If the molecular mass of `Na_(2)S_(2)O_(3)` and `I_(2)` are `M_(1) and M_(2)` respectively, then what will be the equivalent mass of `Na_(2)S_(2)O_(3)` and `I_(2)` in the following reaction
`2S_(2)O_(3)^(2-)+I_(2)rarrS_(2)O_(6)^(2-)+I`

Options

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

Answer

B

Explanation

`overset(+2)(2NaS_(2)O_(4))-=Na_(2)overset(+2.5)(S_(4))O_(6)`
Total increase in O.N of S for two molecules of `Na_(2)S_(2)O_(3)=4xx2.5-2[2(+2)]=2`
`:.` Total increase in O.N of S per molecule of `Na_(2)S_(2)O_(3)=1`
Eq. mass = `M_(1)//1=M_(1)`
Now `I_(2)^(@)rarr2I^(-2)`
Total decrease in O.N per molecule of `I_(2)`
`=0-2 (-1)=2`
`=0-2 (0-1) =2`
`:.` Eq. mass `I_(2) = M_(2)//2`.