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JEE Advanced · Chemistry · 9. Redox Reactions

To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction,
\(\text{MnCl} _2+\text K _2\text S_2\text O _8+\text H _2\text O \rightarrow \text{KMnO} _4~+\) \(\text H _2 \text{SO} _4+ \text{HCl}\) (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. What is the quantity of MnCl2 (in mg) present in the initial solution? (Atomic weights in g mol-1: Mn=55, Cl=35.5)

  1. A 126
  2. B 120
  3. C 134
  4. D 127
Verified Solution

Answer & Solution

Correct Answer

(A) 126

Step-by-step Solution

Detailed explanation


C2O4--+MNO4-H+CO2
meq of C2O4--=meq of MNO4-
2×0.22590=a×5a =0.001 mole =1 milimole
Mass of MnCl2=1×55+71
=126 mg
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