A mixture of pure AgCl and AgBr contains 60.94% Ag by mass. What is the % of AgCl in sample

Asked by Anil | 13th May, 2017, 09:49: PM

Expert Answer:

Let x = mass AgCl
Let y = mass AgBr
Moles Ag in 100 g = 60.94g/ 107.868=0.5649
x + y = 100 g

x / 143.3 + y / 187.8 = 0.5649

187.8 x + 143.3 y = 15202

187.8 (100-y) + 143.3 y = 15202
18780 - 187.8 y + 143.3 y = 15202

3578 = 44.5 y

y = 80.40 g = mass AgBr

100 – 80.40 = 19.6 g

% AgCl = 19.6

Answered by Vaibhav Chavan | 15th May, 2017, 11:52: AM