To measure the quantity of Mncl2 dissolved in an aqueous solution , it was completely converted to KMnO4 using the reaction.

             $MnCl_{2}+K_{2}S_{2}O_{8}+H_{2}O\rightarrow KMnO_{4}+H_{2}SO_{4}+HCl$

  (equation not balanced)

    Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganatic ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is........

[Atomic weights in g mol -1 : Mn =55, Cl= 35.5]

A) 128mg

B) 115mg

C) 126mg

D) 130mg


Option C


$Mn^{2+}\rightarrow MnO_{4}^{-}+5e^{-}$

 $S_{2}O_{8}^{2-}+2e^{-}\rightarrow 2SO_{4}^{2-}$

$2Mn^{2+}+5S_{2}O_{8}^{2-} \rightarrow 2Mn O_{4}^{-}+10SO_{4}^{2-}$



      $2MnO_{4}^{-}+5C_{2}O_{4}^{2-} \rightarrow 2Mn^{2+}+10CO_{2}$

Hence,    $2Mn^{2+}\equiv 5C_{2}O_{4}^{2-}$

             $1MnCl_{2}\equiv 2.5H_{2}C_{2}O_{4}$

Oxalic acid taken   =225 mg

                      $\frac{225}{90}=2.5$ millimoles

  Hence, MnCl2 = 1 millimoles

                        = (55+71)= 126 mg