Correct Answer - D
(d) Since, 22400 mL volume is occupied by 1 mole of `O_(2)` at STP.
Thus, 5600 mL `O_(2)` means `=(5600)/(22400)molO_(2)`
`=(1)/(2) mol O_(2)`
`:. ` Weight of `O_(2)=(1)/(2)xx32=8g`
According to problem,
Equivalent of Ag=Equivalent of `O_(2)`
`=("Weight of Ag")/("Equivalent weight of Ag")`
`=(W_(O_(2)))/("Equivalent weight of " O_(2))`
`((W_(Ag))/(M_(Ag)))/(1)=((W_(O_(2)))/(M_(O_(2))))/(4)`
`:. " " (W_(Ag))/(108)xx1=(8)/(32)xx4`
`[:. 2H_(2)O to O_(2) +4H^(+)+4e^(-)]`
`implies W_(Ag)=108 g`
