A) 11.4mL
B) 12.0 mL
C) 33.3 mL
D) 35.0 mL
E) 25.0 mL
Correct Answer: C
Solution :
In acidic medium,\[MnO_{4}^{-}\]is reduced to\[M{{n}^{2+}}\] \[\overset{+7}{\mathop{Mn}}\,O_{4}^{-}\xrightarrow{{}}M{{n}^{2+}}\] Change in oxidation number\[=7-2=5\] Solution X Solution Y \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] For \[F{{e}^{2+}}\] For \[MnO_{4}^{-}\] \[N\times 25=5\,M\times V\] \[[\because For\,MnO_{4}^{-},N=5\,M]\] in acidic medium] \[25N=5M\times 20\] \[25N=100M\] ...(i) In neutral medium,\[MnO_{4}^{-}\]is reduced to\[Mn{{O}_{2}}\] \[\overset{+7}{\mathop{MnO_{4}^{-}}}\,\xrightarrow{{}}\overset{+4}{\mathop{Mn}}\,{{O}_{2}}\] Change in oxidation number\[=7-4=3\] Solution X Solution Y \[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\] For \[F{{e}^{2+}}\] For \[MnO_{4}^{-}\] \[25\times N=3\text{ }M\times V\] [\[\because \]For\[MnO_{4}^{-},N=3M\] in neutral medium \[25N=3M\times V\] ...(ii) From Eqs (i) and (ii) \[100M=3M\times V\] \[V=\frac{100}{3}=33.3\,mL\]
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