Answer:

Correct Answer: D

Solution:

Since, 22400 mL volume is occupied by 1 mole of $ O _2 $ at STP.

Thus, 5600 mL $ O _2 $ means = $ \frac{5600}{22400}molO _2 $ $ =\frac{1}{4}mol,O _2 $

$ \therefore $ Weight of $ O _2\frac{1}{4}\times 32=8g $ According to problem,

Equivalents of Ag = Equivalents of $ O _2 $ = $ \frac{\text{ }\text{ }~\text{ Weight of Ag}}{\text{Equivalent weight of Ag}} $ = $ \frac{{W_O} _{_2}}{Equivalent,weight,of,{O_2}} $ $ \frac{W _{Ag}}{\frac{M _{Ag}}{VF}}=\frac{{W _{O _2}}}{\frac{{M _{O _2}}}{VF}} $

$ \therefore $ $ \frac{W _{Ag}}{108}\times \frac{8}{32}\times 4 $ $ [\because ,2H _2O\to O _2+4{H^{+}}+4{e^{-}}] $

Þ $ W _{Ag}=108g $