A mixture of pure AgCl and pure AgBr is found to contain 60.94% Ag by mass. What are mass percentages of Cl and Br in the mixture ? (Ag = 108 , Cl = 35 . 5 Br = 80)

Explanation

Let us first calculate the weight ratio of AgCl and AgBr, their wts. being supposed to be x and y, g respectively in the mixture. Apply POAC for Ag atoms
AgCl + AgBr `to` Ag
`1 xx ` moles of AgCl + ` 1 xx ` moles AgBr = 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 mixture contains 19.64 g of AgCl and 80.36 g of AgBr.
Now calculate the amount of Cl in AgCl and the amount of Br in AgBr
Wt. of Cl in 19.64 of AgCl = `(35 . 5)/( 143.5) xx 1964 = 4.85 g`
Wt. of Br in 80.36 g of AgBr `= (80)/(188) xx 80.36 = 34.19 g`
Thus 100 g of the mixture contains 4.85 g of Cl and 34.19 g of Br .
`:.` percentage of Cl = `4.85%`
and percentage of Br = `34.19%`