contain 60.94%Ag by mass. Whatare mass percentages of Cland Br in themixture ?(Ag = 108 , Cl = 35 . 5 Br = 80)
Answer» Solution :Let us first calculate the weight ratio of AgCland AgBr, theirwts. being supposed to be x and y, g respectivelyin themixture. Apply POAC for AG atoms AgCl + AgBr `to` Ag `1 xx ` moles of AgCl + ` 1 xx ` molesAgBr = moles of Ag `(x)/( 143.5)+(y)/(188) ` = moles of Ag. `:.` wt. of Ag in the MIXTURE = `((x)/(143.5)+(y)/(188)) xx 108.g` As GIVEN , `(((x)/(143.5)+(y)/(188))xx 108)/(x + y) = 0.6094` `(x)/(y) = (0.035)/(0.1432)` `:. % ` of AgCl `= ( x xx 100)/( x + y) = (0.035)/( 0 . 035 + 0.1432) xx 100 = 19.64` and `%` of AgBr = 80.36 Thus,a 100-g MIXTURECONTAINS 19.64 g of AgCland 80.36 g ofAgBr. Now calculate the amount of Cl in AgCland the amountof Br in AgBr Wt. of Cl in 19.64 of AgCl = `(35 . 5)/( 143.5) xx 1964 = 4.85 g` Wt. of Brin 80.36 g of AgBr `= (80)/(188) xx 80.36 = 34.19 g` Thus 100 g of themixturecontains 4.85 g of Cl and 34.19 g of Br . `:.`percentage of Cl = `4.85%` and percentage of Br = `34.19%`