MCQ
$40\ ml\ 0.1\ N\ KMnO_4$ is equivalent to $30\ ml\ KHC_2O_4$ solution. How many $ml$ of $0.1\ N\ KOH$ are required to titrate $60\ ml$ of same $KHC_2O_4$ solution ............ $\mathrm{ml}$
  • $40$
  • B
    $30$
  • C
    $28.57$
  • D
    $35.5$

Answer

Correct option: A.
$40$
a
$40 \times 0.1=30 \times m \times 2$

$M=\frac{4}{15 \times 4}=\frac{1}{15}$

$0.1 \times \mathrm{V}=\frac{1}{15} \times 60$

$\mathrm{V}=40\, \mathrm{ml}$

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