If the moleucar wt of `Na_(2)S_(2)O_(3)` and `I_(2)` are `M_(1)` and `M_(2)` respectivly then what will be the equivalent wt of `Na_(2)S_(2)O_(3)` and `I_(2)` in the following reaction
`S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(-)`

Options

A. `M_(1)M_(2)`
B. `M_(1)M_(2)//2`
C. `2M_(1),M_(2)`
D. `Mlt_(1),2 M_(2)`

Answer

b

Explanation

`2Na_(2)S_(2)O_(3)=Na_(2)S_(4)O_(6)`
Total increases is O.N of S for two molecules of `Na_(2)S_(2)O_(3)=4xx2.5-2[2(+2)=2`
`therefore` totoal increae in O.N of S for two molecules of `Na_(2)S_(2)O_(3)` =1
`therefore` Eq mass =`M_(1)//1=M_(1)`
Now `I_(2)rarr2I`
total decrease is O.N per molecule of `I_(2)=0-2(-1)=2`
`therefore` Eq mass of `I_(2)=M_(2)//2`