A) \[\text{20 mL}\]of \[\text{0}\text{.5}\,\text{M }{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
B) \[\text{50}\,\text{mL}\]of \[\text{0}\text{.1}\,\text{M}\,{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
C) \[\text{ }\!\!~\!\!\text{ 50 mL}\]of \[\text{0}\text{.01}\,\text{M}\,{{\text{H}}_{2}}{{C}_{2}}{{O}_{4}}\]
D) \[\text{20}\,\text{mL}\]of \[\text{0}\text{.1 M}\,{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]
Correct Answer: B
Solution :
\[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\xrightarrow{{}}2M{{n}^{2+}}\] \[+\,10C{{O}_{2}}+8{{H}_{2}}O\] \[\text{20}\,\text{mL}\]of \[0.1\,M\,KMn{{O}_{4}}=20\times 0.1=2\,mmol\] \[\because \]\[\text{2}\,\text{mmol}\,\]of \[KMn{{O}_{4}}=5\,mmol\]of \[{{C}_{2}}O_{4}^{2-}\] \[\text{50}\,\text{mL}\]of \[\text{0}\text{.1}\,\text{M}\,{{\text{H}}_{2}}{{C}_{2}}{{O}_{4}}=50\times 0.1=5\,mmol\] Hence,\[20\,mL\]of \[0.1\,M\,KMn{{O}_{4}}\] \[=50\,mL\]of \[0.1\,M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
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