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Question

To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction, MnCl2+K2S2O8+H2OKMnO4+H2SO4+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 permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is (Given atomic weights in gmol1 Mn=55,Cl=35.5)

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Solution

Number of meq of MnCl2 = number of meq of KMnO4 = number of meq of H2C2O4
Number of meq of H2C2O4 = 2×(22590)=5
So number of meq of MnCl2 =5
5×(w126)×103=5 (where w=mass of MnCl2 present)
w=126 g

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