In the reaction `2Na_(a)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+ 2NaI`, the equivalent weight of `Na_(2)S_(2)O_(3) (mol. Wt. =M)` is equal to

In the reaction 2Na_(2)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+2NaI , the oxidation state of S is -

In the reaction : 2Na_2S_2O_3 + I_2 rarr Na_2S_4O_6 +2NaI the equivalent weight of Na_2S_2O_3 is equal to :

In this reaction 2Na_(2)S_(2)O_(3)+I_(2)=Na_(2)SO_(4)O_(6)=2NaI_(2) acts as:

In the reaction: Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl , the equivalent weight of Na_(2)S_(2)O_(3) will be: (M= molecular weight of Na_(2)S_(2)O_(3))

In the section :Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl . the equivalent weight of Na_(2)S_(2)O_(3) will be : (M= molecular weight of Na_(2)S_(2)O_(3) )

In the section :Na_(2)S_(2)O_(3)+4Cl_(2)+5H_(2)O rarr Na_(2)SO_(4)+H_(2)SO_(4)+8HCl . the equivalent weight of Na_(2)S_(2)O_(3) will be : (M= molecular weight of Na_(2)S_(2)O_(3) )