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$35.5$
$M=\frac{4}{15 \times 4}=\frac{1}{15}$
$0.1 \times \mathrm{V}=\frac{1}{15} \times 60$
$\mathrm{V}=40\, \mathrm{ml}$
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