| \[MnO_{4}^{-}+{{C}_{2}}O_{4}^{2-}+{{H}^{+}}\xrightarrow{{}}M{{n}^{2+}}+{{CO}_{2}}+{{H}_{2}}O\] |
| 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 \[{{H}_{2}}{{C}_{2}}{{O}_{4}}\]
Correct Answer: C
Solution :
[c] Equiv. mass of \[KMn{{O}_{4}}\] \[MnO_{4}^{-}=\frac{molar\,mass}{7-2}=\frac{molar\,mass}{5}\] Equiv. mass of oxalic acid \[{{C}_{2}}O_{4}^{2-}=\frac{molar\,mass}{2(4-3)}=\frac{molar\,mass}{2}\] Meq. Of \[KMn{{O}_{4}}=50\times 5\times 0.04=10=\]meq of \[{{H}_{2}}{{C}_{2}}{{O}_{4}}=50\times 2\times 0.1=10.\]Hence [c].
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