A) \[20\,mL\,of0.5M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
B) \[50\,mL\,of\,0.1\,M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
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 :
\[2MnO_{4}^{-}+5{{C}_{2}}O_{4}^{2-}+16{{H}^{+}}\xrightarrow{{}}2M{{n}^{2+}}\] \[+10C{{O}_{2}}+8{{H}_{2}}O\] L of \[0.1\text{ }M\,KMn{{O}_{4}}=20\times 0.1=2\,mmol\] \[\because \] 2 mmol of \[KMn{{O}_{4}}=5\text{ }mmol\text{ }of\text{ }{{C}_{2}}O_{4}^{2-}\] 50 mL of \[0.1\text{ }M\text{ }{{H}_{2}}{{C}_{2}}{{O}_{4}}=50\times 0.1=5\text{ }mmol\] Hence, 20 mL of \[0.1\text{ }M\text{ }KMn{{O}_{4}}\] \[\equiv \,50\,mL\,of0.1\,M\,{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
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