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Question

The equivalent weight of Mohr's salt, $$FeSO_{4}$$. $$(NH_{4})2SO_{4}.6H_{2}O$$ is equal to (i) atomic weight, (ii) its molecular weight, (iii) half its molecular weight, (iv) one third its molecular weight.


Solution

In Mohr's salt oxidation of $$Fe^{2+}$$ to $$Fe^{3+}$$ involves only one electron change.
$$\therefore$$ Equivalent weight= $$\cfrac {\text{Molecular weight}}{1}$$
$$\text{Molecular weight}= 392$$

Chemistry

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