\[KMn{{O}_{4}}\] reacts with oxalic acid as:\[MnO_{4}^{-}+{{C}_{2}}O_{4}^{2-}+{{H}^{+}}\xrightarrow{{}}M{{n}^{2+}}+{{CO}_{2}}+{{H}_{2}}O\]Hence, 50 ml of 0.04 M \[KMn{{O}_{4}}\] is acidic medium is chemically equivalent to

• A. 100 ml of 0.1M \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
• B. 50 ml of 0.2 M \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
• C. 50 ml of 0.1M \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
• D. 25 ml of 0.1M \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]