A) 108 gm
B) 10.8 gm
C) 0.8 gm
D) \[\frac{108}{0.8}gm\]
Correct Answer: B
Solution :
\[\frac{\text{Mass of }{{O}_{2}}\text{ ions}}{\text{Mass of }Ag\text{ ions}}=\frac{\text{Chemical equivalent of }{{O}_{2}}}{\text{Chemical equivalent of }Ag}\] Þ \[\frac{0.8}{m}=\frac{8}{108}\] Þ \[m=10.8\,gm\]
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