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. Molar weight of Mohr’s salt
• B. One-fourth of the molar weight
• C. Half of its molar weight
• D. One-third of the molar weight

Answer 1

The correct option is B One-fourth of the molar weight

Mohr’s salt $FeS{O}_4.(N{H}_4{)}_{2}S{O}_4.6H_{2}O$ has a molar mass of $392\text{g/mol}$
We know that n-factor of salts = total positive charge = total negative charge
Total positive charge = charge of 1 $F{e}^{2+}$ ion and 2 $N{H}_4^{+}=4$
Hence n-factor = 4
$\text{Equivalent weight}=\frac{\text{molar mass}}{\text{n-factor}}=\frac{M}{4}$