question_answer

\[KMn{{O}_{4}}\] reacts with oxalic acid according to the equation : \[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\to 2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O\] Here 20 ml of 0.1 M \[KMn{{O}_{4}}\] is equivalent to [CBSE PMT 1996]

A) \[20ml\] of 0.5 M \[{{C}_{2}}{{H}_{2}}{{O}_{4}}\]

B) \[50ml\] of 0.1 M \[{{C}_{2}}{{H}_{2}}{{O}_{4}}\]

C) \[50ml\] of 0.5 M \[{{C}_{2}}{{H}_{2}}{{O}_{4}}\]

D) \[20ml\] of 0.1 M \[{{C}_{2}}{{H}_{2}}{{O}_{4}}\]