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
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
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