For preparing $250 \mathrm{~mL}$ of $\mathrm{N} / 20$ solution of Mohr's salt, the amount of Mohr's salt needed is

Question

Aicharya · Accepted Answer

The ionic equation for oxidation of Mohr's salt is
$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}$
Now Eq. of Mohr's salt $=\frac{392}{1}=392$ Strength $=$ Normality $\times$ Eq. mass
$=\frac{1}{20} \times 392=19.6 \mathrm{~g} / \mathrm{lit}$
Thus for preparing $250 \mathrm{ml}$ of $\mathrm{N} / 20 \mathrm{Mohr}$ 's salt solution, Mohr's salt needed
$=\frac{19.6}{1000} \times 250=4.9 \mathrm{~g}$ *

For preparing $250 \mathrm{~mL}$ of $\mathrm{N} / 20$ solution of Mohr's salt, the amount of Mohr's salt…

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