An impure sample of Mohr's salt, , weighing 2.5 g was dissolved in dilute . The solution required ex — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Milli-equivalents of $KMnO_4$ used = $50 \times 0.1 = 5$ meq = $0.005$ equivalents. By Law of Equivalence, equivalents of Mohr's salt reacted = 0.005 eq. In Mohr's salt, only $Fe^{2+}$ is oxidized to $Fe^{3+}$, so its n-factor is 1. Thus, equivalent weight = Molar mass = 392. Pure mass = $0.005 \times 392 = 1.96$ g. Percentage purity = $(1.96 / 2.5) \times 100 = 78.4\%$.