How many moles of $MnO_4^-$ are required to completely oxidize $1 \ mol$ of ferrous oxalate in an acidic medium?

  • A
    $7.5$
  • B
    $0.2$
  • C
    $0.6$
  • D
    $0.4$

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$KMnO_4$ reacts with oxalic acid according to the equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \to 2Mn^{2+} + 10CO_2 + 8H_2O$
Here $20 \ mL$ of $0.1 \ M \ KMnO_4$ is equivalent to:

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Oxidising power of chlorine in aqueous solution can be determined by the parameters indicated below:
$\frac{1}{2} Cl_{2(g)}$ $\xrightarrow{\frac{1}{2} \Delta_{diss} H^{\Theta}} Cl_{(g)}$ $\xrightarrow{\Delta_{eg} H^{\Theta}} Cl^{-}_{(g)}$ $\xrightarrow{\Delta_{Hyd} H^{\Theta}} Cl^{-}_{(aq)}$
(using the data,$\Delta_{diss} H_{Cl_2}^{\Theta} = 240 \ kJ \ mol^{-1}$,$\Delta_{eg} H_{Cl}^{\Theta} = -349 \ kJ \ mol^{-1}$,$\Delta_{Hyd} H_{Cl}^{\Theta} = -381 \ kJ \ mol^{-1}$) will be ............. $kJ \ mol^{-1}$.

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