Question

In a quantitative analysis of an iron, an analyst converts $0.40$ g of the ore into its ferrous. This required $40.0$ mL of $0.1N$ solution of $KMn{O}_4$ for titration.

How many moles ($\times {10}^{-4}$) of $KMn{O}_4$ was used for titration? $(Fe=56)$

Answer 1

$40.0$ mL of $0.1N$ solution of $KMn{O}_4$ solution corresponds to $\frac{40.0}{1000}\times 0.1$ $=40\times {10}^{-4}$ geq.
In acidic medium, the equivalent mass of $KMn{O}_4$ is one fifth of its molar mass.
$1$ mole of $KMn{O}_4$ corresponds to $5$ equivalents of $KMn{O}_4$.
$40\times {10}^{-4}$ eq. of $KMn{O}_4$ corresponds to $40\times {10}^{-4}$ eq. $\times \frac{1}{5}=8$ mol.
Hence, $8$ moles of $KMn{O}_4$ were used in the titration.