Question

The equivalent weight of Mohr's salt, $FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O$ is equal to:

• A. its molecular weight
• B. atomic weight
• C. half-its molecular weight
• D. one-third its molecular weight

Answer 1

The correct option is A its molecular weight

In mohr salt, $Fe$ goes form $+3\to +2$ oxidation state, so the valancy factor $=3-2=1$

Hence $Eq.wt=\frac{Mol.Wt}{V.f.}=\frac{342}{1}=342$

Thus, its equivalent weight is similar to its molecular weight.