Q. An ore contains 1.34% of the mineral argentite, $Ag_{2}S,$ by mass. How many gram of this ore would have to be processed in order to obtain 1.00 g of pure solid silver, $Ag?$
A) 74.6 g
B) 85.7 g
C) 107.9 g
D) 134.0 g
Explanation
(b)
$Ag_{2}S \equiv 2Ag$
248g $2 \times 108g$
$2 \times 108$ g Ag is obtained from $Ag_{2}S = 248\ g$
1 g $Ag$ will be obtained from $Ag_{2}S = \frac{248 \times 1}{2 \times 108}$
$= \frac{248}{216}g$
But, the ore contains only 1.34% $Ag_{2}S.$
Thus, 1 $g\ Ag$ is obtained from ore $= \frac{248}{216} \times \frac{100}{1.34}g$
$= 85.68\ g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.