Correct Answer - A
.Intial m moles of `Cu^(2+)=5` ,
m - eq. or m - moles of `H^(+)`
produced `=100xx10^(-2)=1`
`rArr` m - moles of `Cu^(2+)` converted into `Cu=(1)/(2)=0.5`
m-moles of `Cu^(2+)` remaining in solution
`=5-0.5=4.5`
`2Cu^(2+)+4I^(-)rarrCu_(2)I_(2)+I_(2)` and `I_(2)+2NaS_(2)O_(3)rarr 2NaI + Na_(2)S_(4)O_(6)`
m-moles of `Cu^(2+)` remaining = m-moles of `Na_(2)S_(2)O_(3)`
`4.5 = 0.04xxV rArr V=112.5 mL`