A) \[M\]
B) \[\frac{M}{2}\]
C) \[\frac{M}{3}\]
D) \[\frac{3M}{3}\]
Correct Answer: A
Solution :
[a] \[C{{l}_{2}}+O{{H}^{-}}\xrightarrow{{}}C{{l}^{-}}+Cl{{O}^{-}}\] \[\frac{1}{2}\underset{0}{\mathop{C{{l}_{2}}}}\,\xrightarrow{{}}\underset{-1}{\mathop{C{{l}^{-}}}}\,+Cl{{O}^{-}}\,\] oxidation change=1 equivalent mass of \[C{{l}_{2}}\] in reduction half- reaction. \[=\frac{\frac{1}{2}C{{l}_{2}}}{1}=\frac{M}{2}\] \[\frac{1}{2}C\underset{0}{\mathop{{{l}_{2}}}}\,\xrightarrow{{}}\underset{+1}{\mathop{C{{l}^{-}}}}\,+Cl{{O}^{-}}\] reduction change = 1 equivalent mass of \[C{{l}_{2}}\] in oxidation half- reaction. \[=\frac{\frac{1}{2}C{{l}_{2}}}{1}=\frac{M}{2}\] Thus, equivalent mass in overall reaction \[=\frac{M}{2}+\frac{M}{2}=M\]
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