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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 ) |
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Answer» SOLUTION :(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-)) tounderset(+5)((1)/(2) S_(4) O_(6)^(2-))` Equivalent weight of `Na_(2) S_(2) O_(3) . 5H_(2) O = ( 248)/( 1) = 248.` |
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