A) \[60c{{m}^{3}}\]
B) \[80c{{m}^{3}}\]
C) \[90c{{m}^{3}}\]
D) \[120c{{m}^{3}}\]
Correct Answer: B
Solution :
: \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]and \[KMn{{O}_{4}}\] both are good oxidizing agents in acidic medium. The reaction is as follows: \[CrO_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O\] \[MnO_{4}^{2-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O\] Normality of \[C{{r}_{2}}O_{7}^{2-}=Molarity\times no.\,of\] \[electrons\text{ }involved\] \[=0.04\times 6=0.24N\] Normality of \[MnO_{4}^{-}\] = Molarity \[\times \] no. of electrons involved \[=0.03\times 5=0.15N\] \[\underset{({{K}_{2}}C{{r}_{2}}{{O}_{7}})}{\mathop{{{N}_{1}}{{V}_{1}}}}\,=\underset{(KMn{{O}_{4}})}{\mathop{{{N}_{2}}{{V}_{2}}}}\,\] \[\Rightarrow \] \[0.24\times 50=0.15\times {{V}_{2}}\] \[{{V}_{2}}=\frac{0.24\times 50}{0.15}=80c{{m}^{3}}\]
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