A) \[20ml\] of \[0.5\,M\] \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
B) \[50\,ml\] of \[1\,M\,\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
C) \[50\,ml\] of \[0.01\,M\,\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
D) \[20\,ml\] of \[0.1\,M\,\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
Correct Answer: B
Solution :
\[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\to 2M{{n}^{2+}}+10C{{O}_{2}}+8{{H}_{2}}O\] \[20\,ml\]of\[0.1M\,KMn{{O}_{4}}=20\times 0.1=2\,m\,mol\] \[\because \] \[2\,m\,mol\] of \[KMn{{O}_{4}}\equiv \,5\,m\,mol\] of \[{{C}_{2}}O_{4}^{2-}\] \[50\,ml\] of \[0.1M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}=50\times 0.1=5m\,mol\] Hence, \[20\,ml\] of \[1M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}=50\times 0.1=5m\,mol\]You need to login to perform this action.
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