KMnO_(4) reacts with oxalic acid according to the reaction : 2KMnO_(4) +5C_(2)O_(4)^(2-)+16 H^(+) to 2Mn^(2+) + 10 CO_(2) + 7H_(2)O Then, 20 mL of 0.1M KMnO_(4) is equivalent to :

KMnO_(4) reacts with oxalic acid according to the reaction : 2KMnO_(4

1.	KMnO_(4) reacts with oxalic acid according to the reaction : 2KMnO_(4) +5C_(2)O_(4)^(2-)+16 H^(+) to 2Mn^(2+) + 10 CO_(2) + 7H_(2)O Then, 20 mL of 0.1M KMnO_(4) is equivalent to :
Answer» 30 mL of 0.5 M `C_(2)H_(2)O_(4)` (oxalic acid) 50 mL of 0.1 M` C_(2)H_(2)O_(4)`(oxalic acid) 20 mL of `0.5 M C_(2)H_(2)O_(4)` (oxalic acid) 10 mL of `0.1 M C_(2)H_(2)O_(4)` (oxalic acid) Solution :`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2))(H_(2)C_(2)O_(4))` `(0.1xx20)/(2)=(M_(2)V_(2))/(5)` `M_(2)V_(2)=5` (It is possible in case of b.)