MCQ
$KMnO_4$ reacts with oxalic acid according to the equation : 

$2MnO_4^ -  + 5{C_2}O_4^ -  + 16{H^ + } \to 2M{n^{ +  + }} + 10C{O_2} + 8{H_2}O$ 

Here $20\, mL$ of $0.1\, M\, KMnO_4$ is equivalent to

  • A
    $20\, mL$ of $0.5\, M\, H_2C_2O_4$
  • B
    $50\, mL$ of $0.5\, M\, H_2C_2O_4$
  • $50\, mL$ of $0.1\, M\, H_2C_2O_4$
  • D
    $20\, mL$ of $0.1\, M\, H_2C_2O_4$

Answer

Correct option: C.
$50\, mL$ of $0.1\, M\, H_2C_2O_4$
c
Meq of $A =$ Meq of $B$

$0.1\,M\,KMnO_4 = 0.5\,N\,KMnO_4$

$\therefore $ Meq of $KMnO_4 = 20 \times 0.5 = 10$ ($n$ factor $= 5$)

Meq of $50 \,ml$ of $0.1\,M\,H_2C_2O_4 = 50 \times 0.2 = 10$

$(0.1\,M\,H_2C_2O_4 = 0.2\,N\,H_2C_2O_4)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free