1)

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 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

Answer:

Option C

Explanation:

Mn2+MnO4+5e

 S2O28+2e2SO24


2Mn2++5S2O282MnO4+10SO24


 

Also,

      2MnO4+5C2O242Mn2++10CO2

Hence,    2Mn2+5C2O24

             1MnCl22.5H2C2O4

Oxalic acid taken   =225 mg

                      22590=2.5 millimoles

  Hence, MnCl2 = 1 millimoles

                        = (55+71)= 126 mg