KMnO_(4) reacts with oxalic acid according to the equation, 2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)to 2Mn^(2+)+10CO_(2)+8H_(2)O Here, 20 mL of 0.1 M KMnO_(4) is equivalent to :

KMnO_(4) reacts with oxalic acid according to the equation, 2MnO_(4)^

1.	KMnO_(4) reacts with oxalic acid according to the equation, 2MnO_(4)^(-)+5C_(2)O_(4)^(2-)+16H^(+)to 2Mn^(2+)+10CO_(2)+8H_(2)O Here, 20 mL of 0.1 M KMnO_(4) is equivalent to :
Answer» 120 mL of 0.25M `H_(2)C_(2)O_(4)` `150 mL " of" 0.1 M H_(2)C_(2)O_(4)` `50 mL " of" 0.1 M H_(2)C_(2)O_(4)` `50 mL " of" 0.2 M H_(2)C_(2)O_(4)` 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 the option c