Calculate the equivalent weight of the following:
(i) `KMnO_4` in acidic medium
(ii) `KMnO_4` in alkaline medium
(iii) `FeSO_4 (NH_4)_2 SO_4. 6H_(2) O` (converting to `Fe^(3+)` )
(iv) `K_2 Cr_2 O_7` in acidic medium
(v) `H_2 C_2 O_4` (converting to `CO_2` )
(vi) `Na_(2) S_(2) O_(3). 5H_(2) O` ( reacting with `I_2` )

Explanation

(i) `underset(+7)overset("1 mole")(KMnO_4)to underset(+2)(Mn^(2+))" (acidic medium)"`
Equivalent weight of `KMnO_4= ("mol. weight of" KMnO_4)/("change in ON per mole")`
`= (158)/( 5) = 31.6`
(ii) `underset(+7) overset("1 mole") (KMnO_(4)) to underset(+6)( MnO_(4)^(2-)" (alkaline medium)"`
Equivalent weight of `KMnO_4 = ( 158)/(1) = 158`.
(iii) `Fe^(2+) to Fe^(3+)`
Equivalent weight of `FeSO_(4) (NH_(4) )_(2) SO_(4) . 6H_(2) O = ( 392)/( 1) = 392`.
(iv) `underset(+12) overset("1 mole")(K_(2) Cr_(2) O_(7)) to underset(+6) (2Cr^(3+)) `
Equivalent weight of `K_2 Cr_2 O_7 = (294.2)/( 6) = 49.03`.
(v) `underset(+6) overset("1 mole") (H_(2) C_(2) O_(4)) to underset(+8) (2CO_(2))`
Equivalent weight of `H_(2) C_(2) O_(4) = (90)/(2) = 45`.
(vi) `Na_(2) S_(2) O_(3). 5H_(2) O + I_(2) to Na_(2) S_(4) O_(6) + 2I^(-)`
or `underset(+4)(S_(2) O_(3)^(2-)) to underset(+5)((1)/(2) S_(4) O_(6)^(2-))`
Equivalent weight of `Na_(2) S_(2) O_(3) . 5H_(2) O = ( 248)/( 1) = 248.`