To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction,MnCl2 + K2S2O8 + H2O→KMnO4 + H2SO4 + HCl (equation not balanced).Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid(225 g) was added in portions till the color of the permanganate 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).
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answer is 126.
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Detailed Solution
MnCl2amole+K2S2O8+H2O→KMnO4amole+H2SO4+HClC2O4−−+MnO4−→CO2meq of C2O4−−=meq of MnO4− 2×0.225/90=a×5a=1×[55+71]=126mg